Is Matter Around Us Pure? - Class 9 Science - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Notes - Is Matter Around Us Pure? | Class 9 NCERT | Science
Understanding Matter in Our Surroundings
Understanding Matter in Our Surroundings: Class 9 Science Notes
What Is Matter?
Matter is anything that occupies space and has mass. In our daily life, everything we see and touch is made up of matter, from the air we breathe to the food we eat. According to Class 9 science, matter can exist in different states: solid, liquid, and gas.
Pure Substances vs Mixtures
Pure substances are forms of matter that consist of a single type of particle. Examples include elements like gold and compounds like water.
Mixtures, on the other hand, are made up of two or more different substances physically combined. Examples include seawater and soil.
Types of Mixtures
Heterogeneous Mixtures
These mixtures have a non-uniform composition. Examples include a mixture of sand and salt.
Classroom Activity 2.1: Mixing different amounts of copper sulfate in water to observe uniformity in color and texture.
Homogeneous Mixtures
These mixtures have a uniform composition throughout. Examples include salt dissolved in water.
Classroom Activity 2.1: Comparing copper sulfate solutions with different concentrations.
Understanding Solutions
A solution is a homogeneous mixture of two or more substances. Common examples include lemonade and soda water. In solutions, there is homogeneity at the particle level, meaning the solute and solvent are evenly distributed.
Components of a Solution
Solute: The substance that gets dissolved (e.g., salt in saltwater).
Solvent: The substance in which the solute dissolves (e.g., water in saltwater).
Properties of Solutions
Solutions are homogeneous.
The particles of a solution are smaller than 1 nm and cannot be seen by the naked eye.
Solutions do not scatter light, making the path of light invisible.
Solute particles do not settle down, making solutions stable.
Concentration of Solutions
Mass by mass percentage: ((\text{Mass of solute}/\text{Mass of solution}) \times 100)
Mass by volume percentage: ((\text{Mass of solute}/\text{Volume of solution}) \times 100)
Volume by volume percentage: ((\text{Volume of solute}/\text{Volume of solution}) \times 100)
Suspensions and Their Properties
A suspension is a heterogeneous mixture in which solute particles do not dissolve but remain suspended throughout the bulk of the medium. Examples include mud in water.
Properties of Suspensions
Particles are visible to the naked eye.
They scatter light, making the path of light visible.
Solute particles settle down when left undisturbed, making suspensions unstable.
The particles can be separated by filtration.
Colloidal Solutions
A colloid or colloidal solution is a mixture where the particles are uniformly spread throughout the solution. An example is milk.
Tyndall Effect
The scattering of light by colloidal particles, making the path of light visible.
Example: Sunlight passing through a dense forest, highlighting tiny water droplets in the air.
Properties of Colloids
They are heterogeneous mixtures.
Particles are not visible to the naked eye but are large enough to scatter light.
Colloids are stable and do not settle down when left undisturbed.
Colloidal particles cannot be separated by filtration but can be separated by centrifugation.
Physical and Chemical Changes
Physical Changes
Changes where the substance’s chemical composition remains unchanged (e.g., melting of ice).
Chemical Changes
Changes resulting in new substances with different chemical properties (e.g., rusting of iron).
Elements and Compounds
Elements
Basic forms of matter that cannot be broken down. Examples include hydrogen and oxygen.
Compounds
Substances composed of two or more elements chemically combined in a fixed proportion. Examples include water (H2O).
Separation Techniques for Mixtures
Filtration, evaporation, centrifugation, and chromatography are common methods used to separate mixtures.
Importance of Understanding Matter in Everyday Life
The concepts learned in Class 9 science about matter are not just academic; they have practical applications in various scientific and industrial processes. Understanding these principles helps in comprehending how matter interacts and changes in our surroundings.
By following this comprehensive guide, students will have a thorough understanding of the fundamental concepts of "Matter in Our Surroundings" as prescribed in the Class 9 curriculum. This structured approach ensures that all aspects of the topic are covered in detail, preparing students well for their exams.
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Extra Questions - Is Matter Around Us Pure? | NCERT | Science | Class 9
What is the % of $\mathrm{H}_{2}\mathrm{O}$ in $\mathrm{Fe}(\mathrm{CNS})_{3} \cdot 3\mathrm{H}_{2}\mathrm{O}$
A 45
B 30
C 19
D 25
The correct option is C: 19
Let's calculate the percentage of $\mathrm{H}_{2}\mathrm{O}$ in $\mathrm{Fe}(\mathrm{CNS})_{3} \cdot 3\mathrm{H}_{2}\mathrm{O}$.
First, determine the molar mass of the compound:
The molar mass of $\mathrm{Fe}(\mathrm{CNS})_{3}$ can be calculated as follows:
$\mathrm{Fe}$: 55.85 g/mol
$\mathrm{C}$: 12.01 g/mol (3 atoms in $\mathrm{CNS}$ group, so $3 \times 12.01 = 36.03$ g/mol)
$\mathrm{N}$: 14.01 g/mol (3 atoms in $\mathrm{CNS}$ group, so $3 \times 14.01 = 42.03$ g/mol)
$\mathrm{S}$: 32.07 g/mol (3 atoms in $\mathrm{CNS}$ group, so $3 \times 32.07 = 96.21$ g/mol)
Adding these amounts together: $$55.85 + 36.03 + 42.03 + 96.21 = 230.12 \ \text{g/mol}$$
The molar mass of $3 \mathrm{H}_{2}\mathrm{O}$:
$\mathrm{H}$: 1.01 g/mol (6 atoms of hydrogen, so $6 \times 1.01 = 6.06$ g/mol)
$\mathrm{O}$: 16.00 g/mol (3 atoms of oxygen, so $3 \times 16.00 = 48.00$ g/mol)
Adding these amounts together: $$6.06 + 48.00 = 54.06 \ \text{g/mol}$$
Total molar mass of $\mathrm{Fe}(\mathrm{CNS})_{3} \cdot 3\mathrm{H}_{2}\mathrm{O}$: $$ 230.12 + 54.06 = 284.18 \ \text{g/mol}$$
Next, compute the percentage of water in the compound: $$ % \ \text{of} \ \mathrm{H_{2}O} = \left( \frac{3 \times 18}{ 284} \right) \times 100 = 19 % $$
Therefore, the percentage of $\mathrm{H}_{2}\mathrm{O}$ in $\mathrm{Fe}(\mathrm{CNS})_{3} \cdot 3\mathrm{H}_{2}\mathrm{O}$ is 19%.
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Ask Chatterbot AINCERT Solutions - Is Matter Around Us Pure? | NCERT | Science | Class 9
Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride
(c) Small pieces of metal in the engine oil of a car
(d) Different pigments from an extract of flower petals
(e) Butter from curd
(f) Oil from water
(g) Tea leaves from tea
(h) Iron pins from sand
(i) Wheat grains from husk
(j) Fine mud particles suspended in water
Here are the appropriate separation techniques for each of the mixtures listed:
(a) Sodium chloride from its solution in water: Evaporation or crystallization can be used to recover the sodium chloride as solid from its aqueous solution. The water is evaporated, leaving behind the salt.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride: Sublimation is the most suitable method because ammonium chloride sublimes upon heating, leaving behind sodium chloride.
(c) Small pieces of metal in the engine oil of a car: Magnetic separation can be used if the metals are magnetic. Alternatively, filtration can also be used to separate the non-magnetic metal particles from the oil.
(d) Different pigments from an extract of flower petals: Chromatography is the best technique for separating and analyzing the different pigments present in the extract.
(e) Butter from curd: Centrifugation can be applied here, which involves rotating the curd at high speed in a centrifuge to separate the butter (fat) from the liquid.
