An aqueous solution of volume 300 , cm^3 cont | vedclass

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(C) The osmotic pressure formula is $\pi = CRT = \frac{n}{V} RT = \frac{\omega}{M 	imes V} RT$.Rearranging for molar mass $M$: $M = \frac{\omega RT}{

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$40535$

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(C) The osmotic pressure formula is $\pi = CRT = \frac{n}{V} RT = \frac{\omega}{M 	imes V} RT$.Rearranging for molar mass $M$: $M = \frac{\omega RT}{\pi V}$.Given values: $\omega = 0.63 \, g$,$R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$,$T = 300 \, K$,$V = 300 \, cm^3 = 0.3 \, L$,and $\pi = 1.29 \, mbar = 1.29 	imes 10^{-3} \, bar$.Substituting the values: $M = \frac{0.63 	imes 0.083 	imes 300}{1.29 	imes 10^{-3} 	imes 0.3}$.$M = \frac{15.687}{0.000387} \approx 40534.88 \, g \, mol^{-1}$,which rounds to $40535 \, g \, mol^{-1}$.

An aqueous solution of volume $300 \, cm^3$ contains $0.63 \, g$ of protein. The osmotic pressure of the solution at $300 \, K$ is $1.29 \, mbar$. The molar mass of the protein is $......... \, g \, mol^{-1}$. Given : $R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$

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