Equilibrium constant for the following equilibrium is given at `0^(@)C` . `Na_(2)HPO_(4)` . `12H_(2)O(s) hArrNa_(2)HPO_(4)` . `7H_(2)O(s) + 5H_(2)O(g)` `K_(p) = 31.25xx10^(-13)`. At equilibrium what will be partial pressure of water vapour:

Answer

`5xx10^(-3)` atm.

Explanation

`Na_(2)HPO_(4).12H_(2)O(S)hArrNa_(2)HPO_(4).7H_(2)O(S)+5H_(2)O(g) K_(P)=31.25xx10^(-13)`
`K_(P)=(P_(H_(2)O))^(5)`
`=(P_(H_(2)O))^(5)=31.25xx10^(-13)`
`=(P_(H_(2)O))=(3125)^(1//5xx(10^(-10))^(1//5)`
`(P_(H_(2)O))=5xx10^(-3)`