Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?

To decide which 0.10 M aqueous solution gives the largest lowering of the freezing point, we recall the colligative‐property formula for freezing point depression.

We first STATE the formula. The freezing point depression $$\Delta T_f$$ is related to the molality $$m$$ of the solute, the cryoscopic constant $$K_f$$ of the solvent, and the van ’t Hoff factor $$i$$ (number of particles produced per formula unit) by

$$\Delta T_f = i\,K_f\,m.$$

The solvent (water) and the concentration (0.10 M ≈ 0.10 m, because the solutions are dilute) are the same for every option, so $$K_f$$ and $$m$$ are identical for all four cases. Therefore the magnitude of $$\Delta T_f$$ will depend only on the van ’t Hoff factor $$i$$. The larger the value of $$i$$, the greater the product $$i\,K_f\,m$$, and hence the greater the freezing point depression.

Now we determine $$i$$ for each solute:

• Glycine, NH2CH2COOH, is an amino acid. In water it exists largely as the neutral zwitterion but does not dissociate into separate small ions to any significant extent at 0.10 M. Therefore it behaves as a nonelectrolyte. We have

$$\text{glycine (aq)} \longrightarrow \text{one particle (itself)}$$

so $$i = 1.$$

• Hydrazine, N2H4, is also a molecular compound that does not ionize appreciably in water. Thus

$$\text{hydrazine (aq)} \longrightarrow \text{one particle}$$

giving $$i = 1.$$

• Glucose, C6H12O6, is another typical nonelectrolyte that stays intact as molecules in solution. Hence

$$\text{glucose (aq)} \longrightarrow \text{one particle}$$

and again $$i = 1.$$

• Potassium hydrogen sulfate, KHSO4, is an ionic compound. On dissolving it at least undergoes the primary dissociation

$$\text{KHSO}_4 \longrightarrow \text{K}^+ + \text{HSO}_4^-.$$

This reaction creates two ions from one formula unit, so already $$i = 2.$$ In addition, the hydrogen sulfate ion can partially dissociate further:

$$\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}.$$

Because this secondary dissociation is only partial, it does not double the number of particles, but it does raise the effective $$i$$ slightly above 2. We can summarize it symbolically as

$$i = 2 + \alpha,$$

where $$\alpha$$ (fractional degree of second dissociation) is positive. Thus $$i > 2.$$

We now compare all four $$i$$ values:

$$i_{\text{glycine}} = 1,$$

$$i_{\text{hydrazine}} = 1,$$

$$i_{\text{glucose}} = 1,$$

$$i_{\text{KHSO}_4} > 2.$$

Because $$i_{\text{KHSO}_4}$$ is the largest, the product $$i\,K_f\,m$$, and hence $$\Delta T_f$$, is greatest for the 0.10 M KHSO4 solution.

So, KHSO4 produces the largest freezing point depression.

Hence, the correct answer is Option C.