Question

${10}^{-3}$ mol of $CuS{O}_{4.}5H_{2}O$ is introduced in a 1 L vessel maintained at a constant temperature of ${27}^{\circ }C$ containing moist air at relative humidity of 12.5%. The final molar composition of solid mixture is:

(For $CuS{O}_4.5H_{2}O\left(g\right),K_p\left(atm\right)={10}^{-10}$ and take vapor pressure of water at ${27}^{\circ }C$ as 28 torrs)

• A. $CuS{O}_4.5H_{2}O=9.2\times {10}^{-4}$ moles, $CuS{O}_4=8.12\times {10}^{-5}$ moles
• B. $CuS{O}_4.5H_{2}O=8\times {10}^{-4}$ moles, $CuS{O}_4=9.2\times {10}^{-5}$ moles
• C. $CuS{O}_4.5H_{2}O=7.1\times {10}^{-4}$ moles, $CuS{O}_4=9.0\times {10}^{-5}$ moles
• D. $CuS{O}_4.5H_{2}O=8.9\times {10}^{-4}$ moles, $CuS{O}_4=8.12\times {10}^{-5}$ moles

Answer 1

Answer: (A)

$CuS{O}_4.5H_{2}O=9.2\times {10}^{-4}$ moles, $CuS{O}_4=8.12\times {10}^{-5}$ moles

The equilibrium reaction is $CuS{O}_4.5H_{2}O\rightleftharpoons CuS{O}_4+5H_{2}O$.
The expression for the equilibrium constant is $K_p=P_{H_{2}O}^5={10}^{-10}atm$.
$P_{H_{2}O}={10}^{-2}atm$.
The ideal gas equation is $PV=nRT$.
${10}^{-2}\times 1=n\times 0.0821\times 300$.
$n=4.06\times {10}^{-4}$, this is equal to the number of moles of water.
The number of moles of anhydrous copper sulfate is $\frac{4.06\times {10}^{-4}}{5}=8.12\times {10}^{-5}$.
The number of moles of hydrated copper sulfate is $=9.2\times {10}^{-4}$.