Question

each solute is dissolved in 1 l of solvent. which solution will exhibit the smallest increase in boiling point compared to plain water?
0.5 moles aln (dissociates into 2 particles)
2.0 moles k₂so₄ (dissociates into 3 particles)
2.0 moles c₆h₁₂o₆ (dissociates into 1 particle)

Explanation:

Step1: Recall Boiling Point Elevation

Boiling point elevation depends on the number of solute particles, given by \( \Delta T_b = iK_bm \). Here, \( i \) is the van't Hoff factor (particles per formula unit), \( K_b \) is constant for water, and \( m \) is molality (≈molarity for 1 L solvent).

Step2: Calculate Particles for Each Solute

• AlN: \( 0.5 \, \text{mol} \times 2 = 1.0 \, \text{mol} \) of particles.
• \( K_2SO_4 \): \( 2.0 \, \text{mol} \times 3 = 6.0 \, \text{mol} \) of particles.
• \( C_6H_{12}O_6 \): \( 2.0 \, \text{mol} \times 1 = 2.0 \, \text{mol} \) of particles.

Step3: Compare Particle Amounts

Smallest number of particles means smallest \( \Delta T_b \). AlN has 1.0 mol particles, less than \( C_6H_{12}O_6 \) (2.0) and \( K_2SO_4 \) (6.0).

Answer:

0.5 moles AlN (dissociates into 2 particles)