Question

Relative decrease in V.P. of an aqueous solution of NaCl is 0.167. Number of moles of NaCl presentin 180 g of $H_2O$ is

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

Answer 1

Answer: (B)

1 mol

Answer 2

$\frac{\Delta_p}{p} = x_B$ $x_B$ = 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 $n = 0.167 \, n + 0.845$ $0.833 \, n = 0.835 \, \Rightarrow \, n= \frac{0.835}{0.833} = 1$