When 1 g each of compounds AB and 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 ______ (Nearest integer). (Given: Molal boiling point elevation con

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Solution

Q.
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 ${mol}^{- 1}$ ) [2025]

Ans.
(25)

Solution of AB:

$Δ T_{b} = m_{A B} K_{b}$

$Δ T_{b} = \frac{W_{A B}}{M_{A B}} \times \frac{1000}{W_{water}} K_{b}$

$2.7 = \frac{1 \times 1000}{M_{A B} \times 15} \times 0.5$

$M_{A B} = 12.346$

Thus, $M_{A} + M_{B} = 12.346$ ...(I)

Solution of ${AB}_{2}$ :

$Δ T_{b} = m_{A B_{2}} K_{b}$

$Δ T_{b} = \frac{W_{A B_{2}}}{M_{A B_{2}}} \times \frac{1000}{W_{water}} K_{b}$

$1.5 = \frac{1 \times 1000}{M_{A B_{2}} \times 15} \times 0.5$

$M_{A B_{2}} = 22.22 g$

Thus, $M_{A} + 2 M_{B} = 22.22$ ...(II)

(II) – (I) gives us:

$M_{B} = 22.22 - 12.346 = 9.874 g$

Putting this in (I) gives us:

$M_{A} + 9.874 = 12.346$

$M_{A} = 2.472 g = 25 \times 10^{- 1}$

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