Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water.
(K_f of water = 1.86 K kg mol^–1)

We know,

W₁ = 250 g, w₂ = 60 g, mw₂ = 180 g/mol, K_f = 1.86 k kg mol^–1

$$\begin{gathered} ∆T_f = k_f x m \\ Or \\ ∆T_f = k_f x\frac{w_2 x 1000}{m.w_2 x w_1 \left(g\right)} \\ = 1.86 x \frac{60 x 1000}{180 x 250} \\ = \frac{1.80 x 600}{18 x 25} \\ = \frac{1116}{450} \\ ∆T_f = 2.48 K \\ ∆T_f = T_{solvent - T_{solution}} \\ = 2.48 = 273.15 - T_{solution} \\ {T}_{solution } = 270.67 K \end{gathered}$$