A 0.200 m solution of which of the following solutes in water will have the lowest vapor pressure?

• A. CaCl_2
• B. KNO_3
• C. NaCl
• D. C_6H_{12}O_6 (glucose)

Answer: (A)

Verified step by step guidance

1. Understand that the vapor pressure lowering of a solution depends on the number of solute particles dissolved in the solvent, which is related to the colligative property called vapor pressure lowering.
2. Recall that the vapor pressure lowering is proportional to the mole fraction of the solute particles, and for ionic compounds that dissociate, the total number of particles is the product of the molality and the van't Hoff factor (i), which represents the number of ions produced per formula unit.
3. Identify the van't Hoff factor (i) for each solute: glucose (C_6H_{12}O_6) does not dissociate, so i = 1; KNO_3 dissociates into K^+ and NO_3^-, so i = 2; NaCl dissociates into Na^+ and Cl^-, so i = 2; CaCl_2 dissociates into Ca^{2+} and 2 Cl^-, so i = 3.
4. Calculate the effective concentration of particles for each solution by multiplying the molality (0.200 m) by the van't Hoff factor (i), which gives the total molal concentration of particles that affect vapor pressure lowering.
5. Compare these values: the solution with the highest total particle concentration will have the greatest vapor pressure lowering, and thus the lowest vapor pressure. Since CaCl_2 has the highest i value, it produces the most particles and therefore the lowest vapor pressure.