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Question: One mole of anhydrous salt \(AB\) dissolves in water and librates \(21.0\;Jmo{l^{ - 1}}\) of heat . ...

One mole of anhydrous salt ABAB dissolves in water and librates 21.0  Jmol−121.0\;Jmo{l^{ - 1}} of heat . The value of ΔH(Hydration)\Delta {H_{\left( {Hydration} \right)}} of ABAB is 29.4  Jmol−129.4\;Jmo{l^{ - 1}}. The heat of dissolution of hydrated salt AB.2H2O(s)AB.2{H_2}{O_{\left( s \right)}} is:
A. 50.4  Jmol−150.4\;Jmo{l^{ - 1}}
B. 8.4  Jmol−18.4\;Jmo{l^{ - 1}}.
C. −50.4  Jmol−1 - 50.4\;Jmo{l^{ - 1}}
D. −8.4  Jmol−1 - 8.4\;Jmo{l^{ - 1}}

Explanation

Solution

As we know that, the heat of hydration of ions corresponds to the heat that is released by hydration of one mole of ions at a constant pressure.
Energy is released into solution when the water molecules surround the ions; as the water molecules are attracted to and ions; energy is released into the solution. This energy released as a water molecule surrounds the ions is called the hydration energy. Whether the dissolving of a salt is exothermic or endothermic depends on which is greater, the lattice energy or the hydration energy.

Complete step by step solution
We can say that the heat of hydration is calculated, when heat of solution is subtracted by lattice energy of solution. We can write it using this formula.
Heat  of  hydration  =(ΔHSolution−ΔHlattice  energy)Heat\;of\;hydration\; = \left( {\Delta {H_{Solution}} - \Delta {H_{lattice\;energy}}} \right)
Where ΔHSolution\Delta {H_{Solution}} is Heat of solution, and ΔHlattice  energy\Delta {H_{lattice\;energy}} is lattice energy of the solution.
One mole of anhydrous salt ABAB dissolves in water and liberates 21  Jmol−121\;Jmo{l^{ - 1}}.
AB(s)+aq→AB(aq)                          ΔH=−21Jmol AB(s)+xH2O→AB.xH2O,                    ΔH=−29.4Jmol−1 AB(s)+xH2O→AB.x  H2O,                    ΔH=ΔH1 AB.xH2O(S)+aq→AB(aq),                        ΔH=ΔH2 A{B_{\left( s \right)}} + aq \to A{B_{\left( {aq} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\Delta H = - 21Jmol\\\ A{B_{\left( s \right)}} + x{H_2}O \to AB.x{H_2}O,\;\;\;\;\;\;\;\;\;\;\Delta H = - 29.4Jmo{l^{ - 1}}\\\ A{B_{\left( s \right)}} + x{H_2}O \to AB.x\;{H_2}O,\;\;\;\;\;\;\;\;\;\;\Delta H = \Delta {H_1}\\\ AB.x{H_2}{O_{\left( S \right)}} + aq \to A{B_{\left( {aq} \right)}},\;\;\;\;\;\;\;\;\;\;\;\;\Delta H = \Delta {H_2}
Where ΔH2\Delta {H_2} is heat of dissociation
ΔH1+ΔH2=−21 \-29.4+ΔH2=−21 ΔH2=8.4  Jmol−1 \Delta {H_1} + \Delta {H_2} = - 21\\\ \- 29.4 + \Delta {H_2} = - 21\\\ \Delta {H_2} = 8.4\;Jmo{l^{ - 1}}
∴\therefore Heat of dissociation of AB.2H2O(s)    is      8.4  Jmol−1AB.2{H_2}{O_{\left( s \right)}}\;\;is\;\;\;8.4\;Jmo{l^{ - 1}}.

**Therefore, the correct answer is B.

Note:**
Hydration based ions involve both entropy and enthalpy. The multi layered hydration shell of an ion in soluble reduces the net field of an ion, that is an energy release, but the shell is also quite dynamic, that means a higher entropy in comparison to the low entropy of water of crystallization. So hydration is mainly normally an energy release.