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 $1^0$C.

If one litre of air saturated with water vapour at $1^0$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 $1^0$C=7.6 torr.)

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

Answer 1

The correct option is A $6$ mg

$\begin{array}{c}760\text{torr}=1atm\\ 7.6\text{torr}={10}^{-2}\text{atm}\\ {\text{When SrCl}}_2\cdot x2H_{2}O\text{is added, equilibrium established}\end{array}$

${SrCl}_2\cdot 6H_{2}O\text{(s)}\leftrightharpoons Sr{Cl}_2\cdot 2H_{2}O+4H_{2}O\left(g\right)$

$\begin{array}{cc}{K}_P& ={\left[P_{H_{2}O}\right]}^4\\ P_{H_{2}O}& ={(16\times {10}^{-12})}^{1/4}=2\times {10}^{-3}\end{array}$

$\begin{array}{c}\text{Pressure of water vapour absorbed}\\ =\text{Vapour pressure - partial pressure.}\\ ={10}^{-2}\text{at}m-4\times {10}^{-3}\text{atm}\\ =0.08\text{atm}\end{array}$

$\begin{array}{c}\text{A/C to ideal gas equation.}\\ PV=\frac{W}{M}RT\end{array}$

$\begin{array}{c}\text{:)}P\text{is pressure,}V\text{is volume,}M\text{is molar mass}\\ \text{W is weight (mass),}R\text{is constant, T is Temp.}\end{array}$

$\begin{array}{cc}W& =\frac{MPV}{RT}\\ & =\frac{18\times 0.008\times 1}{0.00821\times 274}\\ & =6\times {10}^{-3}g\\ & =6mg\end{array}$