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ChemistryInteger+4 / -1Question 83 of 90

JEE Main 2022 · 28 Jun · Shift 2, Chemistry question 83

2.5 g of protein containing only glycine (C2H5NO2\mathrm{C_2H_5NO_2}) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03×10−35.03\times10^{-3} bar. The total number of glycine units present in the protein is ______.

(Given : R = 0.083 L bar K−1^{-1} mol−1^{-1})

Answer and worked solution

Official answer: 330

Moles of protein. n=πVRT=5.03×10−3×0.50.083×300≈1.01×10−4n=\dfrac{\pi V}{RT}=\dfrac{5.03\times10^{-3}\times0.5}{0.083\times300}\approx1.01\times10^{-4}, so M≈24750M\approx24750 g mol−1^{-1}.

Units. 2475075=330\frac{24750}{75}=330 (using the glycine molar mass 75 as intended).

Answer: 330

Topics:Colligative propertiesAmino acids and proteins
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