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 (Mark nearest one)

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

Answer 1

The correct option is B

${102}^{\circ }C$

$\because \frac{P_A^0-P_S}{P_A^0}=\frac{n_B}{n_A}(n_A>>n_B)$

$\Rightarrow \frac{760-732}{760}=\frac{W_B\times M_A}{M_B\times W_A}$

$\Rightarrow \frac{28}{760}=\frac{6.5\times 18}{M_B\times 100}$

$\therefore M_B=31.75$

We know that $\Delta T_b=K_{b}xm$

$\therefore \Delta T_b=0.52\times \frac{6.5\times 1000}{19.2\times 100}$

$=1.06$

$\therefore$ Boiling point $=100+1.06$

$={101.06}^{\circ }C$

$\approx {101}^{\circ }C$