Question

${10}^{-3}$ mol of $CuS{O}_4\cdot 5H_{2}O$ is introduced in a $1.9L$ vessel maintained at a constant temperature of ${27}^{\circ }C$ containing moist air at relative humidity of $12.5$%. What is the final molar composition of solid mixture?
For $CuS{O}_4.5H_{2}O\left(s\right)\rightleftharpoons CuS{O}_4\left(s\right)+5H_{2}O\left(g\right),K_p\left(atm\right)={10}^{-10}$. Take vapor pressure of water at ${27}^{\circ }C$ as $28$ torrs.

Answer 1

${CuSO}_4.5H_{2}O\left(s\right)\rightleftharpoons {CuSo}_4\left(s\right)+5H_{2}O\left(g\right)$

$K_P={10}^{-10}$

$K_P=P_{H_{2}O}^5,\therefore P_{H_{2}O}^5={10}^{-10}\Rightarrow P_{H_{2}O}={10}^{-12}atm$

$\therefore n_{H_{2}O}=\frac{0.0368atm\times 1.9L}{0.0821lit-atm\times {mol}^{-1}{K}^{-1}\times 300}=0.003$

$\therefore n_{{CuSO}_4}=0.003/5=0.0006$ moles

$n_{{CuSO}_4.{5H}_{2}O}={10}^{-13}-0.0006=0.0004$ moles

$\therefore$ The final mixture contains $0.0004$ moles and $0.0006$ moles of ${CuSO}_4.{5H}_{2}O\left(s\right)$ and ${CuSO}_4\left(s\right)$ respectively.