Question

$83g$ of ethylene glycol dissolved in $625g$ of water. The freezing point of the solution is ________ $K$. (Nearest integer)
[Use : Molal freezing point depression constant of water $=1.86Kkgmo{l}^{-1}.$
Freezing point of water $=273K,$
Atomic masses $=C:12.0u,O:16.0u,H:10u]$

• A. 269
• B. 269.00
• C. 269.0

Answer 1

Answer: (A), (B), (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
i for ethylene glycol is 1.

$\Delta T_f=(T_o-T_s)=i\times Molality\times K_f$
$T_o$ is freezing point of water
$T_s$ is freezing point of solution

$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$273-T_s=1\times \frac{83\times 1000}{62\times 625}\times 1.86$

$T_s=269K$