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Solutions question

2021 · 25 Feb · Shift 2 · Q17
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Solutions question

2021 · 25 Feb · Shift 2 · Q17

JEE MainChemistrySolutionsNumerical+4 / −1
If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ‾\underline{\hspace{2cm}}​ molal. (Rounded off to the nearest integer) [Kb = 0.52 K kg mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 3

  1. Use elevation in boiling point formula

    ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m

    Given:

    • ΔTb=2.5 K\Delta T_b = 2.5\,\text{K}ΔTb​=2.5K
    • Kb=0.52 K kg mol−1K_b = 0.52\,\text{K kg mol}^{-1}Kb​=0.52K kg mol−1
    • Need to find molality mmm

  2. Find van't Hoff factor iii

    Compound ABABAB dissociates as: AB→A+BAB \rightarrow A + BAB→A+B

    If degree of dissociation is α=75%=0.75\alpha = 75\% = 0.75α=75%=0.75,

    For a solute giving 2 ions: i=1+αi = 1 + \alphai=1+α

    Therefore, i=1+0.75=1.75i = 1 + 0.75 = 1.75i=1+0.75=1.75

  3. Substitute into the boiling point equation

    2.5=1.75×0.52×m2.5 = 1.75 \times 0.52 \times m2.5=1.75×0.52×m

    m=2.51.75×0.52m = \frac{2.5}{1.75 \times 0.52}m=1.75×0.522.5​

  4. Calculate

    1.75×0.52=0.911.75 \times 0.52 = 0.911.75×0.52=0.91

    m=2.50.91≈2.747m = \frac{2.5}{0.91} \approx 2.747m=0.912.5​≈2.747

  5. Round to nearest integer

    m≈3 molalm \approx 3\,\text{molal}m≈3molal

Final Answer: 333

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