Question

Elements X and Y form two non-volatile compounds (XY and XY$_2$). When 10 g of XY is dissolved in 50 g of ethanol, the depression in freezing point ($\Delta T_f$) was 5.333 K. When 10 g of XY$_2$ is dissolved in 50 g of ethanol, the $\Delta T_f$ was 2.2857 K. What are the atomic weights of X and Y respectively?
($K_f = 2 \, \text{kg mol}^{-1}$)

• A. 50 u, 50 u
• B. 25 u, 25 u
  \
• C. 75 u, 100 u
• D. 25 u, 50 u

Hint:

Use the depression in freezing point formula to find the molecular mass of the solute, then use stoichiometry to determine atomic masses.

Answer 1

The Correct Option is D

For depression in freezing point: \[ \Delta T_f = \frac{K_f \times m}{M} \] Where \(m\) is the molality, and \(M\) is the molar mass of the solute. Molality \(m\) is given by: \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\text{grams of solute}}{M} \times \frac{1}{50} \] Given data for XY and XY\(_2\) allows us to set up two equations based on the depression in freezing point and solve for the atomic weights of X and Y.