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Assuming 100% ionization, the increasing order of the freezing point of the solution will be

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$0.10 mol {kg}^{- 1} KCl < 0.10 mol {kg}^{- 1} {Na}_{2} {SO}_{4} < 0.10 mol {kg}^{- 1} {Ba}_{3} {({PO}_{4})}_{2}$

$0.10 mol {kg}^{- 1} KCl < 0.10 mol {kg}^{- 1} {Ba}_{3} {({PO}_{4})}_{2} < 0.10 mol {kg}^{- 1} {Na}_{2} {SO}_{4}$

$0.10 mol {kg}^{- 1} {Na}_{2} {SO}_{4} < 0.10 mol {kg}^{1} {Ba}_{3} {({PO}_{4})}_{2} < 0.10 mol {kg}^{1} KCl$

$0.10 mol {kg}^{- 1} {Ba}_{3} {({PO}_{4})}_{2} < 0.10 mol {kg}^{- 1} {Na}_{2} {SO}_{4} < 0.10 mol {kg}^{- 1} KCl$

answer is A.

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Detailed Solution

Assuming complete dissociation ${(- Δ T_{f})}_{{Ba}_{3} {({PO}_{4})}_{2}} > {(- Δ T_{f})}_{{Na}_{2} {SO}_{4}} > {(- Δ T_{f})}_{KCl}$

Hence ${(T_{f})}_{{Ba}_{3} {({PO}_{4})}_{2}} < {(T_{f})}_{{Na}_{2} {SO}_{4}} < {(T_{f})}_{KCl}$

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