1.

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=1/12litre atm "mol"^(-1)K^(-1)]

Answer»

Moles of `CuSO_4. 5H_2O` LEFT in the container is `9xx10^(-3)`
Moles of `CuSO_4. 5H_2O` left in the container is `9.8xx10^(-3)`
Moles of `CuSO_4` left in the container is `10^(-3)`
Moles of `CuSO_4.` left in the container is `2XX10^(-4)`

Solution :`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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