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Q.

The free energy change when 1 mole of NaCl is dissolved in water at 300K is –x KJ, find out value of ‘x’ given-
A) Lattice energy of $N a C l = 778 K J m o l^{- 1}$
B) Hydration energy of $N a C l = - 775 K J m o l^{- 1}$
C) Entropy change at $300 K = 40 J m o l^{- 1}$
(Round off to the nearest integer)

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answer is 9.

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Detailed Solution

$Δ H_{d i s s o l u t i o n} = Δ H_{(i o n i s a t i o n)} + Δ H_{(h y d r a t i o n)} = 778 - 775 = 3 K J m o l^{- 1} = 3000 J m o l^{- 1} Δ S_{d i s s o l u t i o n} = 40 J m o l^{- 1}; Δ G_{d i s s o l u t i o n} = Δ H - T Δ S = 3000 - 300 \times 40 = 9000 J Δ G = - 9 k J .$

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