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Choose the Best Answer · Q22

Q.200 mL of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52×10−32.52 \times 10^{-3} bar. The molar mass of protein will be (R = 0.083 L bar mol−1^{-1} K−1^{-1})

(a) 62.22 Kg mol−1^{-1}
(b) 12444 g mol−1^{-1}
(c) 300 g mol−1^{-1}
(d) none of these
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Step 1. From πV=wMRT\pi V=\dfrac{w}{M}RT, the molar mass is M=wRTπVM=\dfrac{wRT}{\pi V}.

Step 2. Here w=1.26w=1.26 g, R=0.083R=0.083 L bar mol−1^{-1} K−1^{-1}, T=300T=300 K, π=2.52×10−3\pi=2.52\times10^{-3} bar, V=200V=200 mL =0.200=0.200 L.

Step 3. M=1.26×0.083×3002.52×10−3×0.200=31.3745.04×10−4≈62,250M=\dfrac{1.26\times0.083\times300}{2.52\times10^{-3}\times0.200}=\dfrac{31.374}{5.04\times10^{-4}}\approx62{,}250 g mol−1^{-1}. …

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