Question

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 ______ $\times10^{-1}$ (Nearest integer)

(Given: Molal boiling point elevation constant is 0.5 K kg $mol^{-1}$)

Answer 1

25 × 10⁻¹

Answer 2

We use the boiling point elevation formula

$\qquad \Delta T = K_b \cdot m$

with $\qquad m = \frac{\text{number of moles of solute}}{\text{kg of solvent}}$.

Since the compounds do not dissociate (they are molecular compounds) the number of moles is simply $\frac{\text{mass of solute}}{\text{molar mass}}$.

Let the atomic masses be: $\qquad A = x \qquad \text{and} \qquad B = y \quad \text{(g/mol)}$.

For AB the molar mass is x + y. Given 1 g of AB in 15 g water (0.015 kg) and $\Delta T = 2.7 \text{ K}$:

$\qquad 0.5 \cdot \frac{1}{0.015 \cdot (x + y)} = 2.7$
$\qquad \implies \frac{1}{0.015 \cdot (x + y)} = 5.4$
$\qquad \implies x + y = \frac{1}{0.015 \times 5.4} = \frac{1}{0.081} = 12.3457 \qquad \dots (1)$

For $AB_2$ the molar mass is x + 2y. Given 1 g of $AB_2$ gives $\Delta T = 1.5 \text{ K}$:

$\qquad 0.5 \cdot \frac{1}{0.015 \cdot (x + 2y)} = 1.5$
$\qquad \implies \frac{1}{0.015 \cdot (x + 2y)} = 3$
$\qquad \implies x + 2y = \frac{1}{0.015 \times 3} = \frac{1}{0.045} = 22.2222 \qquad \dots (2)$

Subtract (1) from (2):

$\qquad (x + 2y) - (x + y) = y = 22.2222 - 12.3457$
$\qquad \implies y \approx 9.8765 \text{ g/mol}$

Now, from (1):

$\qquad x = (x + y) - y = 12.3457 - 9.8765 \approx 2.4692 \text{ g/mol}$

The question asks for the atomic mass of A in the form (________ × 10⁻¹). Expressing 2.47 as ×10⁻¹ gives 24.7×10⁻¹; rounded to the nearest integer we have

$\qquad 25 \times 10^{-1}$.

Thus, the answer is 25.

Explanation (minimal):

1. For AB:
   $\qquad 0.5 \cdot \frac{1}{0.015(x+y)} = 2.7 \implies x+y = 12.3457$
2. For $AB_2$:
   $\qquad 0.5 \cdot \frac{1}{0.015(x+2y)} = 1.5 \implies x+2y = 22.2222$
3. Subtract to get $y \approx 9.88 \text{ g/mol}$; then $x \approx 2.47 \text{ g/mol}$.
4. Expressed as ×10⁻¹, $x \approx 24.7 \times 10^{-1}$, nearest integer $25 \times 10^{-1}$.