Question

Elements A and B form two non-volatile compounds (AB$_2$ and AB$_4$). When 1 g of AB$_2$ is added to 20 g of C$_6$H$_6$ (molar mass =78 gmol$^{-1}$) the freezing point of C$_6$H$_6$ is lowered by 2.3 K. When 1 g of AB$_4$ is added to 20 g of C$_6$H$_6$, the freezing point of C$_6$H$_6$ was lowered by 1.3 K. The atomic masses of A and B are respectively (K$_f$(C$_6$H$_6$)=5.1 K kg mol$^{-1}$)

• A. 25.59 u, 42.64 u
• B. 42.64 u, 25.59 u
• C. 50.29 u, 31.61 u
• D. 31.61 u, 50.29 u

Hint:

Use $\Delta T_f = K_f \cdot m$ to find molality, then moles. Set up simultaneous equations for molar masses of compounds. Solve for atomic masses by subtraction. Round to match options.

Answer 1

The Correct Option is B

1. Freezing point depression: $\Delta T_f = K_f \cdot m$, where $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$.
2. For AB$_2$: 1 g in 20 g (0.02 kg) C$_6$H$_6$, $\Delta T_f = 2.3$ K, $K_f = 5.1$. Molality $m = \frac{2.3}{5.1} \approx 0.451$. Moles = $0.451 \times 0.02 = 0.00902$. Molar mass $M_1 = \frac{1}{0.00902} \approx 110.86$ g/mol.
3. For AB$_4$: 1 g, $\Delta T_f = 1.3$ K, $m = \frac{1.3}{5.1} \approx 0.255$. Moles = $0.255 \times 0.02 = 0.0051$. Molar mass $M_2 = \frac{1}{0.0051} \approx 196.08$ g/mol.
4. Let atomic masses be A and B. Then AB$_2$: $A + 2B = 110.86$; AB$_4$: $A + 4B = 196.08$.
5. Subtract: $(A + 4B) - (A + 2B) = 2B = 196.08 - 110.86 = 85.22$, so $B \approx 42.61$. Then $A = 110.86 - 2 \times 42.61 \approx 25.64$.
6. Thus, the answer is (2) 42.64 u, 25.59 u.