Chemistry - Liquid Solution Question with Solution | TestHub

$:K_f$ 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, $$\begin{array}{rl}& \Delta T_f=\frac{K_f\times 1000\times w^{\prime }}{W\times m^{\prime }}(\text{ where }{w}^{\prime }=\text{ wt. of glucose, }\\ & m^{\prime }=\text{ molecular wt. of glucose })\\ & =\frac{14.7\times 1000\times 5}{100\times 180}=4.08\end{array}$$ $\therefore$ Freezing point of glucose solution $=273.15-4.08=269.07\mathrm{K}$.