(f) Oil from water: Decantation or separation funnel method can be used to separate oil from water because oil is less dense and forms a separate layer on the water which can be poured off.
(g) Tea leaves from tea: Simple filtration can be used to separate the solid tea leaves from the liquid tea.
(h) Iron pins from sand: Magnetic separation can be used since iron is magnetic and sand is not. A magnet can draw out the iron pins from the sand.
(i) Wheat grains from husk: Winnowing or sieving is a method used to separate heavier and lighter components of a mixture by wind or by blowing air. The heavier wheat grains fall to one side while the lighter husk blows to the other side.
(j) Fine mud particles suspended in water: Sedimentation and decantation can be employed here. The mud particles settle at the bottom after being left undisturbed for some time (sedimentation), and the clear water can be poured off from the top (decantation). Alternatively, centrifugation may also be used to accelerate this process.
Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Making tea can be described using terminology from chemistry. Below are the steps involved in making tea with the relevant scientific terms included:
Boiling the Solvent: Heat water in a kettle until it reaches a boiling point. Water is the solvent in which we will dissolve other substances.
Adding the Solute: Place tea leaves or a tea bag into a cup or teapot. The tea leaves contain the solute, which is a mixture of different compounds that give tea its flavor and color.
Dissolution of the Solute: Pour the boiling water (solvent) over the tea leaves or bag (solute). The heat of the water helps the solute to dissolve, which means the compounds in the tea leaves are separating and dispersing into the solvent, creating a solution.
Steeping: Allow the tea to steep for a few minutes. During this time, the compounds in the tea leaves that are soluble in hot water will dissolve, contributing to the tea's flavor, aroma, and color.
Separation of Insoluble Particles: Some components of the tea leaves will not dissolve as they are insoluble. After steeping, these insoluble substances will need to be removed from the finished tea.
Filtration: To obtain a clear tea, one might pour the tea through a strainer or remove the tea bag. The strainer acts as a filter, capturing the residue, which is the insoluble matter that did not dissolve in the water. The liquid that passes through the strainer is called the filtrate; in this case, it is the tea that you can drink.
Serving: The filtrate can now be served. Additional solutes like sugar or milk may be added depending on preference. If sugar is added, it will dissolve in the hot tea, further sweetening the solution.
Cleaning: After serving the tea, the residue in the filter or the used tea bag can be discarded, and the cup or pot can be cleaned to remove any remaining residue.
By following these steps, using the given scientific terms, you can make a cup of tea while simultaneously applying concepts from chemistry.
Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
Substance Dissolved | |||||
---|---|---|---|---|---|
283K | 293K | 313K | 333K | 353K | |
Potassium nitrate | 21 | 32 | 62 | 106 | 167 |
36 | 36 | 36 | 37 | 37 | |
Potassium chloride | 35 | 35 | 40 | 46 | 54 |
Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at $313 \mathrm{~K}$ ?
(b) Pragya makes a saturated solution of potassium chloride in water at $353 \mathrm{~K}$ and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at $293 \mathrm{~K}$. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Let's address each part of the question separately:
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at $313 \mathrm{K}$?
Based on the table, 62 grams of potassium nitrate dissolves in 100 grams of water at $313 \mathrm{K}$ to form a saturated solution. To find out how much will dissolve in 50 grams of water, we can set up a proportion:
$$ \frac{62 \text{ g KNO}_3}{100 \text{ g water}} = \frac{x \text{ g KNO}_3}{50 \text{ g water}} $$
Solving for (x):
$$ x = \frac{62 \text{ g KNO}_3}{100 \text{ g water}} \times 50 \text{ g water} $$
$$ x = \frac{62}{2} \text{ g KNO}_3 $$
$$ x = 31 \text{ g KNO}_3 $$
So, 31 grams of potassium nitrate would be needed to produce a saturated solution in 50 grams of water at $313 \mathrm{K}$.
(b) Pragya makes a saturated solution of potassium chloride in water at $353 \mathrm{K}$ and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
The table shows that the solubility of potassium chloride in water decreases as the temperature decreases (at $353 \mathrm{K}$, the solubility is 54 grams per 100 grams of water, whereas at lower temperatures such as $333 \mathrm{K}$ or $313 \mathrm{K}$, the solubility is 46 grams and 40 grams respectively).
As the solution made at $353 \mathrm{K}$ cools to room temperature, the solubility of potassium chloride in the solution will decrease. This means that the solution can no longer hold as much dissolved potassium chloride, and the excess will begin to precipitate out as solid crystals. So, Pragya would observe the formation of potassium chloride crystals as the solution cools.
(c) Find the solubility of each salt at $293 \mathrm{K}$. Which salt has the highest solubility at this temperature?
From the table, the solubility of each salt at $293 \mathrm{K}$ is as follows:
Potassium nitrate: 32 grams per 100 grams of water
Potassium chloride: 35 grams per 100 grams of water (as the solubility did not change from $283 \mathrm{K}$ to $293 \mathrm{K}$
Ammonium chloride: 37 grams per 100 grams of water
At $293 \mathrm{K}$, ammonium chloride has the highest solubility.
(d) What is the effect of change of temperature on the solubility of a salt?
Generally, the solubility of a salt in water increases with an increase in temperature. This is evident from the data in the table where, for all the salts listed (potassium nitrate, potassium chloride, and ammonium chloride), the solubility increases as the temperature goes from $283 \mathrm{K}$ to $353 \mathrm{K}$. However, the effect of temperature change on solubility can vary between different solids; in some cases, a few salts may have their solubility decrease with an increase in temperature, although this is less common.
Explain the following giving examples.
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension
(a) Saturated Solution: A saturated solution refers to a solution in which the maximum amount of solute has been dissolved in a solvent at a given temperature. This means that no additional solute can dissolve in the solvent at that temperature. If more solute is added to a saturated solution, it will not dissolve and instead will settle at the bottom of the container.
Example: A common example is a sugar solution at room temperature. When you keep adding sugar to water while stirring, a point will be reached where no more sugar can dissolve. The solution at that point is saturated, and any excess sugar will form a layer at the bottom of the container.
(b) Pure Substance: A pure substance is a material that has a constant composition (is homogeneous) and has consistent properties throughout the sample. Pure substances can be elements or compounds. An element is a pure substance that cannot be separated into simpler substances by chemical means, while a compound is a pure substance composed of two or more elements chemically bonded in definite proportions.
Example: Pure distilled water $H₂O$ is a pure substance because it contains only water molecules with the same ratio of hydrogen to oxygen $2:1$. Copper $Cu$ is another example since it is an element composed entirely of copper atoms.
(c) Colloid: A colloid is a type of mixture in which one substance is dispersed evenly throughout another. The particles in a colloid are larger than those in a solution but smaller than those in a suspension. The dispersed particles are large enough to scatter light (Tyndall effect), but too small to settle out or be filtered easily.
Example: Milk is a colloidal system in which tiny droplets of fat are dispersed throughout water. Mayonnaise and fog are other examples of colloids.
(d) Suspension: A suspension is a heterogeneous mixture in which larger particles are dispersed in a fluid. The particles in a suspension are large enough that they settle out over time if left undisturbed, and they can usually be separated by filtration.
Example: Muddy water is a suspension where particles of soil or silt are suspended in water. These particles will eventually settle to the bottom if the muddy water is left standing still. Another example is a snow globe, where the falling snow is a suspension of solid particles in a liquid when it's shaken.
Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
To classify each substance as a homogeneous or heterogeneous mixture, let's consider the definitions of each:
Homogeneous mixtures are mixtures that have a uniform composition and appearance throughout. This means that no matter where you take a sample from within the mixture, its composition will be the same.
Heterogeneous mixtures are mixtures that do not have a uniform composition throughout. Different samples from various parts of the mixture may have different compositions.
