2.5 , g of protein containing only glycine (C | vedclass

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(B) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration.Given: $\pi = 5.03 	imes 10^{-3} \, bar$,$T = 300 \, K$,$R =

Vedclass · Accepted Answer

$330$

Vedclass · Answer

(B) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration.Given: $\pi = 5.03 	imes 10^{-3} \, bar$,$T = 300 \, K$,$R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$,$V = 0.5 \, L$.Calculating molarity $C$:$C = \frac{\pi}{RT} = \frac{5.03 	imes 10^{-3}}{0.083 	imes 300} \approx 2.02 	imes 10^{-4} \, mol \, L^{-1}$.Moles of protein $= C 	imes V = 2.02 	imes 10^{-4} 	imes 0.5 = 1.01 	imes 10^{-4} \, mol$.Molar mass of protein $(M) = \frac{	ext{mass}}{	ext{moles}} = \frac{2.5}{1.01 	imes 10^{-4}} \approx 24752 \, g \, mol^{-1}$.Molar mass of glycine $(C_2H_5NO_2) = (2 	imes 12) + (5 	imes 1) + 14 + (2 	imes 16) = 75 \, g \, mol^{-1}$.Number of glycine units $= \frac{24752}{75} \approx 330$.

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