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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since the molality $(m)$ and the cryoscopic constan

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$0.1 \ molal \ NH_4Cl$

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(D) The depression in freezing point is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since the molality $(m)$ and the cryoscopic constant $(K_f)$ are the same for all solutions,the depression in freezing point $(\Delta T_f)$ is directly proportional to the van't Hoff factor $(i)$. The freezing point of the solution is $T_f = T_f^0 - \Delta T_f$. Therefore,the solution with the lowest value of $i$ will have the highest freezing point. Compoundvan't Hoff Factor $(i)$$Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$$5$$BaCl_2 ightarrow Ba^{2+} + 2Cl^{-}$$3$$AlCl_3 ightarrow Al^{3+} + 3Cl^{-}$$4$$NH_4Cl ightarrow NH_4^+ + Cl^{-}$$2$ Comparing the values of $i$,$NH_4Cl$ has the lowest van't Hoff factor $(i = 2)$. Thus,it will have the minimum depression in freezing point and consequently the highest freezing point.

Which of the following aqueous solutions has the highest freezing point?

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