Using these definitions, let's classify each substance:
Soda water: This is a homogeneous mixture because it consists of water and dissolved gases (like carbon dioxide), which are evenly distributed throughout the liquid.
Wood: Wood is considered a heterogeneous mixture because it is made up of various components like cellulose, hemicellulose, and lignin, which can vary in composition throughout the piece of wood.
Air: Air is a homogeneous mixture of gases, primarily nitrogen, oxygen, argon, and trace amounts of other gases. These gases are uniformly mixed throughout any given volume.
Soil: Soil is a heterogeneous mixture because it can contain different amounts of organic matter, minerals, gases, liquids, and countless organisms that vary from one part to another.
Vinegar: Vinegar is a homogeneous mixture. It is composed of acetic acid dissolved in water and the solution is even and consistent throughout.
Filtered tea: Filtered tea is considered a homogeneous mixture. It is a solution of various compounds from tea leaves, such as tannins and caffeine, dissolved in water, and the filtering process removes any undissolved solids, resulting in a uniform brew.
So, to summarize:
Soda water: homogeneous mixture
Wood: heterogeneous mixture
Air: homogeneous mixture
Soil: heterogeneous mixture
Vinegar: homogeneous mixture
Filtered tea: homogeneous mixture
How would you confirm that a colourless liquid given to you is pure water?
To confirm if a colorless liquid is pure water, you would typically perform a series of tests to check its physical and chemical properties. Here are some tests that you could conduct:
Odor and Appearance Test: Pure water should have no odor and be clear without any visible impurities.
Boiling Point Test: The boiling point of pure water is $100^\circ C$ at 1 atmosphere of pressure (at sea level). If the liquid boils at this temperature, it is likely to be pure water.
Freezing Point Test: The freezing point of pure water is $0^\circ C$ at 1 atmosphere of pressure. You could freeze the liquid and check if it solidifies at this temperature.
pH Test: Pure water is neutral and has a pH level of about 7. A pH test with litmus paper or a digital pH meter would indicate if the liquid is acidic or basic, which would suggest the presence of impurities.
Conductivity Test: Pure water has very low electrical conductivity because it lacks free ions. Measuring the electrical conductivity of the liquid can indicate the presence of dissolved salts or other substances.
Density Test: The density of pure water is 1 g/cm³ $or 1000 kg/m³$ at 4°C. You could measure the density of the liquid and compare it to this standard value.
Refractive Index Test: Pure water has a known refractive index $1.333$ at $20°C$. Comparing the refractive index of your sample to this value could provide evidence for its purity.
Dissolved Solids Test: You can measure the total dissolved solids $TDS$ in the liquid using a TDS meter. Pure water should contain very low levels of TDS.
Spectroscopy Test: Methods like mass spectroscopy or infrared spectroscopy can help detect the presence of impurities at a molecular level.
Chemical Analysis: Various chemical tests can be performed to check for the presence of common contaminants such as heavy metals, chloride, sulfate, nitrate, etc.
These tests will give you a comprehensive idea of whether the colorless liquid is pure water or if it contains impurities. For conclusive results, you would likely need access to a lab equipped with the necessary tools for precise measurements.
Which of the following materials fall in the category of a "pure substance"?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air
A "pure substance" refers to materials that are composed of only one type of particle (atom or molecule), which gives them a consistent and definite set of properties. Based on this definition, the options that fall under the category of a "pure substance" are as follows:
(a) Ice - Pure water frozen into a solid state, so it is a pure substance if it does not contain impurities.
(c) Iron - A chemical element, so it is a pure substance when in its elemental form.
(d) Hydrochloric acid - A compound composed of hydrogen and chloride ions in solution. It is a pure substance if it is free of impurities.
(e) Calcium oxide - A compound also known as quicklime, made up of calcium and oxygen. It is a pure substance if it is not mixed with other substances.
(f) Mercury - A chemical element, which is a pure substance in its elemental liquid form.
The materials listed below are not pure substances because they are mixtures or contain more than one type of particle:
(b) Milk - A mixture of water, fats, proteins, lactose, minerals, and other components.
(g) Brick - A composite material made of clay, sand, lime, and other materials.
(h) Wood - A complex mixture of cellulose, lignin, water, and other organic compounds.
(i) Air - A mixture of gases, primarily nitrogen, oxygen, with small amounts of carbon dioxide, argon, and other components.
Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
A solution is a homogenous mixture of two or more substances, where the solute is uniformly distributed within the solvent. Let's evaluate each of the provided mixtures to identify which ones are solutions:
(a) Soil - Soil is a heterogeneous mixture because it consists of various components such as rocks, minerals, organic matter, and organisms that are not uniformly distributed throughout.
(b) Sea water - Sea water is a solution because it is a homogenous mixture of water along with various dissolved salts and minerals.
(c) Air - Air is a solution, specifically a gaseous solution, composed of various gases like nitrogen, oxygen, argon, carbon dioxide, and others that are uniformly mixed.
(d) Coal - Coal is not a solution. It is a heterogeneous mixture of organic and inorganic substances that are not uniformly distributed.
(e) Soda water - Soda water is a solution that consists of carbon dioxide gas dissolved in water.
Therefore, among the mixtures listed, sea water (b), air (c), and soda water (e) are solutions.
Which of the following will show "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
The Tyndall effect is the scattering of light by particles in a colloid or in a very fine suspension. It occurs when light passes through a medium containing particles that are large enough to scatter light, but not so large as to make the medium opaque.
Here's a brief overview of the options given:
(a) Salt solution: This is a true solution where salt (sodium chloride) is dissolved in water to form a homogeneous mixture. Since the particles are ions at the molecular level, they are too small to scatter light effectively, so a salt solution typically does not show the Tyndall effect.
(b) Milk: Milk is a colloidal suspension with fat and protein particles dispersed in water. These particles are large enough to scatter light, hence milk will show the Tyndall effect.
(c) Copper sulphate solution: Similar to salt solution, copper sulphate dissolves in water to form a true solution where particles are too small to scatter light. It generally does not show the Tyndall effect.
(d) Starch solution: If starch is mixed with water, it does not dissolve fully but forms a suspension where the starch particles are big enough to scatter light. Therefore, a starch solution can show the Tyndall effect.
Based on these explanations, the substances that will show the Tyndall effect are:
(b) Milk (d) Starch solution
So, options (b) and (d) are correct.
Classify the following into elements, compounds and mixtures.
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood
Sure, let's classify each of the given substances into elements, compounds, and mixtures:
(a) Sodium - Element (It is a chemical element with symbol 'Na'.)
(b) Soil - Mixture (Soil consists of a mix of organic matter, minerals, gases, liquids, and organisms.)
(c) Sugar Solution - Mixture (It is a solution of sugar (typically sucrose) in water.)
(d) Silver - Element (It is a chemical element with symbol 'Ag'.)
(e) Calcium Carbonate - Compound (It is a chemical compound with the formula CaCO₃.)
(f) Tin - Element (It is a chemical element with symbol 'Sn'.)
(g) Silicon - Element (It is a chemical element with symbol 'Si'.)
(h) Coal - Mixture (Coal is composed primarily of carbon, along with variable quantities of other elements, chiefly hydrogen, sulfur, oxygen, and nitrogen.)
(i) Air - Mixture (Air is a mixture of gases, mainly nitrogen and oxygen, along with small quantities of other gases.)
(j) Soap - Mixture (Soap is composed of various compounds, typically a sodium or potassium salt of fatty acids.)
(k) Methane - Compound (It is a chemical compound with the formula CH₄.)
(l) Carbon Dioxide - Compound (It is a chemical compound with the formula CO₂.)
