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Solutions question

2011 · Q67
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Solutions question

2011 · Q67

NEETChemistrySolutionsMCQ+4 / −1
200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 ×\times× 10−-−3 bar. The molar mass of protein will be (R = 0.083 L bar mol−-−1 K−-−1)
  1. A
    51022 g mol−-−1
  2. B
    122044 g mol−-−1
  3. C
    31011 g mol−-−1
  4. D
    61038 g mol−-−1
View written solutionFree

Correct answer: D

Osmotic pressure, π\pi π = CRT

⇒\Rightarrow⇒ π\pi π = nV{n \over V}Vn​RT

⇒\Rightarrow⇒ π\pi πV = wM{w \over M}Mw​RT

M = wRTπV{{wRT} \over {\pi V}}πVwRT​

= 1.26×0.083×3002.57×10−3×2001000{{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times {{200} \over {1000}}}}2.57×10−3×1000200​1.26×0.083×300​

= 1.26×0.083×3002.57×10−3×0.2{{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times 0.2}}2.57×10−3×0.21.26×0.083×300​

= 61038 g mol–1

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