$1.8 \mathrm{~g}$ of glucose (molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in $0.1…
$1.8 \mathrm{~g}$ of glucose (molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in $0.1 \mathrm{~kg}$ of water. The freezing point of the solution ( $\mathrm{in}^{\circ} \mathrm{C}$ ) is
$\left(K_f\right.$ for water $\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
$+0.186$
$-0.372$
$-0.186$
$+0.372$
Solution
Depression in freezing point is given as
$\Delta T=i K_f \times m$
where, $i=$ van't Hoff factor
$\begin{gathered}K_f=\text { molal freezing point depression } \\ \text { constant }\end{gathered}$
$m=$ molality
Molar mass of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$
$=12 \times 6+12 \times 1+6 \times 16=180 \mathrm{~g} / \mathrm{mol}$
$=\frac{1.8}{180} \times \frac{1}{0.1}=0.1 \mathrm{~m}$
$\therefore \Delta T=\mathrm{I} \times 1.86 \mathrm{~K} \mathrm{~kg} / \mathrm{mol} \times 0.1 \mathrm{~m}$
$=0.186^{\circ} \mathrm{C}$
$\therefore$ Freezing point of solution $=$ Freezing point of water - Depression in freezing point
$=0^{\circ} \mathrm{C}-0.186^{\circ} \mathrm{C}=-0.186^{\circ} \mathrm{C}$
