Ncert Exercises & Solutions

2.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Number of moles of liquid A,

n_{A} = \frac{100}{140} mol = 0.714 mol

Number of moles of liquid B, $n_{B} = \frac{1000}{180} mol = 5.556 mol$

Then, mole fraction of A, $x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$

$= \frac{0.714}{0.714 + 5.556}$

= 0.114

And, mole fraction of B, x_B = 1 − 0.114 = 0.886

Vapour pressure of pure liquid B, $p_{B}^{0} = 500 torr$

Therefore, vapour pressure of liquid B in the solution,

$p_{B} = p_{B}^{0} x_{B}$
$= 500 \times 0.886$
$= 443 torr$

Total vapour pressure of the solution, p_total = 475 torr

$∴$ Vapour pressure of liquid A in the solution,

p_A = p_total − p_B = 475 − 443 = 32 torr

Now,

$p_{A} = p_{A}^{0} x_{A}$
$\Rightarrow p_{A}^{0} = \frac{p_{A}}{x_{A}} = \frac{32}{0.114}$
$= 280.7 torr$

Hence, the vapour pressure of pure liquid A is 280.7 torr.

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