Question

Calculate the freezing point of a solution containing 60g of Glucose (Molar Mass =180 g Mol−1) In 250 g of water.

Answer 1

Step 1: Calculation of the number of moles of glucose
$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{glucose}=\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{glucose}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{glucose}}$
$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{glucose}=\frac{60\mathrm{g}}{180{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}$
Number of moles of glucose = 0.333 mol

Step 2: Calculation of molality of glucose
$\mathrm{Molality}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{Kg}}$
$$\begin{gathered} \mathrm{Molality}=\frac{0.333\mathrm{mol}}{250\mathrm{g}\times {\displaystyle \raisebox{1ex}{$1\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{$1000\mathrm{g}$}\right.}} \\ \mathrm{Molality}=1.333\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kg}$}\right. \end{gathered}$$

Step 3: Determination of depression in freezing point
$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$
The value of molal depression constant =$1.86\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{K}$
$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=1.86\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{K}\times 1.33\mathrm{mol}{\mathrm{kg}}^{-1} \\ ∆{\mathrm{T}}_{\mathrm{f}}=2.48\mathrm{K} \end{gathered}$$

Step 4: Calculation of freezing point of glucose solution
The freezing point of pure water = 273.2 K
The freezing point of a solution containing 60g of glucose in 250 g of water. = $273.2\mathrm{K}-2.48\mathrm{K}$
The freezing point of a solution containing 60g of glucose in 250 g of water. = $270.72\mathrm{K}$

Hence the freezing point of a solution containing 60g of Glucose In 250 g of water is 270.72 K