Chemical EquilibriumHard
Question
For the equilibrium SrCl2·6H2O(s) $\rightleftharpoons$SrCl2·2H2O(s) + 4H2O(g), KP = 8.1 × 10−7 atm4 at 27oC. If 1.642 L of air saturated with water vapour at 27oC is exposed to a large quantity of SrCl2·2H2O(s), then what mass of water vapour will be absorbed? Saturated vapour pressure of water at 27oC = 30.4 torr.
Options
A.12 mg
B.6.67 mg
C.9 mg
D.48 mg
Solution
$P_{H_{2}O,eq} = \left( K_{P} \right)^{1/4} = \left( 8.1 \times 10^{- 7} \right) = 0.03$
$P_{H_{2}O,eq},\text{actual} = \frac{30.4}{760} = 0.04\text{ atm}$
∴ Mass of water vapour absorbed = $\frac{(0.09 - 0.03) \times 1.642}{0.0821 \times 300} \times 18 = 0.012\text{ gm}$
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