# Calculate The Freezing Point Of A Solution Containing 60g Of Glucose (Molar Mass =180 G Mol−1) In 250 G Of Water.

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

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

The depression in the freezing point of glucose solution :

$$\Delta T_{f} = K_{f} m = 1.86 Kg / mol * 1.333 mol / kg = 2.48 K$$

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

The freezing point of a solution containing 60g of glucose in 250 g of water. =

$$0 – 2.48 = -2.480^{\circ} C$$

Explore more such questions and answers at BYJU’S.