Lowering of vapour pressure is highest for

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(D) The relative lowering of vapour pressure is a colligative property,which depends on the van't Hoff factor $(i)$.For a solution,$\frac{P^0 - P_s}{P

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$0.1 \ M \ BaCl_2$

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(D) The relative lowering of vapour pressure is a colligative property,which depends on the van't Hoff factor $(i)$.For a solution,$\frac{P^0 - P_s}{P^0} = i 	imes 	ext{mole fraction of solute}$.Since the concentration is the same for all,the lowering of vapour pressure is proportional to the van't Hoff factor $(i)$.$(A)$ Urea: $i = 1$$(B)$ Glucose: $i = 1$$(C)$ $MgSO_4$: $i = 2$ (dissociates into $Mg^{2+}$ and $SO_4^{2-}$)$(D)$ $BaCl_2$: $i = 3$ (dissociates into $Ba^{2+}$ and $2Cl^-$)Since $BaCl_2$ has the highest van't Hoff factor $(i = 3)$,it shows the highest lowering of vapour pressure.

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