Question

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

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

Answer 1

Answer: (A)

${101}^{o}C$

Let us consider the problem.

According to the Raoult's law of partial pressure.

$\frac{p_A^0-p_s}{p_s}=\frac{n_b}{n_a}$

Implies that

$\frac{760-732}{732}=\frac{W_b\times M_a}{M_b\times W_a}$

Implies that

$\frac{28}{732}=\frac{6.5\times 18}{M_b\times 100}$

Implies that

${M_b} = 30.6\]

$\therefore \Delta T_b=0.52\times \frac{6.5\times 1000}{30.6\times 100}=1.10$

Therefore boiling point $=100+1.10$

$={101.1}^{0}C\approx {101}^{0}C$.