CuSO_(4),5H_(2)O(s)hArrCuSO_(4)(s)+5H_(2)O(g)K_(P)=10^(-10) "moles of" CuSO_(4).5H_(2)O(s) is taken in a 2.5L container at 27^(@)C then at equilibrium [Take: R=(1)/(12) litre atm mol^(-1)K^(-1)]

CuSO_(4),5H_(2)O(s)hArrCuSO_(4)(s)+5H_(2)O(g)K_(P)=10^(-10) "moles of"

1.	CuSO_(4),5H_(2)O(s)hArrCuSO_(4)(s)+5H_(2)O(g)K_(P)=10^(-10) "moles of" CuSO_(4).5H_(2)O(s) is taken in a 2.5L container at 27^(@)C then at equilibrium [Take: R=(1)/(12) litre atm mol^(-1)K^(-1)]
Answer» Moles of `CuSO_(4).5H_(2)O` left in the CONTAINER is `9xx10^(-3)` Moles of `CuSO_(4).5H_(2)O` 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 :N//A