Question

For an aqueous solution, freezing point is $-{0.156}^{o}C$. Elevation of the boiling point of the same solution is
Given:
$K_f={1.86}^{o}Cmo{l}^{-1}kg$
$K_b={0.52}^{o}Cmo{l}^{-1}kg$

• A. ${0.186}^{o}C$
• B. ${0.044}^{0}C$
• C. ${1.86}^{o}C$
• D. ${5.12}^{o}C$

Answer 1

The correct option is B

${0.044}^{0}C$

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.

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 the solution.
For a same solution, the relation between depression in freezing point and elevation in boiling point is given by

$\frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$
$\Delta T_b=\frac{K_b}{K_f}\times \Delta T_f$
$\Delta T_b=\frac{0.52}{1.86}\times 0.156\approx {0.044}^{o}C$