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Solutions question

2022 · 26 Jun · Shift 2 · Q19
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Solutions question

2022 · 26 Jun · Shift 2 · Q19

JEE MainChemistrySolutionsNumerical+4 / −1
The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol −-− 1 in 200 mL of water at 27 ∘^\circ∘ C is ‾\underline{\hspace{2cm}}​ Pa. [integer value] (use R = 0.083 L bar mol −-− 1 K −-− 1)
Numerical answer
View written solutionFree

Correct answer: 415

  1. Use the osmotic pressure formula

For a dilute solution,

π=CRT=nVRT\pi = CRT = \frac{n}{V}RTπ=CRT=Vn​RT

where:

  • π\piπ = osmotic pressure
  • nnn = moles of solute
  • VVV = volume of solution
  • R=0.083 L bar mol−1K−1R = 0.083\, \text{L bar mol}^{-1}\text{K}^{-1}R=0.083L bar mol−1K−1
  • T=27∘C=300 KT = 27^\circ\text{C} = 300\,\text{K}T=27∘C=300K
  1. Calculate moles of protein

Given:

  • mass of protein =2.0 g= 2.0\,\text{g}=2.0g
  • molar mass =60 kg mol−1=60000 g mol−1= 60\,\text{kg mol}^{-1} = 60000\,\text{g mol}^{-1}=60kg mol−1=60000g mol−1

So,

n=2.060000=130000=3.33×10−5 moln = \frac{2.0}{60000} = \frac{1}{30000} = 3.33\times 10^{-5}\,\text{mol}n=600002.0​=300001​=3.33×10−5mol
  1. Take solution volume

Given volume =200 mL=0.200 L= 200\,\text{mL} = 0.200\,\text{L}=200mL=0.200L

Thus concentration,

C=nV=3.33×10−50.200=1.667×10−4 mol L−1C = \frac{n}{V} = \frac{3.33\times 10^{-5}}{0.200} = 1.667\times 10^{-4}\,\text{mol L}^{-1}C=Vn​=0.2003.33×10−5​=1.667×10−4mol L−1
  1. Compute osmotic pressure
π=CRT=(1.667×10−4)(0.083)(300)\pi = CRT = (1.667\times 10^{-4})(0.083)(300)π=CRT=(1.667×10−4)(0.083)(300)

First,

0.083×300=24.90.083 \times 300 = 24.90.083×300=24.9

Then,

π=1.667×10−4×24.9=4.15×10−3 bar\pi = 1.667\times 10^{-4} \times 24.9 = 4.15\times 10^{-3}\,\text{bar}π=1.667×10−4×24.9=4.15×10−3bar
  1. Convert bar to pascal

We know,

1 bar=105 Pa1\,\text{bar} = 10^5\,\text{Pa}1bar=105Pa

Therefore,

π=4.15×10−3×105=4.15×102 Pa=415 Pa\pi = 4.15\times 10^{-3} \times 10^5 = 4.15\times 10^2\,\text{Pa} = 415\,\text{Pa}π=4.15×10−3×105=4.15×102Pa=415Pa
  1. Final integer answer
415\boxed{415}415​
  1. Comparison with stored answer

Stored correct answer = 415415415.

Our derived answer matches the stored answer.

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