Enthalpies of solution of BaCl2(s) and BaCl2⋅2H2O(s) are −20.6 kJ/mol and 8.8 kJ/mol, respectively. ΔH Hydration of BaCl2(s) to BaCl2⋅2H2O(s) is

BaCl2(s)+aq→BaCl2(aq), ∆H=-20.6 kJ/mol...(1)BaCl2.2H2O(s)+aq→BaCl2(aq)+2H2O, ∆H=8.8 kJ/molReverse the above reaction and add with (1)BaCl2(aq)+2H2O→BaCl2.2H2Os ∆H=-8.8 kJ/mol BaCl2(s)+aq→BaCl2(aq), ∆H=-20.6 kJ/mol ∆H=-20.6 kJ/mol+-8.8 kJ/mol=-29.4 kJ

Enthalpies of solution of BaCl2(s) and BaCl2⋅2H2O(s) are −20.6 kJ/mol and 8.8 kJ/mol, respectively. ΔH Hydration of BaCl2(s) to BaCl2⋅2H2O(s) is

BaCl2(s)+aq→BaCl2(aq), ∆H=-20.6 kJ/mol...(1)BaCl2.2H2O(s)+aq→BaCl2(aq)+2H2O, ∆H=8.8 kJ/molReverse the above reaction and add with (1)BaCl2(aq)+2H2O→BaCl2.2H2Os ∆H=-8.8 kJ/mol BaCl2(s)+aq→BaCl2(aq), ∆H=-20.6 kJ/mol ∆H=-20.6 kJ/mol+-8.8 kJ/mol=-29.4 kJ

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Enthalpies of solution of $B a C l_{2} (s)$ and $B a C l_{2} \cdot 2 H_{2} O (s)$ are $- 20.6 kJ/mol$ and $8.8 kJ/mol,$ respectively. $Δ H$ Hydration of $B a C l_{2} (s)$ to $B a C l_{2} \cdot 2 H_{2} O (s)$ is

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$- 29.4 kJ$

$- 11.8 kJ$

$29.6 kJ$

$11.8 kJ$

answer is A.

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Detailed Solution

$B a C l_{2} (s) + a q \to B a C l_{2} (a q), ∆ H = - 20.6 k J / m o l$ ...(1)

$B a C l_{2} .2 H_{2} O (s) + a q \to B a C l_{2} (a q) + 2 H_{2} O, ∆ H = 8.8 k J / m o l$

Reverse the above reaction and add with (1)

$B a C l_{2} (a q) + 2 H_{2} O \to B a C l_{2} .2 H_{2} O (s) ∆ H = - 8.8 k J / m o l B a C l_{2} (s) + a q \to B a C l_{2} (a q), ∆ H = - 20.6 k J / m o l ∆ H = (- 20.6 k J / m o l) + (- 8.8 k J / m o l) = - 29.4 k J$