Question

What is molal depression constant?
An aqueous solution freezes at ${-0.385}^{o}C$ if
$K_f=3.85$ $K$ $kg$ ${mol}^{-1}$
and $K_b=0.712$ $K$ $kg$ ${mol}^{-1}$

Answer 1

The depression in freezing point in a solution in which $1$ $gm$ mole of solute dissolved in $1000$ $gm$ of solvent is known as molal depression constant represented by $K_f$
Numerical:
Given value, Freezing point of solution $={-0.385}^{o}C$
$K_f=3.85$ $K$ $kg$ ${mol}^{-1}$
$K_b=0.712$ $K$ $kg$ ${mol}^{-1}$
$\Delta T_b=$ ?
Then, $\Delta T_f=T_1-T_2$
$=0-(-0.385)$
$={0.385}^{o}C$
$K_f=3.85$ $K$ $kg$ ${mol}^{-1}$
$K_b=0.7120$ $K$ $kg$ ${mol}^{-1}$
Molality $m=\frac{\Delta T_f}{K_f}$
$=\frac{0.385}{3.85}=0.1$
Elevation in boiling point $=\Delta T_b$
$\Delta T_b=K_b\times m$
$=0.712\times 0.1={0.0712}^{o}C$