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Consider the chemical equation.
$N a^{+} (g) + C l^{-} (g) \to N a^{+} C l^{-} (s)$
For 1 mole of NaCl(s)
Lattice enthalpy = +788 kJ mol^-1
$Δ_{h y d} H^{0} = - 784 k J m o l^{- 1}$
The enthalpy of solution is calculated as

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$- 6 k J m o l^{- 1}$

$- 8 k J m o l^{- 1}$

$- 4 k J m o l^{- 1}$

$+ 4 k J m o l^{- 1}$

answer is C.

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Detailed Solution

$Δ H_{Solution} = Δ H_{Lattice} + Δ_{Hyd} = + 788 - 784 = + 4 {kJmol}^{- 1}$

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