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 (integer)
330
Solution
Since,
<br/><br/>
$\pi=\mathrm{icR} \mathrm{T}$<br/><br/>
$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$
<br/><br/>
Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$
<br/><br/>
Number of glycine units in protein $=\frac{24751.5}{75}$
<br/><br/>
$=330$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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