2.5 g of protein containing only glycine ( 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 \times 10 - 3 bar. The total number of glycine units present in the protein is \underline\hspace2cm . (Given : R = 0.083 L bar K - 1 mol - 1)

Solutions PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q14 with Solution

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Correct answer: 434

Use osmotic pressure to find molar mass of the protein

For a non-electrolyte,

\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RT

So,

M = \frac{wRT}{\pi V}

Given:

$w = 2.5\,\text{g}$
$R = 0.083\,\text{L bar K}^{-1}\text{mol}^{-1}$
$T = 300\,\text{K}$
$\pi = 5.03 \times 10^{-3}\,\text{bar}$
$V = 500\,\text{mL} = 0.5\,\text{L}$

Substitute:

M = \frac{2.5 \times 0.083 \times 300}{(5.03 \times 10^{-3}) \times 0.5}

First calculate numerator:

2.5 \times 0.083 \times 300 = 62.25

Now denominator:

(5.03 \times 10^{-3}) \times 0.5 = 2.515 \times 10^{-3}

Thus,

M = \frac{62.25}{2.515 \times 10^{-3}} \approx 24751.5\,\text{g mol}^{-1}

So the molar mass of the protein is approximately

M \approx 2.475 \times 10^4\,\text{g mol}^{-1}

Find molar mass of one glycine unit inside the protein

Glycine has formula $C_2H_5NO_2$ . Its molar mass is:

2(12) + 5(1) + 14 + 2(16) = 24 + 5 + 14 + 32 = 75

When glycine units join to form a protein, each peptide bond formation removes one molecule of water. For a protein made of $n$ glycine units, molar mass is

75n - 18(n-1)

Simplify:

75n - 18n + 18 = 57n + 18

Equate with protein molar mass

57n + 18 = 24751.5

57n = 24733.5

n = \frac{24733.5}{57} \approx 433.9

So,

n \approx 434

Comparison with stored answer

My derived answer is $434$ , whereas the stored correct answer is $330$ .

Using the given data and the standard peptide mass relation, $434$ is the consistent result. The stored answer $330$ does not match the calculation.

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