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`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)`]A. Moles of `CuSO_(4).5H_(2)O` left in the container is `9xx10^(-3)`B. Moles of `CuSO_(4).5H_(2)O` 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)` |
Answer» Correct Answer - B::D | |