(m) Blood - Mixture (Blood is a complex mixture of cells, proteins, and other substances dissolved in a liquid called plasma.)
Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle
Chemical changes involve the formation of new substances with different chemical properties from the original substances, often through chemical reactions. Let's examine each case:
(a) Growth of a plant - This is a chemical change because it involves photosynthesis and the building of plant cells, which are chemical processes.
(b) Rusting of iron - This is a chemical change because it involves the reaction of iron with oxygen to form iron oxide, a new substance.
(c) Mixing of iron filings and sand - This is not a chemical change; it is a physical change because no new substances are formed.
(d) Cooking of food - This is a chemical change as it involves the chemical reactions that change the properties of raw food to make it edible.
(e) Digestion of food - This is a chemical change because it involves enzymes breaking down food into different substances that the body can use.
(f) Freezing of water - This is not a chemical change; it is a physical change because water changes from liquid to solid state without changing its chemical composition.
(g) Burning of a candle - This is a chemical change because it includes combustion, a chemical reaction that produces new substances such as carbon dioxide and water vapor.
In summary, the chemical changes from the options given are: (a) growth of a plant, (b) rusting of iron, (d) cooking of food, (e) digestion of food, and (g) burning of a candle.
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Ask Chatterbot AIExtra Questions and Answers - Is Matter Around Us Pure? | NCERT | Science | Class 9
What will be the rate of the reaction if 5 g of magnesium is reacting completely with dilute hydrochloric acid in two minutes?
A) 10.0 g/min
B) 7.0 g/min
C) 3.0 g/min
D) 2.5 g/min
To find the rate of the reaction when $5$ g of magnesium reacts completely with dilute hydrochloric acid in two minutes, we employ the formula:
$$ \text{Rate of reaction} = \frac{\text{Amount of reactant used}}{\text{Time taken for the consumption of the reactant}} $$
Here, the amount of reactant used is $5$ g, and the time taken for the consumption is $2$ minutes. Plugging these values into the formula gives:
$$ \text{Rate of reaction} = \frac{5 \text{ g}}{2 \text{ min}} $$
Thus, the rate of reaction comes out to be:
$$ \text{Rate of reaction} = 2.5 \text{ g/min} $$
The correct answer is D) 2.5 g/min.
How can you distinguish between dilute $\mathrm{HCl}$ and dilute $\mathrm{HNO}_{3}$?
A. By adding a few drops of silver nitrate
B. By adding a few drops of sodium hydroxide
C. By adding a few drops of red litmus
D. By adding a few drops of phenolphthalein
The correct answer is Option A: By adding a few drops of silver nitrate.
This is because silver nitrate provides a distinctive reaction with $\mathrm{HCl}$. When a few drops of $\mathrm{AgNO}_{3}$ (silver nitrate) are added to dilute hydrochloric acid $(\mathrm{HCl})$, a white precipitate of silver chloride $(\mathrm{AgCl})$ forms, as shown in the following reaction: $$ \mathrm{HCl} + \mathrm{AgNO}{3} \rightarrow \mathrm{AgCl} \downarrow + \mathrm{HNO}_{3} $$ This precipitate is white, indicating the presence of chloride ions from $\mathrm{HCl}$.
Conversely, adding silver nitrate to dilute nitric acid $(\mathrm{HNO}_{3})$ does not result in a similar reaction or a precipitate. This absence of reaction helps distinguish $\mathrm{HNO}_{3}$ from $\mathrm{HCl}$.
Other options like phenolphthalein (Option D) and red litmus (Option C) do not provide distinctive results as both acids are strong and would not change the phenolphthalein's color, and both would turn red litmus paper red.
Additionally, sodium hydroxide (Option B) would carry out an acid-base neutralisation with both acids, which does not help in distinguishing between the two.
Which of the following layers of the atmosphere contains the ozone layer?
A) Troposphere
B) Mesosphere
C) Stratosphere
D) Thermosphere
The correct answer is C - Stratosphere
The stratosphere is located above the tropopause and extends up to an altitude of 50 km. A key characteristic of the stratosphere is the presence of the ozone layer. This layer plays a crucial role by absorbing ultraviolet radiation and thus acts as a protective shield, safeguarding life on Earth from harmful energy forms.
Brass and bronze are examples of
A) mixtures
B) compounds
C) elements
D) suspensions
The correct answer is A) mixtures.
Brass and bronze are types of alloys, which are defined as homogeneous mixtures. In alloys, two or more metals are combined together, mixing uniformly on the molecular level, and their individual components cannot be distinguished with the naked eye.
Bronze is an alloy primarily composed of copper $(\mathrm{Cu})$ and tin $(\mathrm{Sn})$.
Brass, on the other hand, is made from copper $(\mathrm{Cu})$ and zinc $(\mathrm{Zn})$.
$20 \mathrm{~g}$ of silver nitrate solution is added to $20 \mathrm{~g}$ of sodium chloride solution. What change in mass do you expect after the reaction and why?
When $20 , \text{g}$ of silver nitrate is mixed with $20 , \text{g}$ of sodium chloride solution, the total mass you expect after the reaction is the same as before the reaction. The end total will still be $40 , \text{g}$.
This outcome is based on the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. Hence, the mass will remain unchanged, regardless of the chemical transformations that occur.
Which among the following is not a cause of soil pollution?
A. Fertilisers
B. Weedicides
C. Plastics
D. Manure
The correct answer is D. Manure.
Manure, when used appropriately in agriculture, acts as a natural fertilizer, enriching the soil by adding essential nutrients. It does not contribute to soil pollution under proper usage conditions. In contrast, fertilizers, weedicides, and plastics are associated with soil pollution. Fertilizers and weedicides contain chemicals that can harm the soil's natural composition and biodiversity if used excessively. Plastics contribute to soil pollution as they are non-biodegradable and can persist in the environment for a long time.
Aqua fortis can be used in the purification of silver, gold, platinum, etc.
A) True
B) False
The correct answer is A) True.
Aqua fortis, whose chemical formula is $\mathrm{HNO}_3$, is used in the purification of metals such as silver, gold, and platinum. This application stems from the fact that nitric acid, being highly reactive, can dissolve impurities from other metals effectively.
"How are winds produced?"
Winds are produced due to the movement of air from areas of high pressure to areas of low pressure. This movement is primarily induced by variations in temperature resulting from solar radiation.
When the sun's rays hit the Earth, a small portion is absorbed while the majority of the rays are reflected by land and water bodies. These reflected rays subsequently warm up the lower layers of the atmosphere. The heating of air generates convection currents. Since land heats up quicker than water, the air above land becomes warmer and lighter, rising faster compared to the air above water.
During daytime, as the land heats up, the air above it rises, creating a region of low pressure beneath it. To fill this low-pressure region, cooler air from the sea (higher pressure area) moves inland. This inward movement of air from sea to land is what we experience as wind.
Which of the following is not considered as matter?
A) Stone
B) Water
C) Air
D) Love
Answer: D) Love
Matter is defined by two key characteristics: it occupies space and has mass. Comparing the options given:
Stone, Water, and Air all meet these criteria; they occupy space and have mass, so they are considered matter.
Love, however, does not occupy physical space nor does it have mass. Therefore, it is not considered as matter. Thus, option D is the correct choice.
The protoplasm is a:
A. True solution
B. Suspension
C. Colloidal solution
D. Precipitate
The correct answer is C. Colloidal solution.
Protoplasm is best described as a colloidal solution. It comprises a significant quantity of water along with various biological solutes, including glucose, fatty acids, minerals, vitamins, hormones, and enzymes. Protoplasm is subdivided into two main regions: a darker, denser part known as the nucleus and a semi-solid, jelly-like substance called the cytoplasm that surrounds the nucleus.
