Which of the following liquids will have the lowest freezing point?

A. aqueous LiF (0.65 m) B. Pure ${H}_{2} O$ C. aqueous sucrose (0.75 m) D. aqueous $C {\mathrm{dI}}_{2}$ (0.39 m) E: aqueous glucose (0.75 m) Thanks in advance

May 31, 2018

It should be D as the true answer, but if we ignore the impossible concentration of LiF, the intended answer is probably A.

Explanation:

According to colligative properties of solutes, depression of freezing point follows the number of particles produced upon solubilization. It depends on the concentration of $C {\mathrm{dI}}_{2}$, which is 0.39 molal.

Theoretically $C {\mathrm{dI}}_{2}$ giving three ions per formula when dissolved in water, should give the second highest freezing depression, at that concentration.

On the other hand $L i F$ is actually sparingly soluble in water (about 0.071 M), and cannot be dissolved at 0.65 molal.

So the answer depends on whether we want to represent physical reality or answer a theoretical question. [Sucrose and glucose give only one particle per molecule.]