Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and $PQ_2$. When 1 g of PQ is dissolved in 50 g of solvent 'A', $\Delta T_b$ was 1.176 K while when 1 g of $PQ_2$ is dissolved in 50 g of solvent 'A', $\Delta T_b$ was 0.689 K. ($K_b$ of 'A' = 5 K kg mol⁻¹). The molar masses of elements P and Q (in g mol⁻¹) respectively, are :

Question

Accepted Answer

C. 1. For a non-volatile, non-ionizable solute, **boiling point elevation** is $\Delta T_b=K_bm$, where molality $m$ depends on the solute's molar mass $M$. With 1 g solute in 50 g (0.05 kg) solvent.
$$
m = \frac{1/M}{0.05} = \frac{20}{M} \;\Rightarrow\; \Delta T_b = 5	imes\frac{20}{M} = \frac{100}{M}
$$
2. Rearranged, this gives the molar mass in terms of the observed elevation.
$$
M = \frac{100}{\Delta T_b}
$$
3. For **PQ**, substitute $\Delta T_b=1.176$ K.
$$
M(PQ) = \frac{100}{1.176} \approx 85
$$
Letting the molar masses of the elements be $x$ (for P) and $y$ (for Q):
$$
x+y = 85 \quad (1)
$$
4. For **$PQ_2$**, substitute $\Delta T_b=0.689$ K.
$$
M(PQ_2) = \frac{100}{0.689} \approx 145 \;\Rightarrow\; x+2y = 145 \quad (2)
$$
5. **Subtract** equation (1) from (2) to isolate $y$.
$$
(x+2y)-(x+y) = 145-85 \;\Rightarrow\; y = 60
$$
6. Substitute back into (1) to find $x$.
$$
x+60 = 85 \;\Rightarrow\; x = 25
$$
Hence, the molar masses of P and Q are 25 and 60 g/mol respectively, so the answer is **option C**.