Ncert Exercises & Solutions

It is given that:

K_H = 1.67 × 10⁸ Pa

$p_{{CO}_{2}}$ = 2.5 atm = 2.5 × 1.01325 × 10⁵ Pa

= 2.533125 × 10⁵ Pa

According to Henry’s law:

$p_{{CO}_{2}} = K_{H} x$
$\Rightarrow x = \frac{p_{{CO}_{2}}}{K_{H}}$
$= \frac{2.533125 \times 10^{5}}{1.67 \times 10^{8}}$

= 0.00152

We can write, $n_{{CO}_{2}}$ is negligible as compared to $n_{H_{2} O}$

[Since, $x = \frac{n_{{CO}_{2}}}{n_{{CO}_{2}} + n_{H_{2} O}} \approx \frac{n_{{CO}_{2}}}{n_{H_{2} O}}$ ]

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water

$= \frac{500}{18} mol of water$

= 27.78 mol of water

Now, $\frac{n_{{CO}_{2}}}{n_{H_{2} O}} = x$

$\frac{n_{{CO}_{2}}}{27.78} = 0.00152$
$n_{{CO}_{2}} = 0.042 mol$

Hence, quantity of CO₂ in 500 mL of soda water = (0.042 × 44)g = 1.848 g

2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.