Question

Relative decrease in V.P. of an aqueous solution of NaCl is 0.167. Number of moles of NaCl presentin 180 g of $H_{2}O$ is

• A. 2 mol
• B. 1 mol
• C. 3 mol
• D. 4 mol

Answer 1

Answer: (B)

$\frac{{\Delta }_p}{p}=x_B$
$x_B$ = 0.167
No. of moles of water = $\frac{180}{18}$ = 10 mol
Let no. of moles of NaCl = $n$
Van't Hoff factor i for NaCl = 2
$\therefore$ $x_B=\frac{2n}{2n+10}$
$\frac{2n}{2n+10}=0.167\Rightarrow \frac{n}{n+5}$ = 0.167
$n=0.167n+0.845$
$0.833n=0.835\Rightarrow n=\frac{0.835}{0.833}=1$