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Solutions question

2021 · 25 Feb · Shift 1 · Q21
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Solutions question

2021 · 25 Feb · Shift 1 · Q21

JEE MainChemistrySolutionsNumerical+4 / −1
1 molal aqueous solution of an electrolyte A2B3A_2B_3A2​B3​ is 60% ionised. The boiling point of the solution at 1 atm is ‾\underline{\hspace{2cm}}​ K. (Rounded off to the nearest integer) [Given Kb for (H2OH_2OH2​O) = 0.52 K kg mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 375

  1. Use elevation in boiling point formula

For a solution, ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m where:

  • iii = van’t Hoff factor
  • Kb=0.52 K kg mol−1K_b = 0.52\ \text{K kg mol}^{-1}Kb​=0.52 K kg mol−1
  • m=1m = 1m=1 molal
  1. Find van’t Hoff factor

Electrolyte: A2B3A_2B_3A2​B3​

On complete ionisation: A2B3→2A3++3B2−A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}A2​B3​→2A3++3B2− So total ions formed =5= 5=5.

Given degree of ionisation: α=60%=0.6\alpha = 60\% = 0.6α=60%=0.6

For an electrolyte giving ν\nuν ions, i=1+(ν−1)αi = 1 + (\nu - 1)\alphai=1+(ν−1)α

Here ν=5\nu = 5ν=5, so i=1+(5−1)(0.6)=1+2.4=3.4i = 1 + (5-1)(0.6) = 1 + 2.4 = 3.4i=1+(5−1)(0.6)=1+2.4=3.4

  1. Calculate elevation in boiling point

ΔTb=iKbm=3.4×0.52×1=1.768 K\Delta T_b = i K_b m = 3.4 \times 0.52 \times 1 = 1.768\ \text{K}ΔTb​=iKb​m=3.4×0.52×1=1.768 K

  1. Find boiling point of solution

Normal boiling point of water at 1 atm is 373 K373\ \text{K}373 K

Hence, Tb=373+1.768=374.768 KT_b = 373 + 1.768 = 374.768\ \text{K}Tb​=373+1.768=374.768 K

Rounded to nearest integer: 375 K\boxed{375\ \text{K}}375 K​

  1. Comparison with stored answer

Stored correct answer = 375375375

Our derived answer matches the stored answer.

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