Question

Anhydrous calcium chloride is often used as a desiccant. In the presence of excess of $CaC{l}_2$, the amount of the water taken up is governed by $K_P=6.4\times {10}^{85}$ for the following reaction at room temperature, $CaC{l}_2\left(s\right)+6H_{2}O\left(g\right)\rightleftharpoons CaC{l}_2.6H_{2}O\left(s\right).$ What is the equilibrium vapour pressure of water in a closed vessel that contains $CaC{l}_2\left(s\right)?$

• A. $5\times {10}^{-13}atm$
• B. $5\times {10}^{-15}atm$
• C. $5\times {10}^{-14}atm$
• D. $5\times {10}^{-11}atm$

Answer 1

The correct option is B $5\times {10}^{-15}atm$

Given reaction,
$CaC{l}_2\left(s\right)+6H_{2}O\left(g\right)\rightleftharpoons CaC{l}_2.6H_{2}O\left(s\right)$
$K_p=6.4\times {10}^{85}{K}_p=\frac{1}{(p_{H_{2}O}{)}^6}$
$\Rightarrow \frac{1}{(p_{H_{2}O}{)}^6}=6.4\times {10}^{85}$
$\Rightarrow (p_{H_{2}O}{)}^6=\frac{1}{6.4\times {10}^{85}}$
$\Rightarrow (p_{H_{2}O}{)}^6=\frac{1}{6.4\times {10}^{84}\times 10}$
$\Rightarrow (p_{H_{2}O}{)}^6=\frac{1}{64\times {10}^{84}}$
$\Rightarrow (p_{H_{2}O}{)}^6=\frac{1}{(2{)}^6\times ({10}^{14}{)}^6}$
$\Rightarrow p_{H_{2}O}=\frac{1}{\left(2\right)\times \left({10}^{14}\right)}$

$p_{H_{2}O}=0.5\times {10}^{-14}atm$
$p_{H_{2}O}=5\times {10}^{-15}atm$