Questions and Answers NCERT Science Class 9, Chapter 2 “Is Matter Around Us Pure”

Questions and Answers NCERT Science Class 9, Chapter 2 “Is Matter Around Us Pure”

Chapter 2: Is Matter Around Us Pure

Questions and Answers

In-text Questions

1. What is meant by a substance?

– A substance is a kind of matter that cannot be separated into other kinds of matter by any physical process. It has a definite set of properties.

1. List the points of differences between homogeneous and heterogeneous mixtures.

– Homogeneous Mixture:
– Has a uniform composition throughout.
– Components are not visible separately.
– Example: Salt dissolved in water.

• Heterogeneous Mixture:
• Does not have a uniform composition.
• Components can be seen separately.
• Example: Sand in water.

Exercise Questions

1. What is meant by a pure substance?

– A pure substance consists of a single type of particles. All elements and compounds are pure substances.

1. List the points of differences between homogeneous and heterogeneous mixtures.

– This question is similar to the in-text question above.

1. How are sol, solution, and suspension different from each other?

– Solution:
– Homogeneous mixture.
– Particle size: Less than 1 nm.
– Example: Salt in water.
– Sol:
– Colloidal solution (heterogeneous).
– Particle size: 1-100 nm.
– Example: Milk.
– Suspension:
– Heterogeneous mixture.
– Particle size: More than 100 nm.
– Example: Sand in water.

1. To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

– Solution:
[
\text{Concentration} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100
]
[
\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent} = 36 \text{ g} + 100 \text{ g} = 136 \text{ g}
]
[
\text{Concentration} = \frac{36 \text{ g}}{136 \text{ g}} \times 100 = 26.47\%
]

1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?

– This mixture can be separated by the process of simple distillation, as the boiling points of the two liquids differ significantly.

1. Name the technique to separate:

– Butter from curd: Centrifugation
– Salt from seawater: Evaporation
– Camphor from salt: Sublimation

1. What type of mixtures are separated by the technique of crystallisation?

– Crystallisation is used to separate solid-liquid mixtures, where the solid is soluble in the liquid. It helps in obtaining pure solid particles from a solution.

1. Classify the following as chemical or physical changes:

– Cutting of trees: Physical change
– Melting of butter in a pan: Physical change
– Rusting of almirah: Chemical change
– Boiling of water to form steam: Physical change
– Passing of electric current through water and the water breaking down into hydrogen and oxygen gases: Chemical change
– Dissolving common salt in water: Physical change
– Making a fruit salad with raw fruits: Physical change
– Burning of paper and wood: Chemical change

1. Try segregating the things around you as pure substances or mixtures.

– Pure Substances: Gold, water, sodium chloride, oxygen.
– Mixtures: Air, milk, seawater, steel.

1. Which separation techniques will you apply for the separation of the following?

– Sodium chloride from its solution in water: Evaporation
– Ammonium chloride from a mixture containing sodium chloride and ammonium chloride: Sublimation
– Small pieces of metal in the engine oil of a car: Filtration
– Different pigments from an extract of flower petals: Chromatography
– Butter from curd: Centrifugation
– Oil from water: Separating funnel
– Tea leaves from tea: Filtration
– Iron pins from sand: Magnetic separation
– Wheat grains from husk: Winnowing
– Fine mud particles suspended in water: Filtration

1. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue.

– To make tea:
1. Boil water (solvent).
2. Add tea leaves (solute) to the boiling water to make a tea solution.
3. Add sugar (solute) and stir to dissolve it.
4. Add milk and bring the mixture to a boil.
5. Filter the tea using a sieve. The insoluble tea leaves remain as residue on the sieve.
6. The filtrate is the tea, ready to be served.

1. Pragya tested the solubility of three different substances at different temperatures and collected the following data (results are given in the table as grams of substance dissolved in 100 grams of water to form a saturated solution).

– Solubility Data:

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 g of water at 313 K?
– At 313 K, the solubility of potassium nitrate is 62 g per 100 g of water.
[
\text{Mass needed for 50 g water} = \frac{62 \text{ g}}{100 \text{ g}} \times 50 \text{ g} = 31 \text{ g}
]

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and then cools it to 293 K. What would she observe as the solution cools? Explain.
– As the solution cools, potassium chloride will precipitate out of the solution because its solubility decreases with a decrease in temperature. The solubility at 353 K is 54 g/100 g of water, while at 293 K, it is 35 g/100 g of water. The excess potassium chloride will form crystals.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
– Solubility at 293 K:
– Potassium nitrate: 32 g/100 g water
– Sodium chloride: 36 g/100 g water
– Potassium chloride: 35 g/100 g water
– Sodium chloride has the highest solubility at 293 K.

(d) What is the effect of change of temperature on the solubility of a salt?
– The solubility of salts typically increases with an increase in temperature. However, the extent of this increase varies for different salts.

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