JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Solutions

2.5 g of protein containing only glycine

(C_{2} H_{5} NO_{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 1 0^{- 3}

bar. The total number of glycine units present in the protein is ______
(Given :

R = 0.083 L

bar

K^{- 1} mol^{- 1}

)

Answer: 330

Solution

Using the osmotic pressure equation

π = C R T

5.03 \times 1 0^{- 3} = C \times 0.083 \times 300

C = \frac{5.03 \times 1 0 ^{- 3}}{24.9} = 0.202 \times 1 0^{- 3}

M

Moles of protein =

0.202 \times 1 0^{- 3} \times 0.5 = 1.01 \times 1 0^{- 4}

mol

Molar mass of protein:

M = \frac{2.5}{1.01 \times 1 0 ^{- 4}} = 24752

g/mol

Molar mass of glycine = 75 g/mol

Number of glycine units =

\frac{24752}{75} = 330.03 \approx

330

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer

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