The freezing point of aqueous solution contains $5 \%$ by mass urea, $1.0 \%$ by mass $\mathrm{KCl}$ and $10…

The freezing point of aqueous solution contains $5 \%$ by mass urea, $1.0 \%$ by mass $\mathrm{KCl}$ and $10 \%$ by mass of glucose is
$\left({\mathrm{K}}_{\mathrm{fH}_{2} \mathrm{O}}=1.86 \mathrm{~K}ight.\mathrm{~molality}$ $\left.^{-1}ight)$
  1. $290.2 \mathrm{~K}$
  2. $285.5 \mathrm{~K}$
  3. $269.93 \mathrm{~K}$
  4. $250 \mathrm{~K}$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=\Delta \mathrm{T}_{\mathrm{f}}$ for glucose $+\Delta \mathrm{T}_{\mathrm{f}}$ for $\mathrm{KCl}+\Delta \mathrm{T}_{\mathrm{f}}$
for urea
$=\frac{1000 \times 1.86 \times 10}{100 \times 180}+\frac{1000 \times 1.86 \times 1 \times 2}{74.5 \times 100}+$ $\frac{1000 \times 1.86 \times 5}{100 \times 60}=3.069^{\circ}$
$\therefore$ freezing point $=273-3.069=269.93 \mathrm{~K}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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