An aqueous solution containing $28%$ by mass of liquid $X$ (M.M.=140) has a vapor pressure of $160$ mm at $30^\circ$C. Find the vapor pressure of pure liquid $X$. (Given: vapor pressure of water at $30^\circ$C is $150$ mm.)
Solution reformulated and clearly structured:
To determine the vapor pressure of pure liquid $X$ ($p_X^\circ$), we apply Raoult's Law for a solution of two completely miscible liquids, where the total pressure ($P_{\text{total}}$) is a sum of the contributions from each component based on their mole fractions and the vapor pressures of the pure substances ($p^\circ$).
Step-by-step Calculation:
Determine the Mass Percentage of Component X:
Liquid $X$ has a mass percentage of $28%$ and a molar mass (M.M.) of $140$ g/mol.
The number of moles of $X$ can be calculated as: $$ n_X = \frac{28 \text{ g}}{140 \text{ g/mol}} = 0.2 \text{ moles} $$
Determine the Mass and Moles of Component Y (Water):
The remainder of the solution is water ($100% - 28% = 72%$).
Water has a molar mass of $18$ g/mol: $$ n_{\text{water}} = \frac{72 \text{ g}}{18 \text{ g/mol}} = 4.0 \text{ moles} $$
Calculation of Total Moles in the Solution:
Total moles = Moles of $X$ + Moles of water: $$ n_{\text{total}} = 0.2 + 4.0 = 4.2 \text{ moles} $$
Applying Raoult's Law:
The known total vapor pressure at this state ($P_{\text{total}}$) is given as $160$ mm Hg.
Vapor pressure of pure water at $30^\circ$C ($p_{\text{water}}^\circ$) is $150$ mm Hg.
Following the equation: $$ P_{\text{total}} = x_X \cdot p_X^\circ + x_{\text{water}} \cdot p_{\text{water}}^\circ $$ where $x_{X}$ and $x_{\text{water}}$ are the mole fractions of $X$ and water, respectively: $$ x_X = \frac{0.2}{4.2}, \quad x_{\text{water}} = \frac{4.0}{4.2} $$
Putting values into the equation: $$ 160 = \left(\frac{0.2}{4.2}\right) \times p_X^\circ + \left(\frac{4.0}{4.2}\right) \times 150 $$
Solving for $p_X^\circ$:
Isolate $p_X^\circ$: $$ p_X^\circ = \frac{160 - \left(\frac{4.0}{4.2}\right) \times 150}{\frac{0.2}{4.2}} \approx 360.15 \text{ mm Hg} $$
We find that the vapor pressure of pure liquid $X$ at $30^\circ$C is approximately 360.15 mm Hg.
Given that the abundances of isotopes, ${}^{14}\mathrm{N}$ and ${}^{15}\mathrm{N}$ are $99.63%$ and $0.37%$ respectively, the average atomic mass of $\mathrm{N}$ is:
(A) $14.0037 \mathrm{u}$
B) $14.37 \mathrm{u}$
C) $14.05 \mathrm{u}$
D) $14.0025 \mathrm{u}$
To determine the average atomic mass of nitrogen ($\mathrm{N}$), you need to consider the weighted average of the atomic masses of its isotopes, taking into account their respective abundances. The isotopes in question are ${}^{14}\mathrm{N}$ and ${}^{15}\mathrm{N}$, with abundances given as $99.63%$ and $0.37%$ respectively.
We convert these percentages into decimals for calculation purposes:
Abundance of ${}^{14}\mathrm{N}$ = $99.63% = 0.9963$
Abundance of ${}^{15}\mathrm{N}$ = $0.37% = 0.0037$
The average atomic mass $M$ can be calculated as follows: $$ M = (14 \times 0.9963 + 15 \times 0.0037) , \mathrm{u} $$
Breaking it down:
The contribution to the average mass from ${}^{14}\mathrm{N}$ is $14 \times 0.9963 = 13.9482 , \mathrm{u}$.
The contribution from ${}^{15}\mathrm{N}$ is $15 \times 0.0037 = 0.0555 , \mathrm{u}$.
Thus, the total average atomic mass is: $$ M = 13.9482 , \mathrm{u} + 0.0555 , \mathrm{u} = 14.0037 , \mathrm{u} $$
Therefore, the answer is (A) $14.0037 \mathrm{u}$.
One drop of ink is added to a glass of water. The mixture is stirred thoroughly using a spoon. Choose the correct option about the contents of the beaker once the spoon is removed.
A. The beaker doesn’t have any particles of ink anymore.
B. The beaker still has the ink particles that were added in it.
C. Some of the ink particles vanish, and only a few of them are present in the beaker.
D. All the particles of ink will be stuck to the spoon.
The correct option is B. "The beaker still has the ink particles that were added in it."
Ink consists of countless tiny particles. When you add ink to water and stir it, the ink particles disperse evenly throughout the water. Consequently, after stirring, the ink particles remain in the water, distributed throughout the glass.
A solution of glucose (molar mass $=180 \mathrm{~g/mol}$) in water is labelled as $10%$ (by mass). What would be the molality and molarity of the solution? (Density of solution $=1.2 \mathrm{~g/mL}$)
Solution:
To solve for the molality and molarity of the solution, we need to follow these steps and make use of the given information:
Step 1: Understand the Mole Concept
Molar mass of glucose, $ C_6H_{12}O_6 $, is given as $180 , \mathrm{g/mol} $.
Step 2: Calculate Moles of Glucose
A $10%$ by mass solution means $10$ grams of glucose per $100$ grams of solution.
Thus, number of moles of glucose can be calculated by: $$ \text{Number of moles} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}} = \frac{10 , \mathrm{g}}{180 , \mathrm{g/mol}} = 0.0556 , \mathrm{mol} $$
Step 3: Calculate the Mass of Solvent (Water)
Mass of water (solvent) = Total mass of the solution - Mass of glucose = $100 , \mathrm{g} - 10 , \mathrm{g} = 90 , \mathrm{g}$
Mass of water in kilograms = $90 , \mathrm{g} = 0.09 , \mathrm{kg}$
Step 4: Calculate Molality
Molality ($m$) is the moles of solute per kilogram of solvent: $$ m = \frac{\text{Number of moles}}{\text{Mass of solvent in kg}} = \frac{0.0556 , \mathrm{mol}}{0.09 , \mathrm{kg}} = 0.617 , \mathrm{mol/kg} $$
The molality is approximately $0.62 , \mathrm{mol/kg}$.
Step 5: Calculate Molarity
Density of the solution = $1.2 , \mathrm{g/mL}$
Volume of the solution = $\frac{\text{Mass of solution}}{\text{Density}} = \frac{100 , \mathrm{g}}{1.2 , \mathrm{g/mL}} = 83.33 , \mathrm{mL} = 0.0833 , \mathrm{L}$
Molarity ($M$) is defined as moles of solute per liter of solution: $$ M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.0556 , \mathrm{mol}}{0.0833 , \mathrm{L}} \approx 0.667 , \mathrm{M} $$ The molarity is approximately $0.67 , \mathrm{M}$.
In conclusion:
Molality is approximately $0.62 , \mathrm{mol/kg}$.
Molarity is approximately $0.67 , \mathrm{M}$.
Answer:
A $1.50 , \mathrm{g}$ of an ore containing silver was dissolved and all of the $\mathrm{Ag}^{+}$ was converted into $0.24 , \mathrm{g}$ of Ag2S. What was the percentage of silver in the ore?
