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Solutions question

2019 · 10 Apr · Shift 2 · Q8
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Solutions question

2019 · 10 Apr · Shift 2 · Q8

JEE MainChemistrySolutionsMCQ+4 / −1
1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, ΔTb(A)ΔTb(B){{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}ΔTb​(B)ΔTb​(A)​, is :
  1. A
    5 : 1
  2. B
    1 : 0.2
  3. C
    10 : 1
  4. D
    1 : 5
View written solutionFree

Correct answer: D

  1. For elevation in boiling point, ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m where:

    • i=1i=1i=1 for a non-electrolyte,
    • KbK_bKb​ is the ebullioscopic constant,
    • mmm is the molality.

  2. The same solute mass is dissolved in the same mass of solvent in both cases:

    • solute = 1 g1\,\text{g}1g in each case,
    • solvent = 100 g100\,\text{g}100g in each case.

    Since it is the same non-electrolyte solute and same solvent mass, the molality is proportional to moles of solute per kg of solvent, which is the same for both solutions.

    Hence, mA=mBm_A = m_BmA​=mB​ and also iA=iB=1i_A=i_B=1iA​=iB​=1.

  3. Therefore, ΔTb(A)ΔTb(B)=Kb(A)Kb(B)\frac{\Delta T_b(A)}{\Delta T_b(B)} = \frac{K_b(A)}{K_b(B)}ΔTb​(B)ΔTb​(A)​=Kb​(B)Kb​(A)​

  4. Given: Kb(A):Kb(B)=1:5K_b(A):K_b(B) = 1:5Kb​(A):Kb​(B)=1:5

    So, ΔTb(A)ΔTb(B)=1:5\frac{\Delta T_b(A)}{\Delta T_b(B)} = 1:5ΔTb​(B)ΔTb​(A)​=1:5

  5. Thus the correct option is: 1:5\boxed{1:5}1:5​

  6. Checking options:

    • A: 5:15:15:1 ❌
    • B: 1:0.2=5:11:0.2 = 5:11:0.2=5:1 ❌
    • C: 10:110:110:1 ❌
    • D: 1:51:51:5 ✅
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