Question

An ideal solution contains two volatile liquids $A$ ($P^o$ $=$ $100$ torr) and $B$($P^o$ $=$ 200 torr) and mixture contains $1$ mole of $A$ and $4$ moles of $B$. If $1$ mole non-volatile and non-electrolyte solid solute is added in original mixture then what will be the new vapour pressure of solution?

• A. $180$ torr
• B. $150$ torr
• C. $210$ torr
• D. None

Answer 1

Answer: (B)

$150$ torr

The mole fractions of two liquids A and B are $\frac{1}{1+4}=0.2$ and $\frac{4}{1+4}=0.8$ respectively.

The vapour pressure of a pure mixture of A and B is given by the following expression.

$\text{Vapour pressure of pure mixture of A and B}=\text{mole fraction of A}\times \text{partial pressure of A}+\text{mole fraction of B}\times \text{partial pressure of B}$.

Substitute values in the above expression.

$\text{vapour pressure of pure mixture of A and B}=0.2\times 100+0.8\times 200=180torr$

Mole fraction of non volatile solute $=\frac{1}{1+5}=0.167$

Lowering of vapour pressure $\Delta p=p_1^0{x}_2=180\times 0.167=30torr$

The vapour pressure of the solution $=180-30=150torr$.

Option B is correct.