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Solutions question

2023 · 10 Apr · Shift 2 · Q12
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Solutions question

2023 · 10 Apr · Shift 2 · Q12

JEE MainChemistrySolutionsNumerical+4 / −1
An aqueous solution of volume 300 cm3300 \mathrm{~cm}^{3}300 cm3 contains 0.63 g0.63 \mathrm{~g}0.63 g of protein. The osmotic pressure of the solution at 300 K300 \mathrm{~K}300 K is 1.29 mbar. The molar mass of the protein is ‾\underline{\hspace{2cm}}​g mol−1\mathrm{g} ~\mathrm{mol}^{-1}g mol−1 Given : R = 0.083 L bar K −1^{-1}−1 mol −1^{-1}−1
Numerical answer
View written solutionFree

Correct answer: 40535

  1. Use the osmotic pressure relation

For a dilute solution,

πV=nRT\pi V = nRTπV=nRT

where

n=wMn = \frac{w}{M}n=Mw​

So,

πV=wMRT\pi V = \frac{w}{M}RTπV=Mw​RT

Hence,

M=wRTπVM = \frac{wRT}{\pi V}M=πVwRT​
  1. Given data
  • Mass of protein: w=0.63 gw = 0.63\,\text{g}w=0.63g
  • Volume of solution: 300 cm3=0.300 L300\,\text{cm}^3 = 0.300\,\text{L}300cm3=0.300L
  • Temperature: T=300 KT = 300\,\text{K}T=300K
  • Osmotic pressure: 1.29 mbar=1.29×10−3 bar1.29\,\text{mbar} = 1.29 \times 10^{-3}\,\text{bar}1.29mbar=1.29×10−3bar
  • Gas constant: R=0.083 L bar K−1 mol−1R = 0.083\,\text{L bar K}^{-1}\text{ mol}^{-1}R=0.083L bar K−1 mol−1
  1. Substitute into the formula
M=(0.63)(0.083)(300)(1.29×10−3)(0.300)M = \frac{(0.63)(0.083)(300)}{(1.29\times 10^{-3})(0.300)}M=(1.29×10−3)(0.300)(0.63)(0.083)(300)​
  1. Calculate numerator
0.63×0.083×300=15.6870.63 \times 0.083 \times 300 = 15.6870.63×0.083×300=15.687
  1. Calculate denominator
1.29×10−3×0.300=3.87×10−41.29\times 10^{-3} \times 0.300 = 3.87\times 10^{-4}1.29×10−3×0.300=3.87×10−4
  1. Calculate molar mass
M=15.6873.87×10−4M = \frac{15.687}{3.87\times 10^{-4}} M=3.87×10−415.687​ M≈40534.88 g mol−1M \approx 40534.88\,\text{g mol}^{-1}M≈40534.88g mol−1

Thus,

M≈40535 g mol−1M \approx 40535\,\text{g mol}^{-1}M≈40535g mol−1
  1. Comparison with stored answer

Stored correct answer = 405354053540535

This matches the derived answer.

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