Of the following aqueous solutions, the one with the lowest freezing point is (a) \(0.010 \mathrm{mgSO}_{4} ;\) (b) \(0.011 \mathrm{m}\) \(\mathrm{NaCl} ;(\mathrm{c}) 0.018 \mathrm{m} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} ;(\mathrm{d}) 0.0080 \mathrm{m} \mathrm{MgCl}_{2}\).

Freezing point depression is a colligative property of solutions calculated by \( \Delta T = k_f * m * i \), where \( \Delta T \) is the change in temperature (the freezing point depression), \( k_f \) is the freezing point depression constant, \( m \) is the molality of the solution, and \( i \) is the van't Hoff factor (the number of ion particles per formula unit of solute).

We can look at each of the options given: (a) \(0.010 \, m \, MgSO_{4}\), van't Hoff factor \( i = 3 \) (because it dissociates into one \(Mg^{2+}\) ion and one \(SO_{4}^{2-}\) ion), and thus the freezing point depression \(\Delta T = i * m = 0.010 * 3 = 0.03\). (b) \(0.011 \, m \, NaCl\), van't Hoff factor \( i = 2 \) (because it dissociates into one \(Na^{+}\) ion and one \(Cl^{-}\) ion), and thus the freezing point depression \(\Delta T = i * m = 0.011 * 2 = 0.022\). (c) \(0.018 \, m \, CH_{3}CH_{2}OH\), van't Hoff factor \( i = 1 \), (because it does not dissociate into ions), and thus the freezing point depression \(\Delta T = i * m = 0.018 * 1 = 0.018\). (d) \(0.0080 \, m \, MgCl_{2}\), van't Hoff factor \( i = 3 \) (because it dissociates into one \(Mg^{2+}\) ion and two \(Cl^{-}\) ions), and thus the freezing point depression \(\Delta T = i * m = 0.0080 * 3= 0.024\).

The solution with the lowest freezing point will be the one with the highest freezing point depression according to the formula of freezing point depression derived from colligative properties. Thus, the solution with the highest total after multiplying by the van't Hoff factor and molality, will have the lowest freezing point. Comparing the values of \(\Delta T\), we can see that the highest value belongs to the \(MgSO_{4}\) solution.