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(A) The freezing point depression is given by $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor and $m$ is the molality.Since $0

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$4$

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(A) The freezing point depression is given by $\Delta T_f = i 	imes K_f 	imes m$,where $i$ is the van't Hoff factor and $m$ is the molality.Since $0.10 \, M \, C_{2}H_{5}OH$ is a non-electrolyte,its van't Hoff factor $i = 1$. Thus,its effective concentration is $1 	imes 0.10 = 0.10 \, M$.For the other solutions:$(i)$ $Ba_{3}(PO_{4})_{2}$ dissociates into $5$ ions $(3Ba^{2+} + 2PO_{4}^{3-})$,so $i = 5$. Effective concentration = $5 	imes 0.10 = 0.50 \, M$.$(ii)$ $Na_{2}SO_{4}$ dissociates into $3$ ions $(2Na^{+} + SO_{4}^{2-})$,so $i = 3$. Effective concentration = $3 	imes 0.10 = 0.30 \, M$.$(iii)$ $KCl$ dissociates into $2$ ions $(K^{+} + Cl^{-})$,so $i = 2$. Effective concentration = $2 	imes 0.10 = 0.20 \, M$.$(iv)$ $Li_{3}PO_{4}$ dissociates into $4$ ions $(3Li^{+} + PO_{4}^{3-})$,so $i = 4$. Effective concentration = $4 	imes 0.10 = 0.40 \, M$.Since all these solutions have an effective concentration greater than $0.10 \, M$,they all exhibit a greater depression in freezing point,meaning their freezing points are lower than that of $0.10 \, M \, C_{2}H_{5}OH$.Therefore,the total number of such solutions is $4$.

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