Question

Two liquid X and Y on mixing form ideal solution.At 30 degree C , the vapour pressure of the solution containing 3 moles of X and one moles of Y is 550 mm.But when 4 moles of X and 1 mole of Y are mixed, the vapour pressure of the solution is 560 mm. Calculate the vapour pressures of pure liquids X and Y at this temp.

Answer 1

The vapour pressure of the solution containing 3 moles of X and 1 mole of Y is 550 mm Hg.

Mole fraction of X,

• $\sf{\chi_{_X}=\dfrac{3}{3+1}=\dfrac{3}{4}}$

Mole fraction of Y,

• $\sf{\chi_{_Y}=\dfrac{1}{3+1}=\dfrac{1}{4}}$

The vapour pressure of the solution is given by,

$\longrightarrow\sf{P=P^0_X\,\chi_{_X}+P^0_Y\,\chi_{_Y}}$

where $\sf{P^0_X}$ and $\sf{P^0_Y}$ are partial pressures of pure liquids X and Y respectively.

$\longrightarrow\sf{550=\dfrac{3}{4}\,P^0_X+\dfrac{1}{4}\,P^0_Y}$

$\longrightarrow\sf{3P^0_X+P^0_Y=2200}$

$\longrightarrow\sf{P^0_Y=2200-3P^0_X\quad\quad\dots(1)}$

The vapour pressure of the solution containing 4 moles of X and 1 mole of Y is 560 mm Hg.

Mole fraction of X,

• $\sf{\chi_{_X}=\dfrac{4}{4+1}=\dfrac{4}{5}}$

Mole fraction of Y,

• $\sf{\chi_{_Y}=\dfrac{1}{4+1}=\dfrac{1}{5}}$

The vapour pressure of the solution is given by,

$\longrightarrow\sf{P=P^0_X\,\chi_{_X}+P^0_Y\,\chi_{_Y}}$

$\longrightarrow\sf{560=\dfrac{4}{5}\,P^0_X+\dfrac{1}{5}\,P^0_Y}$

$\longrightarrow\sf{4P^0_X+P^0_Y=2800}$

From (1),

$\longrightarrow\sf{4P^0_X+2200-3P^0_X=2800}$

$\longrightarrow\underline{\underline{\sf{P^0_X=600\ mm\,Hg}}}$

Again from (1),

$\longrightarrow\sf{P^0_Y=2200-3\times600}$

$\longrightarrow\underline{\underline{\sf{P^0_Y=400\ mm\,Hg}}}$

Hence vapour pressures of pure liquids X and Y are 600 mm Hg and 400 mm Hg respectively.