JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 14)
2.5 g of protein containing only glycine (C2H5NO2) 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 ____________.
(Given : R = 0.083 L bar K$$-$$1 mol$$-$$1)
Answer
330
Explanation
Since,
$$ \pi=\mathrm{icR} \mathrm{T} $$
$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$
Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$
Number of glycine units in protein $=\frac{24751.5}{75}$
$$ =330 $$
$$ \pi=\mathrm{icR} \mathrm{T} $$
$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$
Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$
Number of glycine units in protein $=\frac{24751.5}{75}$
$$ =330 $$
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