Question

Suppose that 98.0g of a non electrolyte is dissolved in 1.00kg of water. The freezing point of this solution is found to be -0.465. What is the molecular mass of the solute?

Answer 1

Answer:

$\large \boxed{\text{392 u}}$

Explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 \, ^{\circ} C \cdot kg \cdot mol}^{-1}} = \text{0.250 mol/kg}$

2. Calculate the moles of solute

$\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}$

3. Calculate the molecular mass

$\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is \large \boxed{\textbf{392 u}} }$