Use the freezing point depression formula

The depression in freezing point is given by $\Delta T_f = i K_f m$ where:

• $i$ = van’t Hoff factor
• $K_f$ = cryoscopic constant of water
• $m$ = molality

Since all the given solutions are 0.10 M aqueous solutions, for dilute solutions we compare mainly through the van’t Hoff factor $i$. The larger the number of solute particles produced in water, the larger the freezing point depression.

Examine each solute

(A) Hydrazine, $N_2H_4$

Hydrazine is a nonelectrolyte / very weak base in water and does not dissociate significantly into ions. Thus, $i \approx 1$

(B) Glucose

Glucose is a nonelectrolyte. It dissolves as molecules only. $i = 1$

(C) Glycine

Glycine exists predominantly as a zwitterion in aqueous solution: $^+H_3NCH_2COO^-$ But this is still one solute entity in solution, not dissociation into multiple independent particles like a strong electrolyte. Hence effectively, $i \approx 1$

(D) $KHSO_4$

Potassium hydrogen sulfate dissociates in water as $KHSO_4 \rightarrow K^+ + HSO_4^-$ So at least it gives $i \approx 2$

Also, $HSO_4^-$ can further partially dissociate: $HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$ This makes the effective number of particles even greater than 2 in water.

Compare van’t Hoff factors

• Hydrazine: $i \approx 1$
• Glucose: $i = 1$
• Glycine: $i \approx 1$
• $KHSO_4$: $i > 1$, approximately $2$ or slightly more effective due to further ionization

Therefore, $KHSO_4$ produces the largest number of dissolved particles and hence the largest freezing point depression.

Final answer

$\boxed{\text{D: } KHSO_4}$

Comparison with stored correct answer

Stored correct answer = D

My derived answer also = D, so they agree.