For the equilibrium.
${SrCl}_{2} \cdot 6 H_{2} O (s) ⇌ {SrCl}_{2} \cdot 2 H_{2} O (s) + 4 H_{2} O (g)$
the equilibrium constant $K_{P} = 16 \times 10^{- 12} {atm}^{4}$ at $1^{\circ} C$ . If one litre of air saturated with water vapour at $1^{\circ} C$ is exposed to a large quantity of ${SrCl}_{2} \cdot 2 H_{2} O (s)$ , what weight of water vapour will be absorbed? Saturated vapour pressure of water at $1^{\circ} C$ =7.6 torr.

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3.25 mg

8.5 mg

6.4 mg

2.3 mg

answer is A.

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Detailed Solution

Here, the pressure on the water molecule,

$P_{H_{2} O}^{4} = 16 \times 10^{- 12}$

$P_{H_{2} O} = 2 \times 10^{- 3} atm$

Vapor pressure $= \frac{7.6}{760} = 10^{- 2} atm$

$w = \frac{(ΔP) V}{RT} \times M = \frac{[10^{- 2} - 2 \times 10^{- 3}] \times 1}{0 .0821 \times 274} \times 18 = 6 .4 mg$

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