Question

Two elements A and B form compounds having molecular formulae $AB_2$ and $AB_4$, when dissolved in 20 g of $C_6H_6$. 1.0 g of $AB_2$ lowers the freezing point by 2.3 K whereas 1.0 g of $AB_4$ lowers it by 1.3 K. The molal depression constant for benzene is $5.1\, K\, kg\, mol^{-l}$. The atomic masses of A and B are, respectively

• A. 26, 42.64
• B. 31.72, 47.02
• C. 13.11, 24.25
• D. 19.17, 35.01

Answer 1

Answer: (A)

Suppose atomic masses of A and B are x and y respectively.
Molar mass of $AB_2 = x + 2y$
$(x + 2y) \,g$ of $AB_2 = 1\, mol$
$1 \, g$ of $AB_{2} = \frac{1}{x+2y} \, mol$
Molar mass of $AB_{4} = x + 4y$
$\left(x + 4y\right) \,g$ of $AB_{4} = 1 \,mol$
$1 \, g$ of $AB_{4} = \frac{1}{x+4y} \,mol$
$m_{\left(x+4y\right)} = \frac{1/\left(x+2y\right)}{20} \times 1000 = \frac{50}{x+2y}$
$ΔT_{f} = K_{ f} ×m$
$2.3 = 5.1 \times \frac{50}{x+2y}$
or $\quad x + 2y = 110.87 \quad\dots\left(i\right)$
Similarly, for the second case
$m_{\left(AB_4\right) = }= \frac{1/\left(x+4y\right)}{20} \times 1000 = \frac{50}{x+4y}$
$ΔT_{f} = K_{f} × m$
$1.3 =5.1 \times \frac{50}{x+4y}$
$x + 4y = 196.15\quad\dots\left(ii\right)$
Solving eqs $\left(i\right)$ and $\left(ii\right)$, we get
$x = 25.59; \,y = 42.64$