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

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

Step-by-step explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

`\Delta T_(f) = -K_(f)b\\b = -(\Delta T_(f))/( K_(f)) = -\frac{-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 & = & \frac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b * {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} * \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}`

3. Calculate the molecular mass

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