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 $K_p=16\times {10}^{-12}at{m}^4$ at $1^{0}C$, what weight of water vapour will be absorbed ? Saturated vapour pressure of water at $1^{0}C=7.6tor$

• A. 6 mg
• B. 3.25 mg
• C. 2.3 mg
• D. 8.5 mg

Answer 1

The correct option is A 6 mg

$Sr{Cl}_2.6H_{2}O\left(s\right)\rightleftharpoons Sr{Cl}_2.2H_{2}O\left(g\right)+4H_{2}O\left(g\right)$ at $1^{0}C$

We know that in $K_P$ of reaction only gases are involved

So Here $K_P=P_{H_{2}O}^4$

$P_{H_{2}O}={\left(K_P\right)}^{1/4}={(16\times {10}^{-12})}^{1/4}$

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

$2\times {10}^{-3}\left(pressureduetoremainingwater\right)=\underset{}{\left(saturatedvapourpressure\right)}-\underset{}{pressureofwatervapourabsorbed}$

$P^0$ $x$

$x=\left(\frac{7.6}{760}\right)-(2\times {10}^{-3})=8\times {10}^{-3}atm$

we know that $PV=nRT$

$PV=\frac{W}{M}RT\Rightarrow W=\frac{P\times V\times M}{R\times T}$

$=\frac{8\times {10}^{-3}\times 1\times 18}{0.0821\times 274}=0.0069\approx 6mg$