Chapter 2: Is Matter around Us Pure?

Answers are a) decantation, b) Sublimation, c) Filtration and evaporation, d) decantation and evaporation

Explanation
a) Decantation method is used to separate the mixture of Mercury and water. Here Mercury is heavier than water hence it forms a separate layer which can be easily separated in a separating funnel.

b) Potassium chloride and ammonium chloride are separated by sublimation method because ammonium chloride is a sublimate, sublimes leaving behind the potassium chloride.

c) Common salt, water and sand are separated by filtration and evaporation processes. Common salt, water and sand are filtrated to separate the sand from the salt solution. Then, salt solution is heated to evaporate the water leaving behind salt.

(d) Kerosene is separated from salt solution in water by decantation using a separating funnel. Kerosene being heavier, forms a separate layer. Salt from water is further removed by evaporation.

The answer is: tube (a)

Explanation

Beads in tube A provide an increased surface area for cooling of the vapours that come in contact with them. Hence, it is a good condenser.

Answer is crystallization

Explanation:

Salt can be recovered from its solution by crystallization. Crystallization is a more efficient process as it removes soluble impurities, which cannot be done by evaporation.

If we consider the seawater on the surface, it comprises water and salts hence it is a homogenous mixture. If we consider the seawater from the deep sea, it consists of salts, water, mud, decayed plants etc, which will be a heterogeneous mixture.

Fractional Distillation can be used to separate acetone from the mixture of salt and water.

Explanation

There are considerable differences in the boiling points of acetone (56°C) and water (100°C). When the solution is heated, acetone evaporates first. The water is collected in the distillation flask. The vapours of acetone are then condensed to obtain acetone.

a) Sodium potassium chloride will separate from the saturated solution when the temperature of the solution is reduced from 60°C to room temperature. The solubility of the solid state is affected by the change in temperature.
b) When aqueous sugar solution is heated, first water gets evaporated up to some extent then sugar gets charred.
c) Iron combines with sulphur and forms iron sulphide (FES).

The size of the particles in suspension is relatively larger than the size of the particles in a solution. Moreover, in suspension, molecular interaction is weaker to keep the molecules in suspended form. Hence the particles settle down. Similarly, in colloidal solution, molecular interaction is strong hence they remain in suspended form.

Both smoke and fog have gas as the dispersion medium (continuous phase). But the difference lies in the dispersed phase. The dispersed phase in fog is liquid, whereas, in smoke, it is solid (particulate matter).

Answers: Physical properties are a) and c). Chemical properties are b) and d)

Explanation

a) Composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements. It is a chemical property because no new compound is formed as steel is an alloy and alloy and is a homogeneous mixture of two or more metals or of metallic elements with non-metallic elements.

b) It is a chemical property because zinc reacts with HCL to give out zinc chloride and hydrogen gas.

c) The cutting knife will not involve any chemical reaction and did not form a new compound hence it is a physical property.

d) It is chemical property as the new compound is formed by the interaction of metal oxides with alkalis.

The answer is student C because both B and A have made the solution of 150 ml, whereas student C prepared the required quantity.

Explanation

Students A and B prepared 150 ml solution, so student c made the desired solution because he added water to make 100 ml solution and from the calculation,

%w/v = 100 × weight of sub (solute)

volume of solution

∴ %50 = 100 × weight of sub / 100 ml

weight of sub = 50 ×100 ml / 100

∴  weight of sub = 50 g

Here the 50g NaOH required for 50% w/v 100ml solution of NaOH