Chapter 2: Is Matter Around Us Pure?- Exercise- 2

Science

(www.olympiadsuccess.com)

Chapter 2: Is Matter Around Us Pure?

Class: IX

Exercise 2

Question 1

Differentiate between homogeneous and heterogeneous mixtures with examples.

Answer 1

A mixture that has uniform composition throughout is called as homogeneous mixture.

For example, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloy, and air have uniform compositions throughout the mixtures.

A mixture that has non-uniform composition throughout is called as heterogeneous mixture.

For example, composition of mixtures of sodium chloride and iron fillings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water are not uniform throughout the mixtures.

Question 2

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

Answer 2

Sol is a heterogeneous mixture. The size of the sol particles is too small to seen by naked eye. Also, they seem to be spread uniformly throughout the mixture. The Tyndall effect can be seen in this mixture.

examples: mud, milk of magnesia

Solution is a homogeneous mixture. Particles are smaller than 1 nm, they are not visible to naked eye. In this mixture, the solute particles dissolve and spread uniformly throughout. The Tyndall effect is not seen in this mixture.

example: sugar in water, iodine in alcohol, salt in water, alloy

Suspensions are heterogeneous mixtures. In this mixture, the solute particles are visible to the naked eye. The Tyndall effect is observed in this mixture.

For example: chalk powder and water, wheat flour and water

Question 3

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.

Answer 3

Mass of sodium chloride (solute) = 36 g (Given)

Mass of water (solvent) = 100 g (Given)

Then, mass of solution = Mass of solute + Mass of solvent

= (36 + 100) g

= 136 g

Therefore, concentration (mass by mass percentage) of the solution

= (Mass of solute / Mass of solvent ) × 100 %

= (36 / 136) × 100 %