MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Vapour pressure lowering is a colligative property and depends only on the number of solute particles present.Explanation:Maximum vapour pressure corresponds to minimum lowering of vapour pressure.Minimum lowering occurs when the number of solute particles in solution is least.For $1 \, \mathrm{M}$ solutions, count the particles produced per formula unit:$\mathrm{NaCl} \rightarrow \mathrm{Na}^{+} + \mathrm{Cl}^{-}$, so particles $= 2$.$\mathrm{CaCl}_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{Cl}^{-}$, so particles $= 3$.$\mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{Cl}^{-}$, so particles $= 4$.Sucrose, $\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$, does not dissociate in water, so particles $= 1$.Since sucrose gives the least number of solute particles, it causes the least lowering of vapour pressure.Therefore, it shows the maximum vapour pressure at $300 \, \mathrm{K}$.Answer:Option D: $1 \, \mathrm{M} \, \mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$