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$Solute + H_2O \rightleftharpoons solution; \Delta H=\pm x$ For the given equation, Solubility(S) of a solute in the solvent (say $H_2O$) is given by $S = A e^{-\frac{\Delta H}{RT}}$ For a given solution, variation of log S with temperature is shown graphically. Hence, Solute is:

(a) $CuSO_4.5H_2O$

(b) NaCl

(c) Sucrose

(d) CaO

Given, Solubility(S) of a solute in a solvent,
$S = A e^{\frac{-\Delta H}{RT}}$

Taking logarithm on both sides we get,
$\Rightarrow log S = log A + log e^{\frac{-\Delta H}{RT}}$
$\Rightarrow log S = log A - \frac{\Delta H}{RT}$
$\Rightarrow log S = (\frac{-\Delta H}{R}) \frac{1}{T} + log A$

As slope of line is +ve
$\Rightarrow \frac{-\Delta H}{R}$ is +ve $\Rightarrow \Delta H$ is -ve
i.e., the dissolution is exothermic.

As we know CaO (lime) dissolves in water with the evolution of heat. (Dissolution of other choices are endothermic).

edited Nov 23, 2013