Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and $$PQ_{2}$$. When 1g 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 50g of solvent 'A'.$$\Delta T_{b}$$ was 0.689 K ($$K_{b}$$ of 'A' =5K kg $$mol^{-1}$$) The molar masses of elements P and Q (in g  $$mol^{-1}$$ )  respectively, are:

Elements P and Q form compounds PQ and PQ₂. We need to find their molar masses from boiling point elevation data.

$$ \Delta T_b = K_b \times m = K_b \times \frac{\text{mass of solute}/M}{\text{mass of solvent (in kg)}} $$

$$ 1.176 = 5 \times \frac{1/M_{PQ}}{0.050} = \frac{5}{0.050 \times M_{PQ}} = \frac{100}{M_{PQ}} $$

$$ M_{PQ} = \frac{100}{1.176} \approx 85 \text{ g/mol} $$

$$ 0.689 = 5 \times \frac{1/M_{PQ_2}}{0.050} = \frac{100}{M_{PQ_2}} $$

$$ M_{PQ_2} = \frac{100}{0.689} \approx 145 \text{ g/mol} $$

$$M_{PQ} = P + Q = 85$$ ... (i)

$$M_{PQ_2} = P + 2Q = 145$$ ... (ii)

Subtracting (i) from (ii): $$Q = 60$$ g/mol

Substituting back: $$P = 85 - 60 = 25$$ g/mol

The correct answer is Option (3): P = 25, Q = 60.