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Solutions question

2021 · 27 Jul · Shift 1 · Q16
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Solutions question

2021 · 27 Jul · Shift 1 · Q16

JEE MainChemistrySolutionsNumerical+4 / −1
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 ×\times× 10 −-− 3 bar. The molar mass of the biopolymer is ‾×\underline{\hspace{2cm}}\times​× 104 g mol −-− 1. (Round off to the Nearest Integer) [Use : R = 0.083 L bar mol −-− 1 K −-− 1]
Numerical answer
View written solutionFree

Correct answer: 15

  1. Use the osmotic pressure formula

For a dilute solution,

π=CRT=nVRT=wMVRT\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RTπ=CRT=Vn​RT=MVw​RT

where:

  • π=2.42×10−3 bar\pi = 2.42 \times 10^{-3}\,\text{bar}π=2.42×10−3bar
  • w=1.46 gw = 1.46\,\text{g}w=1.46g
  • V=100 mL=0.100 LV = 100\,\text{mL} = 0.100\,\text{L}V=100mL=0.100L
  • T=300 KT = 300\,\text{K}T=300K
  • 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
  • M=M =M= molar mass

So,

M=wRTπVM = \frac{wRT}{\pi V}M=πVwRT​
  1. Substitute the values
M=1.46×0.083×300(2.42×10−3)×0.100M = \frac{1.46 \times 0.083 \times 300}{(2.42 \times 10^{-3}) \times 0.100}M=(2.42×10−3)×0.1001.46×0.083×300​
  1. Calculate numerator
1.46×0.083=0.121181.46 \times 0.083 = 0.121181.46×0.083=0.12118 0.12118×300=36.3540.12118 \times 300 = 36.3540.12118×300=36.354
  1. Calculate denominator
(2.42×10−3)×0.100=2.42×10−4(2.42 \times 10^{-3}) \times 0.100 = 2.42 \times 10^{-4}(2.42×10−3)×0.100=2.42×10−4
  1. Find molar mass
M=36.3542.42×10−4=150223.14 g mol−1M = \frac{36.354}{2.42 \times 10^{-4}} = 150223.14\,\text{g mol}^{-1}M=2.42×10−436.354​=150223.14g mol−1 M≈1.502×105 g mol−1M \approx 1.502 \times 10^5\,\text{g mol}^{-1}M≈1.502×105g mol−1
  1. Express in the required form

We need

M=‾×104 g mol−1M = \underline{\hspace{1cm}} \times 10^4\,\text{g mol}^{-1}M=​×104g mol−1

So,

1.502×105=15.022×1041.502 \times 10^5 = 15.022 \times 10^41.502×105=15.022×104

Rounded to the nearest integer:

151515

Final Answer

15\boxed{15}15​
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