Question

Freezing point of water is $0^{\circ }C$. At what temperature will one molal solution of $NaCl$ freeze, if $NaCl$ is taken to be completely dissociated?
$(\text{Molal depression constant of water}=1.86C^{o}kgmo{l}^{-1})$

• A. $$-{3.72}^{o}C$$
• B. $$-{1.86}^{o}C$$
• C. $$0^{o}C$$
• D. $${3.72}^{o}C$$

Answer 1

The correct option is A $$-{3.72}^{o}C$$

Freezing point depression of a solution is given by,
$△T_f=i\times K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of the solution.
$i$ is van't Hoff factor

Given,
$K_f=1.86C^{o}kgmo{l}^{-1}$
$\text{Molality}=1m$
$NaCl\to N{a}^{+}+C{l}^{-}\left(100\%\text{dissociation}\right)$
$\therefore$
$i=2$
then,
$△T_f=i\times K_f\times m$
$△T_f=2\times 1.86\times 1={3.72}^{o}C$
$△T_f={3.72}^{o}C$
$\text{Freezing point of NaCl solution}=(0-3.72{)}^{o}C$
$T_f=-{3.72}^{o}C$