# Question #0c0c9

Dec 30, 2016

${K}_{2} S {O}_{4}$ will cause the greatest lowering of the freezing point, as this solution would contain the greatest number of particles.

#### Explanation:

The amount of freezing point depression will depend on the total concentration of all particles in a solution.

If you prepared 0.1 M solutions of each solution, this only refers to the amount of solute you mixed per litre of solution.

The difference among your four choices lies in noticing that urea and glucose are molecular compounds and do not dissociate into smaller units upon dissolving.

NaCl dissociates into two ions, but ${K}_{2} S {O}_{4}$ dissociates into three particles - two ${K}^{+}$ ions and one $S {O}_{4}^{2 -}$ ion.

So, ${K}_{2} S {O}_{4}$ will lower the freezing point by the greatest amount. (In theory, it has three times the effect of urea or glucose and 1.5 times the effect of NaCl.)