Question

The vapour pressure of a solvent at $293$K is $100$ mm Hg. Then the vapour pressure of a solution containing $1$ mole of a strong electrolyte $\left(A{B}_2\right)$ in $99$ moles of the solvent at $293$ K is:

(Assume complete dissociation of solute)

• A. $103$ mm Hg
• B. $99$ mm Hg
• C. $97$ mm Hg
• D. $101$ mm Hg

Answer 1

Answer: (C)

$97$ mm Hg

Let us consider the problem:

$\frac{({P_A}^0-P^0)}{{P_A}^0}=i\frac{\left(n_B\right)}{(n_A+n_B)}$

Since,

$\frac{\left(100-P_A\right)}{100}=3\left[\frac{1}{\left(99+1\right)}\right]$

Since,

$100-P_A=3$

Hence the solution is $P_A=97mm$ of $Hg$.