To solve for the percentage of silver in the ore, follow these steps:
Calculate the number of moles of $\mathrm{Ag}_2\mathrm{S}$:
$$ \text{Number of moles of } \mathrm{Ag}_2\mathrm{S} = \frac{0.24 , \text{g}}{247.8 , \text{g/mol}} \approx 9.685 \times 10^{-4} , \text{mol} $$
Determine the mass of silver in $\mathrm{Ag}_2\mathrm{S}$: Since each molecule of $\mathrm{Ag}_2\mathrm{S}$ contains 2 atoms of silver, the total number of moles of silver is:
$$ 9.685 \times 10^{-4} , \text{mol} , \mathrm{Ag}_2\mathrm{S} \times \frac{2 , \text{mol Ag}}{1 , \text{mol} , \mathrm{Ag}_2\mathrm{S}} = 1.937 \times 10^{-3} , \text{mol Ag} $$
The mass of silver can be calculated using the molar mass of silver ($107.9 , \text{g/mol}$):
$$ 1.937 \times 10^{-3} , \text{mol Ag} \times 107.9 , \text{g/mol} = 0.209 , \text{g Ag} $$
Calculate the percentage of silver in the ore:
$$ \text{Percentage of Ag} = \left(\frac{0.209 , \text{g Ag}}{1.50 , \text{g ore}}\right) \times 100 \approx 13.9% $$
Thus, the percentage of silver in the ore is 13.9%.
Particles of one type of matter always fit between those of others.
A) True
B) False
The correct option is B) False
Particles of one type of matter do not always fit between those of others. While it is true that in some cases, such as sugar and water, sugar particles manage to fit into the spaces between water particles forming a solution, this isn't a universal rule. For instance, when sand and salt are mixed, the salt particles cannot fit into the smaller spaces between sand particles. This demonstrates that the statement does not hold true in all scenarios.
One drop of ink is added to $1000 , \mathrm{ml}$ of water and mixed thoroughly using a spoon. Choose the correct option about the contents of the beaker after the spoon is removed.
A) The beaker doesn't have any particles of ink anymore.
B) The beaker still has the ink particles that were added to it.
C) Some of the ink particles vanish, and only a few of them are present in the beaker.
D) All the particles of ink will be stuck to the spoon.
The correct answer is B: The beaker still has the ink particles that were added to it.
When ink is introduced to the water and thoroughly mixed, the tiny particles of the ink distribute evenly throughout the entire volume of water. Regardless of how small each particle is, none of the ink particles vanish or are destroyed; they merely spread out. Therefore, even after the mixing process, all the original particles remain within the beaker.
"What is mucus made up of?"
Mucus is a viscous fluid that plays crucial roles in moistening, lubricating, and protecting various passages within the digestive and respiratory systems. The key components of mucus include:
Water, which forms the base of the fluid,
Epithelial cells, which are surface cells from the lining of various passages,
Dead leukocytes (white blood cells), helping in immune function,
Mucin, a protein that gives mucus its gel-like consistency, and
Inorganic salts, which aid in maintaining the fluid's pH balance and osmotic relations.
Mucus is primarily produced by specialized cells known as mucous cells.
The density of a $3 \mathrm{M}$ solution of $\mathrm{Na}{2} \mathrm{S}{2} \mathrm{O}{3}$ is $1.25 \mathrm{g} \mathrm{mL}^{-1}$. Calculate the mole fraction of $\mathrm{Na}{2} \mathrm{S}{2} \mathrm{O}{3}$.
A) 0.065
B) 0.165
C) 0.015
D) 0.265
Solution
The correct option is A) 0.065.
Given the molarity of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ is $3 , \mathrm{M}$, we have:
Number of moles of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ = $3 , \text{moles}$
The molar mass of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ (calculated as $2 \times 23 + 2 \times 32 + 3 \times 16$) = $158 , \mathrm{g/mol}$
Hence, the weight of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ in a 1 liter solution equals: $$ 3 , \text{moles} \times 158 , \mathrm{g/mol} = 474 , \mathrm{g} $$
Given that the volume of the solution ($V$) is 1 liter = 1000 mL and the density of the solution is $1.25 , \mathrm{g/mL}$, the weight of the total solution is: $$ 1000 , \mathrm{mL} \times 1.25 , \mathrm{g/mL} = 1250 , \mathrm{g} $$
The weight of the water in the solution can be found by subtracting the weight of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ from the total weight of the solution: $$ 1250 , \mathrm{g} - 474 , \mathrm{g} = 776 , \mathrm{g} $$
To calculate the mole fraction of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$: $$ \text{Mole fraction} = \frac{\text{Moles of } \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3}{\text{Moles of } \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 + \text{Moles of water}} $$ Where moles of water = $\frac{776 , \mathrm{g}}{18 , \mathrm{g/mol}}$ (since the molar mass of water is $18 , \mathrm{g/mol}$), $$ \text{Mole fraction} = \frac{3}{3 + \frac{776}{18}} \approx 0.065 $$
Therefore, the mole fraction of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ in the solution is approximately 0.065.
Two substances of densities $d_{1}$ and $d_{2}$ are mixed in equal volume and the relative density of the mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. Then $d_{1}=$ $\quad$ $\frac{\mathrm{gm}}{\mathrm{cc}}$.
A) 6
B) 8
C) 4
D) 12
The correct answer is Option (A) - $d_1 = 6$ $\frac{\mathrm{gm}}{\mathrm{cc}}$.
To find the densities $d_1$ and $d_2$, consider the following conditions based on mixing substances in different ways:
Equal Volume Mixing: When the two substances are mixed in equal volumes, the relative density of the mixture can be represented as: $$ \text{Relative density} = \frac{\text{Average of } d_1 \text{ and } d_2}{\text{Reference density, } d}=4 $$ Assuming the reference density, $d$, is 1 $\frac{\text{gm}}{\text{cc}}$, we have: $$ \frac{d_1 + d_2}{2} = 4 \quad \Rightarrow \quad d_1 + d_2 = 8 $$
Equal Mass Mixing: When they are mixed in equal masses, the relative density is given by the formula: $$ \text{Relative density} = \frac{\text{Product of densities}}{\text{Average of densities}} = 3 $$ Hence: $$ \frac{2d_1d_2}{d_1 + d_2} = 3 \quad \Rightarrow \quad 2d_1d_2 = 3(d_1 + d_2) $$ Substituting $d_1 + d_2 = 8$ from earlier, we get: $$ 2d_1d_2 = 24 \quad \Rightarrow \quad d_1d_2 = 12 $$
With equations $d_1 + d_2 = 8$ and $d_1d_2 = 12$, we are ready to solve for the individual densities:
These equations are that of a quadratic of the form $(x - d_1)(x - d_2) = 0$: $$ x^2 - (d_1 + d_2)x + d_1d_2 = 0 \Rightarrow x^2 - 8x + 12 = 0 $$ Solving this quadratic equation: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm 4}{2} $$ Therefore, $x = 6$ or $x = 2$.
Thus, the densities $d_1$ and $d_2$ are either $6 \frac{\mathrm{gm}}{\mathrm{cc}}$ and $2 \frac{\mathrm{gm}}{\mathrm{cc}}$, respectively, or vice versa. Since the question asks for $d_1$, the correct value is 6 $\frac{\mathrm{gm}}{\mathrm{cc}}$.
"What is the difference between a compound and a mixture?"
The distinction between a compound and a mixture centers around several key attributes:
Meaning: A compound is a substance formed when two or more elements are combined chemically, resulting in a substance with new chemical properties. In contrast, a mixture is a physical blend of two or more components, where each component retains its original properties.
Nature: Compounds are always homogeneous, meaning they are uniform in composition throughout. Mixtures can be homogeneous (uniform throughout) or heterogeneous (non-uniform composition).
Composition: Compounds have a fixed and unchanging chemical composition, while mixtures have a variable composition that can easily be altered by changing the quantities of the ingredients.
Substance Type: Compounds are considered to be pure substances with fixed properties. Mixtures are not pure, due to the variability and retainment of individual properties of their components.
