Question

If 10 grams of solid dissolved in 200 grams of water and the resulting solution has a freezing point $-{3.72}^{0}C$ then determine the molar mass of non-ionizing solid?

• A. $25$ g$/$mol
• B. $50$ g$/$mol
• C. $100$ g$/$mol
• D. $150$ g$/$mol
• E. $1,000$ g$/$mol

Answer 1

The correct option is A $25$ g$/$mol

Let M g/mol be the molar mass of the non-ionizing solid.

The number of moles of solid $=\frac{10\text{g}}{\text{M g/mol}}=\frac{10}{\text{M}}$ M
Mass of water $=200$ g $=0.200$ kg
The molality of solid will be the number of moles of solute present in 1 kg of solvent.

The molality $m=\frac{10}{\text{M}\times 0.200}=\frac{50}{\text{M}}$ m
The freezing point of pure water is 0 deg C. The freezing point of the solution is -3.72 deg C.

The depression in the freezing point, $\Delta T_f={3.72}^{o}C$
The molal depression in the freezing point constant $K_f=1.86$ K/m
$\Delta T_f=K_{f}m$

$3.72=1.86\times \frac{50}{\text{M}}$

$\text{M}=\frac{1.86\times 50}{3.72}=25$ g/mol