each solute is dissolved in 1 l of solvent. which solution will exhibit the smallest increase in boiling…

each solute is dissolved in 1 l of solvent. which solution will exhibit the smallest increase in boiling point compared to plain water?\n0.5 moles aln (dissociates into 2 particles)\n2.0 moles k₂so₄ (dissociates into 3 particles)\n2.0 moles c₆h₁₂o₆ (dissociates into 1 particle)

each solute is dissolved in 1 l of solvent. which solution will exhibit the smallest increase in boiling point compared to plain water?\n0.5 moles aln (dissociates into 2 particles)\n2.0 moles k₂so₄ (dissociates into 3 particles)\n2.0 moles c₆h₁₂o₆ (dissociates into 1 particle)

Answer

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)