Question

$K_f$ for water is ${1.86}^{\circ }Ckgmo{l}^{-1}.$ What is the molality of a solution which freezes at $-{0.192}^{\circ }C$? Assuming no change in the solute by raising the temperature, at what temperature will the solution boil?
$K_b$ for $H_{2}O={0.515}^{\circ }Ckgmo{l}^{-1})$

• A. ${102.053}^{o}C$
• B. ${100.053}^{o}C$
• C. ${100.15}^{o}C$
• D. ${101.153}^{o}C$

Answer 1

The correct option is B ${100.053}^{o}C$

Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of solution.
Freezing point of the pure water is $0^{\circ }$
Thus, the depression in freezing point of water is ${0.192}^{\circ }C$
Molality $=\frac{△T_f}{K_f}$
$=\frac{0.192}{1.86}=0.103m$
Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.
$△T_b=0.515\times 0.103={0.053}^{o}C$
Assuming the boiling point of pure water to be ${100}^{\circ }C$, the boiling point of the solution will be ${100.053}^{o}C.$