Question

An aqueous solution containing a non-volatile solute boils at ${101}^{\circ }C.$ What is the freezing point of the same solution?
Given
$(K_f={1.86}^{\circ }C/m$ and $K_b={0.51}^{\circ }C/m)$

• A. ${3.647}^{\circ }C$
• B. $-{3.647}^{\circ }C$
• C. $-{0.364}^{\circ }C$
• D. $+{0.364}^{\circ }C$

Answer 1

The correct option is B $-{3.647}^{\circ }C$

$\Delta T_b=K_b\times m$
Here m is molality
$m=\frac{\Delta T_b}{K_b}=\frac{101-100}{0.51}=\frac{1}{0.51}$
Also,
$\Delta T_f=K_f\times m$
$m=\frac{\Delta T_f}{K_f}$
Put the value of m
$\Rightarrow \frac{1}{0.51}=\frac{\Delta T_f}{1.86}$

$\Delta T_f={3.647}^{\circ }C{T}_f^0-T_f={3.647}^{\circ }C0-T_f={3.647}^{\circ }C{T}_f=-{3.647}^{\circ }C$