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(B) The freezing point depression is given by $\Delta T_f = i 	imes K_f 	imes m$. The freezing point is $T_f = T_f^0 - \Delta T_f$. Thus, a higher v

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$d > c > a > b$

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(B) The freezing point depression is given by $\Delta T_f = i 	imes K_f 	imes m$. The freezing point is $T_f = T_f^0 - \Delta T_f$. Thus, a higher value of the van't Hoff factor $(i)$ multiplied by molarity $(M)$ results in a lower freezing point.Calculate the product $i 	imes M$ for each solution:$(a) \ CuSO_4 ightarrow Cu^{2+} + SO_4^{2-}$, so $i = 2$. Product $= 2 	imes 0.075 = 0.150 \ M$.$(b) \ (NH_4)_2SO_4 ightarrow 2NH_4^+ + SO_4^{2-}$, so $i = 3$. Product $= 3 	imes 0.060 = 0.180 \ M$.$(c) \ 	ext{Urea is a non-electrolyte}$, so $i = 1$. Product $= 1 	imes 0.14 = 0.140 \ M$.$(d) \ MgCl_2 ightarrow Mg^{2+} + 2Cl^-$, so $i = 3$. Product $= 3 	imes 0.04 = 0.120 \ M$.Comparing the products $(i 	imes M)$: $(b) \ 0.180 > (a) \ 0.150 > (c) \ 0.140 > (d) \ 0.120$.Since freezing point decreases as the product $i 	imes M$ increases, the order of decreasing freezing points is $(d) > (c) > (a) > (b)$.

Arrange the following solutions in order of decreasing freezing points:
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ \text{urea}$ $(d) \ 0.04 \ M \ MgCl_2$

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Arrange the following solutions in order of decreasing freezing points:$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$$(c) \ 0.14 \ M \ \text{urea}$ $(d) \ 0.04 \ M \ MgCl_2$