Question

At ${100}^{\circ }C$ the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_b=0.52$, the boiling point of this solution will be

• A. ${102}^{\circ }C$
• B. ${103}^{\circ }C$
• C. ${101}^{\circ }C$
• D. ${100}^{\circ }C$

Answer 1

The correct option is C

${101}^{\circ }C$

Given $W_B=6.5g,W_A=100g,$
$p^s=732mm,K_b=0.52,T_b^0={100}^{\circ }C,p^{\circ }=760mm$
$\frac{p^{\circ }-p^s}{p^0}=\frac{n_2}{n_1}\Rightarrow \frac{760-732}{760}=\frac{n_2}{\frac{100}{18}}$
$\Rightarrow n_2=\frac{n_2\times 1000}{760\times 18}=0.204mol$
$\Delta T_b=K_b\times m$
$T_b-{100}^{\circ }C=K_b\times \frac{n_2\times 1000}{w_A\left(g\right)}$
$T_b-{100}^{\circ }C=\frac{0.25\times 0.2046\times 1000}{100}=1.06$
$T_b=100+1.06={101.06}^{\circ }C$