Question

If relative decrease in V.P is 0.4 for a solution containing 1 mol $NaCl$ in 3 mol of H${}_2$O. Calculate % ionization of $NaCl$.

• A. $100$
• B. $95$
• C. $90$
• D. $85$

Answer 1

The correct option is A $100$

The relative lowering in vapour pressure is equal to the mole fraction of the solute.

$\frac{\Delta P}{P}=x=0.4$

But $x=\frac{n}{n+N}$

Substitute $x=0.4$ and $N=3$ in the above expression.

$0.4=\frac{n}{n+3}$ or $n=2$

Thus, 1 mole of $NaCl$ on ionization gives 2 moles of ions. The percentage of ionization is 100%.

Option A is correct.