Question

1 mole $A_X{B}_Y$ when added to 1.08 kg of water lowers its vapour pressure by 35 torr to 700 torr. If further 1.184 kg of methanol is added, then the vapour pressure increases by 37 torr. Consider only few vapors are formed, and $A_X{B}_Y$ is a non-volatile solute. If the vapour pressure of pure methanol is $a\times {10}^2$ torr, the value of $a$ is :

• A. $8$
• B. $9$
• C. $7$
• D. $10$

Answer 1

The correct option is A $8$

The relative lowering in the vapor pressure is equal to the mole fraction of solute.
$\frac{p^0-p}{p^0}=x$......(1)

The number of moles of solute $A_x{B}_y$, water and methanol are 1 mole
$\frac{1.08\times 1000}{18}=60$ moles and $\frac{1.184\times 1000}{32}=37$ moles respectively.

When only solute $A_x{B}_y$ is added, the equation (1) becomes

$\frac{735-700}{735}=\frac{1}{1+60}=0.0164$......(2)

When methanol is also added, the equation (1) becomes

$\frac{p^0-737}{p^0}=\frac{1}{1+60+37}=0.0102$......(2).

Hence, $P^0=745$ torr.

But this corresponds to a mixture of methanol and water.

Let $p^{\prime }$ be the partial pressure of methanol in this mixture.

Then $p^{\prime }\times \frac{37}{97}+735\times \frac{60}{97}=745$

Hence, $P^{\prime }=761\approx 800$ torr $=8\times {10}^2$ torr $=a\times {10}^2$ torr.

Hence, $a=8$.