Question

A solution containing 6.35 g of a nonelectrolyte dissolved in 500 g of water freezes at $-{0.465}^{\circ }C$ Determine the molecular weight of the solute. $\left[K_{f}forwater{1.86}^{\circ }C/m\right]$

• A. $25.4$ g/mol
• B. $50.8$ g/mol
• C. $76.2$ g/mol
• D. $90.2$ g/mol

Answer 1

The correct option is B $50.8$ g/mol

The depression in the freezing point $=\Delta K_f=0-(-0.465)=0.465$

$\Delta T_f=K_f\times m$
$0.465=1.86m$
$m=0.25m$

Let $M$ be the molecular weight of the solute.

Number of moles is the ratio of mass to molecular weight.

Number of moles of solute $=\frac{6.35}{M}$

The molality of a solution is the number of moles of solute in $1000$ g of water.

Molality $=\frac{6.35\times 1000}{M\times 500}=0.25$

$⟹M=50.8g/mol$