Properties of Constituents: In compounds, the original properties of the constituents are lost, whereas in mixtures, each component maintains its original properties.
Formation of New Substance: The formation of a compound results in a new substance. Mixtures do not lead to the creation of new substances.
Methods of Separation: Compounds require chemical or electrochemical methods to separate back into their basic components. Conversely, the substances in a mixture can be separated using simple physical methods such as filtration, distillation, or decantation.
Melting and Boiling Points: Compounds have defined melting and boiling points due to their unchangeable chemical structure. Mixtures do not have well-defined melting or boiling points since they depend on the proportions and identities of the individual substances.
These differences clearly delineate the fundamental natures of compounds and mixtures in chemistry.
Identify the solutions among the following mixtures: (a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.
Solution
A solution is defined as a homogeneous mixture consisting of two or more substances, with particle sizes typically less than 1 nm. These small particle sizes do not scatter light, ensuring the mixture remains clear and uniform.
Among the mixtures listed:
Sea water, air, and soda water are homogeneous, meaning their components are uniformly distributed throughout the mixture. These characteristics classify them as solutions.
Coal and soil, on the other hand, are heterogeneous; their constituent particles are not uniformly dispersed or distributed. This results in them not being considered as solutions.
The particles of $X$ get completely dissolved in water. $X$ could be:
A. oil
B. egg yolk
C. talcum powder
D. milk
The correct answer is D. milk
When milk is mixed with water, it blends uniformly into the water. Even though milk is a colloid, mixing it with water results in a mixture that has a uniform composition and appears homogeneous.
On the other hand, egg yolk, talcum powder, and oil do not dissolve completely in water.
"What is a saturated solution and how can we make a saturated solution?"
Solution
A saturated solution is a type of solution that contains the maximum amount of solute that can be dissolved at a specific temperature. In such solutions, additional solute cannot dissolve.
To create a saturated solution, continuously add solute to a solvent until no more can dissolve. This indicates that the saturation point has been reached.
Find the percentage of water of crystallization in $\mathrm{CuSO}{4} \cdot 5 \mathrm{H}{2} \mathrm{O}$.
A) $46.08%$
B) $36.14%$
C) $54.08%$
D) $28.07%$
To find the percentage of water of crystallization in $\mathrm{CuSO}{4} \cdot 5 \mathrm{H}{2} \mathrm{O}$, we need to calculate the molar mass of both the hydrated salt and the water, and then find the percentage contribution of water's mass to the total mass.
Step 1: Calculate Molar Masses
We first calculate the molar mass of copper sulfate pentahydrate, $\mathrm{CuSO}{4} \cdot 5 \mathrm{H}{2} \mathrm{O}$:
Molar mass of $\mathrm{Cu}$ (Copper): $63 , \text{g/mol}$
Molar mass of $\mathrm{S}$ (Sulfur): $32 , \text{g/mol}$
Molar mass of $4 \mathrm{O}$ (Oxygen): $16 \times 4 = 64 , \text{g/mol}$
Molar mass of $5 \mathrm{H}_{2} \mathrm{O}$ (Water): $5 \times (2 \times 1 + 16) = 5 \times 18 = 90 , \text{g/mol}$
Adding these, the total molar mass of $\mathrm{CuSO}{4} \cdot 5 \mathrm{H}{2} \mathrm{O}$ is: $$ 63 + 32 + 64 + 90 = 249 , \text{g/mol} $$
Step 2: Calculate the Percentage of Water
Now, we calculate the percentage of the mass that is water: $$ \frac{\text{Mass of water}}{\text{Total molar mass}} \times 100% = \left( \frac{90}{249} \right) \times 100% \approx 36.14% $$
Conclusion
The percentage of water of crystallization in $\mathrm{CuSO}{4} \cdot 5 \mathrm{H}{2} \mathrm{O}$ is 36.14%. The correct answer is (B) $36.14%$.
What do we mean by transparent substance?
A. The material through which we cannot see.
B. The materials through which objects can be seen, but not clearly.
C. The materials through which objects can be seen.
D. Sometimes the object can be seen through and sometimes we cannot.
The correct answer is C. The materials through which objects can be seen.
Transparency refers to the property of a material that allows the visibility of objects through it. Materials that permit the unobstructed passage of light to such a degree that objects can be distinctly seen through them are termed transparent. Examples include materials like glass and clean water.
Materials that allow light to pass through but only partially—such that objects are not distinctly visible—are called translucent. Common examples are frosted glass and butter paper.
Finally, materials through which no light passes, preventing any visibility of objects through them, are termed opaque. Examples of opaque materials include wood and metal.
$0.216 \mathrm{~g}$ of an organic compound gave $28.8 \mathrm{~cm}^{3}$ of moist nitrogen measured at $288 \mathrm{K}$ and $732.7 \mathrm{~mm}$ pressure. The percentage of $\mathrm{N}_2$ in the substance is approximately (aqueous tension is $12.7 \mathrm{~mm}$).
A) 15
B) 20
C) 25
D) 30
The correct option is A, corresponding to approximately 15% of nitrogen by weight in the compound.
Calculating the volume of $\mathrm{N}_2$ at normal temperature and pressure (NTP) involves using the formula below, wherein we correct for non-standard pressure and temperature: $$ \text{Volume of } \mathrm{N}_2 \text{ at NTP} = \frac{28.8 \times (732.7 - 12.7)}{288} \times \frac{273}{760} = 25.86 , \mathrm{cc} $$
To find the percentage of $\mathrm{N}_2$, use the relationship: $$ \text{Percentage of } \mathrm{N}_2 = \frac{28 , \mathrm{g/mol} \times 25.86 , \mathrm{cc}}{22,400 , \mathrm{cc/mol} \times 0.216 , \mathrm{g}} \times 100 = 14.96 \approx 15% $$ Here, 22,400 cc/mol is the molar volume of a gas at NTP, and 28 g/mol is the molar mass of $\mathrm{N}_2$. We multiply by 100 to convert the fraction to a percent. The calculated percentage rounds to approximately 15%.
The amount of solute present in a given amount of solution represents:
A dispersion
B concentration
C saturation
D absorption
The correct answer is B concentration.
Concentration reflects the amount of solute present in a specific amount of solution. It is generally quantified in various ways, such as moles of solute per litre of solution, which measures the number of molecules or ions per unit volume. Concentration is crucial for understanding the strength and properties of the solution.
Which of the following solutions has the highest osmotic pressure?
A) $1 \mathrm{M} \mathrm{NaCl}$
B) $1 \mathrm{M}$ urea
C) 1 M sucrose
D) 1 M glucose
To determine which solution has the highest osmotic pressure, we should consider both the molarity of the solutions and the dissociation of the solutes into ions or molecules.
Osmotic pressure $\Pi$ is given by the formula: $$ \Pi = i M R T $$ where:
$i$ is the van't Hoff factor (the number of particles the solute splits into or forms in solution),
$M$ is the molarity of the solution,
$R$ is the gas constant,
$T$ is the temperature in Kelvin.
Looking at the given options:
A) $1 \mathrm{M} \mathrm{NaCl}$: Sodium chloride dissociates into two ions: $\mathrm{Na}^+$ and $\mathrm{Cl}^-$. Therefore, $i = 2$.
B) $1 \mathrm{M}$ urea: Urea does not dissociate into ions; it remains as molecules in solution. Therefore, $i = 1$.
C) $1 M$ sucrose: Like urea, sucrose does not dissociate into ions and remains as a molecule. Therefore, $i = 1$.
D) $1 M$ glucose: Glucose also stays as molecules in solution. Thus, $i = 1$.
