The solubility of a solute in water depends on temperature as S = A e^{-Δ H/RT}, where Δ H is the heat of the process: solute + H₂O ⇌ solution. For a given solution, the variation of log S with 1/T is shown graphically. The solute should be

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log S = log A - Δ H/(2.303 RT) The slope of log S against 1/T is -Δ H/(2.303 R)

The graph falls as 1/T increases, so the slope is negative and Δ H > 0

Dissolution is endothermic, so solubility rises with temperature

CaO, Ca(OH)₂ and Li₂SO₄ dissolve with release of heat, and their solubility falls on heating

KNO₃ dissolves endothermically, so the solute is KNO₃, option (c).

Chemistry - Chemical Equilibrium Question with Solution | TestHub

Options:(select one or more)

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