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

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The enthalpy of dissolution of ${BaCl}_{2} (s) and {BaCl}_{2} \cdot 2 H_{2} O (s)$ are -20.6 and 8.8 kl mol^-1 respectively. The enthalpy of hydration for BaCl₂, that is,

${BaCl}_{2} (s) + 2 H_{2} O \to {BaCl}_{2} \cdot 2 H_{2} O (s) is$

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29.4 kJ

- 29.4 kJ

- 11.8 kJ

38.2 kJ

answer is B.

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Detailed Solution

It is given that

${BaCl}_{2} (s) + aq \to {BaCl}_{2} (aq); Δ H = - 20.6 kJ {mol}^{- 1}$ . . . . (1)

Also,

${BaCl}_{2} \cdot 2 H_{2} O (s) + aq \to {BaCl}_{2} (aq); Δ H = + 8.8 KJ {mol}^{- 1}$ . . . . (2)

Equation ( 1) can be split in two steps:

$\begin{array}{ll} {BaCl}_{2} (s) + 2 H_{2} O (l) \to {BaCl}_{2} \cdot 2 H_{2} O (s); Δ H_{1} = ? \\ {BaCl}_{2} \cdot 2 H_{2} O (s) + aq \to {BaCl}_{2} (aq); Δ H_{2} = - 8.8 KJ \\ Or, Δ H_{1} + Δ H_{2} = - 20.6 \\ Δ H_{1} + 8.8 = - 20.6 \\ Or, Δ H_{1} = - 29.4 \end{array}$