Question

Calculate the freezing point of a solution containing $60g$ of glucose (Molar mass $=180gmo{l}^{-1}$) in $250g$ of water.

($Kf$ of water $=1.86Kkgmo{l}^{-1}$).

Answer 1

The number of moles of glucose $=\frac{\text{Mass of glucose}}{\text{Molecular weight of glucose}}=\frac{\text{60 g}}{\text{180 g/mol}}=\text{0.333 mol}$

The molality of glucose solution $=\frac{\text{Number of moles of glucose}}{\text{Mass of solvent (in kg)}}=\frac{\text{0.333 mol}}{\text{250 g}\times \frac{\text{1 kg}}{\text{1000 g}}}=\text{1.333 mol/kg}$

The depression in the freezing point of glucose solution :

$\Delta T_f=K_{f}m=\text{1.86K kg /mol}\times \text{1.333 mol/kg}=\text{2.48 K}$

The freezing point of pure water is $0^{o}C$

The freezing point of glucose solution $=0-2.48=-{2.48}^{o}C$