Question

$1000g$ of $1m$ aqueous solution of sucrose is cooled and maintained at $-{3.534}^{\circ }C.$ What amount of ice will separate out at this temperature?
$K_f\left(H_{2}O\right)=1.86K\cdot kg\cdot mo{l}^{-1}.$

• A. ​​​​​​$451g$
• B. ​​​​​​$301g$
• C. ​​​​​​$400g$
• D. ​​​​​​$353g$

Answer 1

The correct option is D ​​​​​​$353g$

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 solution.

Let $m^{\prime }$ be the molality of the solution after the ice separates out at $-{3.534}^{o}C.$
Thus, we have
$△T_f=K_f\cdot m^{\prime }$
$\therefore m^{\prime }=\frac{△T_f}{K_f}=\frac{3.534}{1.86}=1.9$
Let us now calculate the amount of ice separated.
Initially the molality of solution is $1m$ and weight of solution is $1000g$
$1mol$ of sucrose is dissolved in $1000g$ of $H_{2}O$
or $342g$ of sucrose is dissolved in $1000g$ of $H_{2}O.$
$\therefore$ 1342 g of solution contained 342 g of sucrose.
$\therefore 1000$ g of solution contained $\frac{342}{1342}\times 1000=254.84g.$
Amount of $H_{2}O=1000-254.84=745.16g$
Now, when ice separates out, the molality is $1.9m$ and the weight of sucrose remains the same as before.
$\because (1.9\times 342)$ g of sucrose is present in 1000 g of $H_{2}O$
$\therefore 254.84$ g of sucrose should be in $\frac{1000\times 254.84}{1.9\times 342}$
$=392.18gof{H}_{2}O$
Thus, the amount of ice separated $=745.16-392.18$
$=352.98=353g$