Question

A solution of water (Kf=1.86 ∘C/m) and glucose freezes at − 4.75 ∘C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 ∘C.

Answer 1

Answer:

$\large \boxed{\text{2.55 mol/kg}}$

Explanation:

1.Calculate the freezing point depression

ΔTf = 0.00 °C - (-4.75 °C) = = 0.00 °C + 4.75 °C = 4.75 °C

2. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = K_{f}b\\b = \dfrac{\Delta T_{f}}{ K_{f}} = \dfrac{4.75 \, ^{\circ}\text{C}}{\text{1.86 \, ^{\circ} C \cdot kg \cdot mol}^{-1}} = \textbf{2.55 mol/kg}\\\\\text{The molal concentration of sucrose is \large \boxed{\textbf{2.55 mol/kg}} }$