Solubility (S) of a solution in a solvent (say `H_(2))` is dependent on temperature as given by `S=Ae^(-Delta//RT)` where `DeltaH` is enthalpy of solution, Solute `+H_(2)OhArr ` solution`, DeltaH=+- x`
For a given solution, variation of log S with temperature is shown graphically. Hence, solute is

Options

A. `CuSO_(4).5H_(2)O`
B. `NaCl`
C. Sucrose
D. `CaO`

Answer

D

Explanation

`S=Ae^(-DeltaH//RT)`
`log S=log A +log e^(-DeltaH//RT)`
`log S =log A -(Delta H)/(RT)`
`log S=((-Delta H)/(R))(1)/(T)+logA`
As slope of line 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 ).