The enthalpy of dissolution of BaCl2(s) and BaCl2⋅2H2O(s) are -20.6 and 8.8 kl mol-1 respectively. The enthalpy of hydration for BaCl2, that is,BaCl2(s)+2H2O→BaCl2⋅2H2O(s) is

It is given thatBaCl2(s)+aq→BaCl2(aq);ΔH=−20.6kJmol−1                    . . . . (1)Also,BaCl2⋅2H2O(s)+aq→BaCl2(aq);ΔH=+8.8KJmol−1         . . . . (2)Equation ( 1) can be split in two steps:    BaCl2(s)+2H2O(l)→BaCl2⋅2H2O(s); ΔH1=?    BaCl2⋅2H2O(s)+aq→BaCl2(aq); ΔH2=−8.8KJ Or,     ΔH1+ΔH2=−20.6    ΔH1+8.8=−20.6 Or,     ΔH1=−29.4