Question

For the equilibrium:

$SrC{l}_2.6H_{2}O\left(s\right)\rightleftharpoons SrC{l}_2.2H_{2}O\left(s\right)+4H_{2}O\left(g\right)$

the equilibrium constant $K_p$ = 16 x 10${}^{-12}$ atm${}^4$ at $1^{o}C$. If one litre of air saturated with water vapour at $1{}^{o}C$ is exposed to a large quantity of $SrC{l}_2.2H_{2}O\left(s\right)$, what weight of water vapour will be absorbed? Saturated vapour pressure of water at $1^{o}C$ a 7.6 torr.

• A. 6.4 mg
• B. 3.25 g
• C. 2.3 g
• D. 8.5 g

Answer 1

The correct option is A 6.4 mg

$SrC{l}_2.6H_{2}O\left(s\right)\rightleftharpoons SrC{l}_2.2H_{2}O\left(s\right)+4H_{2}O\left(g\right)$

$K_p=[P_{H_{2}O}{]}^4=16\times {10}^{-12}at{m}^4$

$P_{H_{2}O}=4.\sqrt{16\times {10}^{-12}}=2\times {10}^{-3}atm$

Volume $=1$ litre , temp = $1^{\circ }C$

Pressure of water vapour = vapour pressure - partial pressure

$=(7.6\times \frac{1}{760})-2\times {10}^{-3}$

(since $760torr=1atm$)

$=0.008atm$

$PV=nRT$

$PV=\frac{w}{m}RT$

$w=\frac{MPV}{RT}=\frac{18\times 0.008\times 1}{0.0821\times 274}=6.4\times {10}^{-3}g=6.4mg$

So, the correct option is $A$.