Question

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x°C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y × 10⁻² °C.

Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.

Use: Molal elevation constant (k_b = 0.5 K kg mol⁻¹, Boiling point of pure water as 100°C)

1. The value of ‘x’ is ______.
2. The value of ‘y’ is ______.

Answer 1

1. The value of ‘x’ is 100.1.
2. The value of ‘y’ is 2.5.

Explanation:

i. We know that

ΔT_b = i K_b m

Here (i) for AgNO₃ = 2

ΔT_b = 2 × 0.5 × 0.1

= 0.1

(T_s − T₀)_A = 0.1

(T_s)_A = 100.1°C

∴ x = 100.1

ii. Now ‘A’ of equal volume is mixed with 0.1 molal BaCl₂ solution to get solution B. Both react to give AgCl.

0.1 mole of AgNO₃ is present in 1000 g of solvent or 1017 g or 1017 ml of solution.

\[\ce{\underset{0}{\underset{0.1 V}{2AgNO3_{(aq)}}} + \underset{0.05 V}{\underset{0.1 V}{BaCl2_{(aq)}}} -> \underset{0.1}{\underset{0}{2AgCl_{(s)}}} + \underset{0.05 V}{\underset{0}{Na(NO3)2_{(aq)}}}}\]

`Delta T_b = ((0.05 V xx 3)/(2 V) + (0.05 V xx 3)/(2 V)) xx 0.5`

= 0.075

(T_s)_B = 100.075°C

(T_s)_A − (T_s)_B = 100.1 − 100.075

= 0.025°C

= 2.5 × 10⁻² °C

Therefore, x = 100.1 and y = 2.5