`CuSO_4. 5H_2O(s)hArrCuSO_4(s)+5H_2O(g)K_P=10^(-10)("atm").10^(-2)`moles of `CuSO_4. 5H_2O(s)` is taken in a 2.5 L container at `27^@C` then at equili
`CuSO_4. 5H_2O(s)hArrCuSO_4(s)+5H_2O(g)K_P=10^(-10)("atm").10^(-2)`moles of `CuSO_4. 5H_2O(s)` is taken in a 2.5 L container at `27^@C` then at equilibrium [Take: `R=1/12`litre atm `"mol"^(-1)K^(-1)`]
A. Moles of `CuSO_4. 5H_2O` left in the container is `9xx10^(-3)`
B. Moles of `CuSO_4. 5H_2O` left in the container is `9.8xx10^(-3)`
C. Moles of `CuSO_4` left in the container is `10^(-3)`
D. Moles of `CuSO_4.` left in the container is `2xx10^(-4)`
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Correct Answer - B,D
`10^(-10)"atm"^5=P_(H_2O)^5 " " implies P_(H_2O)=10^(-2) "atm" " " n=(PV)/(RT)=(10^(-2)xx2.5)/(1/12xx300)=10^(-3)`
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