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Solutions question

2019 · 10 Jan · Shift 2 · Q15
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Solutions question

2019 · 10 Jan · Shift 2 · Q15

JEE MainChemistrySolutionsMCQ+4 / −1
Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is
  1. A
    Kb = Kf
  2. B
    Kb = 0.5 Kf
  3. C
    Kb = 1.5 Kf
  4. D
    Kb = 2 Kf
View written solutionFree

Correct answer: D

  1. Use the colligative property formulas

For a non-electrolyte like glucose, van’t Hoff factor i=1i=1i=1.

  • Elevation in boiling point: ΔTb=Kbm\Delta T_b = K_b mΔTb​=Kb​m
  • Depression in freezing point: ΔTf=Kfm\Delta T_f = K_f mΔTf​=Kf​m

Here mmm is molality, not molarity.

  1. First statement: 1 molar solution of glucose has boiling point elevation 2 K

Given: ΔTb=2 K\Delta T_b = 2\,\text{K}ΔTb​=2K for a 1 molar solution.

For dilute aqueous-type solutions, 111 molar is taken approximately equal to 111 molal in such standard problems. So, m≈1m \approx 1m≈1

Hence, ΔTb=Kb⋅1=2\Delta T_b = K_b \cdot 1 = 2ΔTb​=Kb​⋅1=2 Kb=2K_b = 2Kb​=2

  1. Second statement: 2 molal solution has freezing point depression 2 K

Given: ΔTf=2 K\Delta T_f = 2\,\text{K}ΔTf​=2K for m=2m = 2m=2

Using ΔTf=Kfm\Delta T_f = K_f mΔTf​=Kf​m we get 2=Kf⋅22 = K_f \cdot 22=Kf​⋅2 Kf=1K_f = 1Kf​=1

  1. Compare KbK_bKb​ and KfK_fKf​

We found: Kb=2,Kf=1K_b = 2, \qquad K_f = 1Kb​=2,Kf​=1

Therefore, Kb=2KfK_b = 2K_fKb​=2Kf​

  1. Check options
  • A: Kb=KfK_b = K_fKb​=Kf​ ❌
  • B: Kb=0.5KfK_b = 0.5K_fKb​=0.5Kf​ ❌
  • C: Kb=1.5KfK_b = 1.5K_fKb​=1.5Kf​ ❌
  • D: Kb=2KfK_b = 2K_fKb​=2Kf​ ✅

So the correct option is D.

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