Use boiling point elevation formula

$\Delta T_b = K_b m$

where $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

Given:

• $K_b = 0.5\ \mathrm{K\,kg\,mol^{-1}}$
• mass of solvent $=15\,\text{g}=0.015\,\text{kg}$
• mass of each solute $=1\,\text{g}$

For compound $AB$

Boiling point elevation is $2.7\,\text{K}$.

$2.7 = 0.5 \times m$ $m = \frac{2.7}{0.5} = 5.4\,\text{mol kg}^{-1}$

Now, $m = \frac{(1/M_{AB})}{0.015}$

where $M_{AB}$ is molar mass of $AB$.

So, $5.4 = \frac{1}{0.015 M_{AB}}$ $M_{AB} = \frac{1}{5.4 \times 0.015} = \frac{1}{0.081} \approx 12.3457$

Hence, $M_{AB} = A + B \approx 12.3457$

For compound $AB_2$

Boiling point elevation is $1.5\,\text{K}$.

$1.5 = 0.5 \times m$ $m = \frac{1.5}{0.5} = 3.0\,\text{mol kg}^{-1}$

Also, $3.0 = \frac{(1/M_{AB_2})}{0.015}$ $M_{AB_2} = \frac{1}{3.0 \times 0.015} = \frac{1}{0.045} \approx 22.2222$

Hence, $M_{AB_2} = A + 2B \approx 22.2222$

Solve for atomic masses of $A$ and $B$

We have: $A + B = 12.3457$ $A + 2B = 22.2222$

Subtracting, $B = 22.2222 - 12.3457 = 9.8765$

Therefore, $A = 12.3457 - 9.8765 = 2.4692$

Match the asked format

Atomic mass of $A$ is written as $\underline{\hspace{2cm}} \times 10^{-1}$

Since $A \approx 2.4692 = 24.692 \times 10^{-1}$

Nearest integer $= 25$.

Final answer

$\boxed{25}$