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)`

Question

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)`

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
A. ` 50.4 Jmol ^(-1)`
B. `8.4 Jmol^(-1)`
C. `-50.4 Jmol ^(-1)`
D. `-8.4Jmol^(-1)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - b
`(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)`