Question

For the equilibrium $SrC{l}_2\cdot 6H_{2}O\left(s\right)\rightleftharpoons SrC{l}_2\cdot 2H_{2}O\left(s\right)+4H_{2}O\left(g\right)$ the equilibrium constant $K_P=16\times {10}^{-12}at{m}^4$ at $1^{\circ }C$. If one litre of air saturated with water vapour at $1^{\circ }C$ is exposed to a large quantity of $SrC{l}_2\cdot 2H_{2}O\left(s\right)$, what weight of water vapour will be absorbed? Saturated vapour pressure of water at $1^{\circ }C=7.6torr$.

• A. $7.2mg$
• B. $3.25mg$
• C. $2.3mg$
• D. $8.5mg$

Answer 1

The correct option is A $7.2mg$

$\begin{array}{c}7.6torr=1atm\\ 7.6torr=\frac{1}{100}atm\Rightarrow {10}^{-2}atom\end{array}$

When $\underset{2H_{2}O}{SrC{l}_2}$ is added equilibrium is

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

$\Delta P={10}^{-2}-2\times {10}^{-3}\Rightarrow {10}^{-2}\left(1-{10}^{-1}\right)\Rightarrow 0.9\times {10}^{-2}atm$

almost 90 % of water vapour are absorbed.

$Pv=nRT$

${10}^{-2}\times 1=n\times 21\times 274$

$n=\frac{{10}^{-2}}{274\times 0.0221}\Rightarrow 0.044\times {10}^{-2}mol$

$90\%ofn=0.9\times 0.44\times {10}^{-2}molabsorbed.$

$\Rightarrow 0.9\times 0.44\times {10}^{-2}\times 18\Rightarrow 0.72\times {10}^{-2}g$

$\Rightarrow 7.2mg$

$7.2mgisabsorbed.$