Question

One mole of anhydrous salt AB dissolves in water and librates $21.0J{mol}^{-1}$ of heat. The value of ${\Delta }_{Hydration}$ AB is $-29.4J{mol}^{-1}$. The heat of dissolution of hydrated salt $AB.2H_2{O}_{\left(s\right)}$ is :

• A. $50.4J{mol}^{-1}$
• B. $8.4J{mol}^{-1}$
• C. $-50.4J{mol}^{-1}$
• D. $-8.4J{mol}^{-1}$

Answer 1

The correct option is B $8.4J{mol}^{-1}$

One mole of anhydrous salt $AB$ dissolves in water and liberated $21$ J/mol

$A{B}_{\left(s\right)}+aq\to {AB}_{\left(aq\right)}△H=-21J㏖\to \left(1\right)$

$A{B}_{\left(s\right)}+x{H}_{2}O\to AB.x{H}_{2}O,△H=-29.4$ J/mol

${AB}_{\left(s\right)}+x{H}_{2}O\to AB.x{H}_2{O}_{\left(s\right)};\Delta H={\Delta H}_1$

$AB.{xH}_2{O}_{\left(s\right)}+aq\to {AB}_{\left(aq\right)};\Delta H={\Delta H}_2$

where ${\Delta H}_2$ is heat of dissociation

${\Delta H}_1+{\Delta H}_2=-21$

$-29.4+{\Delta H}_2=-21$

${\Delta H}_2=8.4$ J/mol

$\therefore$ heat of dissociation of $AB.2H_2{O}_{\left(s\right)}$ is $8.4$ $J{mol}^{-1}$.