A solution of glucose (molar mass =180 g mol^-1 ) in water has a boiling point of 100.20^° C. Calculate the freezing point of the same solution.…

Tb of glucose solution = 100.20°C Δ Tb =Tb - Tob = 100.20 o – 100 oC = 0.20 oC or 0.20 K △Tb = Kb.m m = 0.20 = 0.390 mol/kg Or -0.725 oC

CBSE 2025 · Region 1 · Set 1 · Q22 · 3 marks

A solution of glucose (molar mass $\displaystyle =180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in water has a boiling point of $\displaystyle 100.20^{\circ} \mathrm{C}$. Calculate the freezing point of the same solution. Molal constants for water $\displaystyle \mathrm{K}_{\mathrm{f}}$ and $\displaystyle \mathrm{K}_{\mathrm{b}}$ are $\displaystyle 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $\displaystyle 0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ respectively.

SolutionsColligative Properties and Determination of Molar MassApplyshort_answermedium