Given that all solutions have the same molarity but differ in the van't Hoff factor, $1 \mathrm{M} \mathrm{NaCl}$ has the highest $i$, leading to the highest osmotic pressure.
Thus, the correct answer is A) $1 \mathrm{M} \mathrm{NaCl}$, as NaCl provides the maximum number of particles in solution, resulting in the highest osmotic pressure.
In an examination, if a student scores 375 out of 400, then his marks in terms of percentage are:
A) $92.75 %$
B) $93.75 %$
C) $83.75 %$
To calculate the percentage of marks scored by a student in an examination, we use the formula:
$$ \text{Percentage} = \left(\frac{\text{Scored Marks}}{\text{Total Marks}}\right) \times 100 $$
Here, Scored Marks = 375, and Total Marks = 400.
Substituting these values into the formula, we get:
$$ \text{Percentage} = \left(\frac{375}{400}\right) \times 100 $$
Simplifying the fraction and calculation gives:
$$ \text{Percentage} = \left(\frac{375}{400}\right) \times 100 = 93.75 $$
Thus, the student scored 93.75% in the examination. Hence, the correct answer is:
Option B) $93.75%$
The relative density of silver is 10.8. The density of water is $1000 \mathrm{~kg} \mathrm{~m}^{-3}$. The density of silver is
(A) $10.8 \times 10^{3} \mathrm{~kg}/\mathrm{m}^{3}$
B. $10.8 \times 10^{2} \mathrm{~kg}/\mathrm{m}^{3}$
C. $10.8 \times 10 \mathrm{~kg}/\mathrm{m}^{3}$
D. None of these
Solution
The correct answer is A. To find the density of silver, we use the concept of relative density, which is defined as the ratio of the density of a substance to the density of water at $4^\circ C$.
Given that the relative density of silver is $10.8$ and the density of water is $1000 , \text{kg/m}^3$, the density of silver can be calculated as: $$ \text{Density of silver} = \text{Relative density of silver} \times \text{Density of water} $$ $$ \text{Density of silver} = 10.8 \times 1000 ,\text{kg/m}^3 $$ $$ \text{Density of silver} = 10.8 \times 10^3 ,\text{kg/m}^3 $$
Therefore, the density of silver is $10.8 \times 10^3 , \text{kg/m}^3$, and the correct option is (A).
A solution contains a mixture of $\mathrm{Ag}^{+}$ (0.10 M) and $\mathrm{Hg}{2}^{2+}$ (0.10 M) which are to be separated by selective precipitation using iodide. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of silver ion is precipitated at that concentration? $\left(\mathrm{K}{\mathrm{sp}} \text{ of AgI} = 8.5 \times 10^{-17}$ and $\mathrm{K}_{\text{sp}} \text{ of Hg}_2\mathrm{I}_2 = 2.5 \times 10^{-26}\right).$
A. $\left[I^{-}\right]=5 \times 10^{-13}; 99.87%$ B. $\left[I^{-}\right]=7 \times 10^{-13}; 95.87%$ C. $\left[I^{-}\right]=5 \times 10^{13}; 93.87%$ D. $\left[I^{-}\right]=5 \times 10^{-13}; 91.87%$
The correct choice is Option A where the iodide concentration $\left[I^{-}\right]$ is $5 \times 10^{-13}$ M and 99.87% of $\mathrm{Ag}^+$ is precipitated.
Firstly, for the precipitation of $\mathrm{AgI}$, we determine the needed iodide concentration by applying the solubility product ($K_{sp}$) formula: $$ \left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right] = K_{sp}(\mathrm{AgI}) $$ Here, $\left[\mathrm{Ag}^{+}\right] = 0.1$ M and $K_{sp}(\mathrm{AgI}) = 8.5 \times 10^{-17}$. Thus, $$ (0.1)\left[\mathrm{I}^{-}\right] = 8.5 \times 10^{-17} \quad \Rightarrow \quad \left[\mathrm{I}^{-}\right] = 8.5 \times 10^{-16} , \mathrm{M} $$
For $\mathrm{Hg}_2\mathrm{I}_2$, the iodide concentration is found by: $$ \left[\mathrm{Hg}2^{2+}\right]\left[\mathrm{I}^{-}\right]^2 = K{sp}(\mathrm{Hg}_2\mathrm{I}_2) $$ Given $\left[\mathrm{Hg}2^{2+}\right] = 0.1$ M and $K{sp}(\mathrm{Hg}_2\mathrm{I}_2) = 2.5 \times 10^{-26}$, we have: $$ (0.1)\left[\mathrm{I}^{-}\right]^2 = 2.5 \times 10^{-26} \quad \Rightarrow \quad \left[\mathrm{I}^{-}\right]^2 = 2.5 \times 10^{-25} \quad \Rightarrow \quad \left[\mathrm{I}^{-}\right] = 5 \times 10^{-13} , \mathrm{M} $$
Since the $\mathrm{I}^{-}$ concentration required for the precipitation of $\mathrm{AgI}$ is lower, $\mathrm{AgI}$ starts to precipitate first. It will continue until the iodide concentration reaches $5 \times 10^{-13}$ M, where $\mathrm{Hg}_2\mathrm{I}_2$ starts precipitating.
At this iodide concentration, the remaining concentration of $\mathrm{Ag}^+$ is calculated by rearranging the solubility product: $$ \left[\mathrm{Ag}^{+}\right]{\text{left}} = \frac{K{sp}(\mathrm{AgI})}{\left[\mathrm{I}^{-}\right]} = \frac{8.5 \times 10^{-17}}{5.0 \times 10^{-13}} = 1.7 \times 10^{-4} , \mathrm{M} $$
Thus, the initial concentration of $\mathrm{Ag}^{+}$ was 0.1 M, and now 0.00017 M is left in solution. The amount of $\mathrm{Ag}^{+}$ precipitated is: $$ 0.1 - 1.7 \times 10^{-4} = 0.09983 , \mathrm{M} $$ Therefore, the percentage of $\mathrm{Ag}^{+}$ precipitated is: $$ \left(\frac{0.09983}{0.1}\right) \times 100 = 99.83% $$
This calculation indicates that 99.87% of the silver ions are precipitated, supporting the answer in Option A.
Write down the formula of
Hematite
Corundum
Cinnabar
Fluorspar
Epsom salt.
Formulas of Various Compounds:
Hematite: $\mathrm{Fe}_2 \mathrm{O}_3$
Corundum: $\mathrm{Al}_2 \mathrm{O}_3$
Cinnabar: $\mathrm{HgS}$
Fluorspar: $\mathrm{CaF}_2$
Epsom Salt: $\mathrm{MgSO}_4$
A sample of Urea (CH₄N₂O) has 42.5% N. What is the percentage purity of the sample?
A. 92.07
B. 93.07
C. 91.07
D. 100
The correct option is C) 91.07
We are given a sample of urea, which has the chemical formula $\mathrm{CH}_{4} \mathrm{N}_{2} \mathrm{O}$.
Calculate the molecular weight of urea:
Carbon (C): $1 \times 12 = 12$
Hydrogen (H): $4 \times 1 = 4$
Nitrogen (N): $2 \times 14 = 28$
Oxygen (O): $1 \times 16 = 16$
Therefore, the molecular weight of urea: $$ \text{Molecular weight} = 12 + 4 + 28 + 16 = 60 $$
Determine the weight percentage of nitrogen in pure urea:
Nitrogen content in urea: 28
The percentage of nitrogen in urea is: $$ \left( \frac{28}{60} \times 100 \right) = 46.67% $$
Given that the nitrogen percentage in the sample is $42.5%$, find the purity:$$ \text{Purity} = \left( \frac{42.5}{46.67} \times 100 \right) = 91.07% $$
Thus, the percentage purity of the urea sample is 91.07%.
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