Question

The ratio of $\frac{\text{Δ} \text{T}_{\text{b}}}{\text{K}_{\text{b}}}$ of 10 g AB₂ and 14 g A₂B per 100 g of solvent in their respective solutions (AB₂ and A₂B both are non-electrolyte) in both the cases is 1 mol/kg. Hence, the atomic weight of A and B are respectively:

• A. 100, 40
• B. 60, 20
• C. 20, 60
• D. None of these

Answer 1

Answer: (B)

$\text{Δ}\left(\text{T}\right)_{\text{b}}=\left(\text{K}\right)_{\text{b}}\left(\frac{\left(\text{W}\right)_{\text{solute}}}{\left(\text{W}\right)_{\text{solvent}}} \times \frac{1}{\left(\text{M}\right)_{\text{solute}}}\right)\times 1000$
$1 = \frac{1 0 0 0 \times 1 0}{1 0 0 \times \text{M}_{\text{AB}_{2}}} \Rightarrow \text{M}_{\text{AB}_{2}} = 1 0 0$
Similarly, $\text{M}_{\text{A}_{2} \text{B}} = 1 4 0$
$100=\text{M}_{\text{A}}+2\text{M}_{\text{B}}$
and $140=2\text{M}_{\text{A}}+\text{M}_{\text{B}};$
$\text{M}_{\text{A}}=60,\text{M}_{\text{B}}=20$
Alternative solution
$\frac{\Delta \text{T}_{\text{b}}}{\text{K}_{\text{b}}} = 1 = \frac{1 0}{\frac{\text{M}_{\text{A}} + 2 \text{M}_{\text{B}}}{\text{0.1}}}$
$\text{M}_{\text{A}} + 2 \text{M}_{\text{B}} = 1 0 0$ ........ (1)
$\frac{\Delta \text{T}_{\text{b}}}{\text{K}_{\text{b}}} = 1 = \frac{1 4}{\frac{\text{2M}_{\text{A}} + \text{M}_{\text{B}}}{\text{0.1}}}$
$2 \text{M}_{\text{A}} + \text{M}_{\text{B}} = 1 4 0$ ........ (2)
$\text{M}_{\text{A}} = \text{60, M}_{\text{B}} = 2 0$