1.

one mole of an anydrose salt AB dissolves in water with the evolution of 21.0mol^(-1)of heat. If the heat of hydration of AB is -29.4 J mol^(-1) , then the heat of dissociation of hydrated salt AB is

Answer»

` 50.4 Jmol ^(-1)`
`8.4 Jmol^(-1)`
`-50.4 Jmol ^(-1)`
`-8.4Jmol^(-1)`

Solution :`(i)AB(s)+(AQ)toAB(aq),DELTAH=-21 J mol^(-1)`
`(ii) AB(s)+xH_(2)O to AB.xH_(2)O(s),DeltaH=-29.4 J mol ^(-1)`
REQUIRED equation is `AB.xh_(2)O(s)+ (aq) to AB(aq), DeltaH=?`
EQ.(i) is equivalent to
` AB(s)+ xH_(2)O to AB.x H_(2)O(s), DeltaH=DeltaH_(1)`
`AB.xH_(2)O(s) + (aq)to AB(aq), DeltaH=DeltaH_(2)`
DeltaH_(1) + DeltaH_(2)=-21`
` -29.4 + DeltaH_(2)=-21 or DeltaH_(2) =8.4 mol^(-1)`


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