Q64 / 180 · AIIMS 2006 · Chemistry
A $5 \%$ solution (by mass) of cane sugar in water has freezing point of $271 \mathrm{~K}$ and freezing point of pure water is $273.15 \mathrm{~K}$. The freezing point of a $5 \%$ solution (by mass) of glucose in water is
- A$271 \mathrm{~K}$
- B$273.15 \mathrm{~K}$
- Correct answerC$269.07 \mathrm{~K}$
- D$277.23 \mathrm{~K}$.
Solution
$: K_f$ for water $=\frac{\Delta T_f \times W \times m}{1000 \times w}$
(where $W=$ wt. of water, $w=$ wt. of cane sugar,
$m=$ molecular wt. of cane sugar)
$$
=\frac{2.15 \times 100 \times 342}{1000 \times 5}=14.7
$$
Now, for $5 \%$ glucose,
$$
\begin{aligned}
& \Delta T_f=\frac{K_f \times 1000 \times w^{\prime}}{W \times m^{\prime}}\left(\text { where } w^{\prime}=\right.\text { wt. of glucose, } \\
&\left.m^{\prime}=\text { molecular wt. of glucose }\right) \\
&=\frac{14.7 \times 1000 \times 5}{100 \times 180}=4.08
\end{aligned}
$$
$\therefore$ Freezing point of glucose solution $=273.15-4.08=269.07 \mathrm{~K}$.
Hence, the correct answer is option: 'C'.