From the data given below derive which of the following reactant is most effective drying agent at `0^(@)C`. Given `P_(H_(2)O)^(@)=4.58 mm` at `0^(@)C`.
i. `SrCl_(2).6H_(2)O(s) hArr SrCl_(2).2H_(2)O(s)+4H_(1)O(g), K_(p)=6.9xx10^(-12) atm^(4)`
ii. `Na_(2)SO_(4).10H_(2)O(s) hArr Na_(2)SO_(4)(s)+10H_(2)O(g), K_(p)=4.08xx10^(-25) atm^(10)`
iii. `Na_(2)HPO_(4). 12H_(2)O(s) hArr Na_(2)HPO_(4). 7H_(2)O(s)+5H_(2)O(g), K_(p)=5.525xx10^(-13) atm^(5)`

Explanation

For (i) `K_(p)=(P_(H_(2)O))^(4)`
`:. P_(H_(2)O)=root(4)(sqrt(K_(p)))=root(4)(sqrt(6.9xx10^(-12)))=1.62xx10^(-3) "atm"`
For (ii) `K_(p)=(P_(H_(2)O))^(10)`
`P_(H_(2)O)=root(10)(sqrt(K_(P)))=root(10)(sqrt(4.08xx10^(-25)))=3.64xx10^(-3) "atm"`
For (iii) `K_(p)=(P_(H_(2)O))^(5)`
`:. P_(H_(2)O)=root(5)(sqrt(K_(P)))=root(5)(sqrt(5.25xx10^(-13)))=3.5xx10^(-3) "atm"`
All these reactant can be used as drying agent since their `P_(H_(2)O)` is lesser than `V.P.` of `H_(2)O` in atmosphere, i.e., `4.58 mm` or `6.02xx10^(-3) "atm"`. The efficiency order is
`SrCl_(2).6H_(2)O gt Na_(2)HPO_(4). 12H_(2)O gt Na_(2)SO_(4). 10 H_(2)O`