Question

For the equilibrium

\[\ce{SrCl2 * 6 H2O (s) <=> SrCl2 * 2 H2O (s) + 4H2O (g)}\]

the equilibrium constant K_p = 16 × 10^-12 atm⁴ at 1° C. If one litre of air saturated with water vapour at 1°C is exposed to a large quantity of \[\ce{SrCl2 · 2H2O(s)}\], what weight of water vapour will be absorbed? Saturated vapour pressure of water at 1°C = 7.6 torr.

Options

• 6 g

• 6.4 mg

• 2.3 g

• 8.5 g

Answer 1

6.4 mg

Explanation:

PH₂O₄ = 16 × 10^-12

⇒ PH₂O = 2 × 10^-3 atm

P_s at 1°C = `7.6/760 = 1 xx 10^-2` atm

Water absorbed = `((1 xx 10^-2 - 2 xx 10^-3) xx 1)/(0.082 xx 274)`

= `3.57 xx 10^-4` mol

mass = `3.5 xx 10^-4 xx 18`

= `6.4 xx 10^-3` g

= 6.4 mg