When 1 g each of compounds AB and $$AB_2$$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is _________ $$\times 10^{-1}$$.
(Nearest integer)
(Given : Molal boiling point elevation constant is 0.5 K kg mol$$^{-1}$$)

The boiling-point elevation of a dilute solution is given by
$$\Delta T_b = K_b \, m$$
where $$K_b$$ is the molal elevation constant and $$m$$ is the molality of the solution.

Case 1: 1 g of $$AB$$ in 15 g (0.015 kg) of water raises the boiling point by 2.7 K.
For water, $$K_b = 0.5 \text{ K kg mol}^{-1}$$, so

$$m_{\!AB} = \frac{\Delta T_b}{K_b} = \frac{2.7}{0.5}=5.4 \text{ mol kg}^{-1}$$

Molality is also $$m = \dfrac{w/M}{W}$$, hence
$$5.4 = \frac{1/M_{AB}}{0.015} \;\;\Rightarrow\;\; M_{AB} = 12.34 \text{ g mol}^{-1}$$

Case 2: 1 g of $$AB_2$$ in the same quantity of water raises the boiling point by 1.5 K.

$$m_{AB_2} = \frac{1.5}{0.5}=3.0 \text{ mol kg}^{-1}$$

$$3.0 = \frac{1/M_{AB_2}}{0.015} \;\;\Rightarrow\;\; M_{AB_2} = 22.22 \text{ g mol}^{-1}$$

Let the atomic masses be $$M_A$$ and $$M_B$$. Then

$$M_A + M_B = 12.34 \;\;-(1)$$
$$M_A + 2M_B = 22.22 \;\;-(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$M_B = 22.22 - 12.34 = 9.88 \text{ g mol}^{-1}$$

Substitute in $$(1)$$ to find $$M_A$$:

$$M_A = 12.34 - 9.88 = 2.46 \text{ g mol}^{-1}$$

The atomic mass of A is therefore $$2.46 \text{ amu} \approx 2.5 \text{ amu}$$.
Expressed as $$\times 10^{-1}$$, this is
$$2.5 \text{ amu} = 25 \times 10^{-1} \text{ amu}$$

Hence the required nearest integer is 25.