300 mL of an aqueous solution of a protein c | vedclass

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Question

(C) The formula for osmotic pressure is $\pi = iCRT$,where $C = \frac{n}{V}$.Given: $\pi = 5.04 	imes 10^{-3} \ bar$,$V = 300 \ mL = 0.3 \ L$,$w = 2.

Vedclass · Accepted Answer

$41.5 	imes 10^3$

Vedclass · Answer

(C) The formula for osmotic pressure is $\pi = iCRT$,where $C = \frac{n}{V}$.Given: $\pi = 5.04 	imes 10^{-3} \ bar$,$V = 300 \ mL = 0.3 \ L$,$w = 2.52 \ g$,$T = 300 \ K$,$i = 1$ (for protein).Using $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$ (since pressure is in $bar$):$\pi = \frac{w}{M 	imes V} 	imes R 	imes T$$5.04 	imes 10^{-3} = \frac{2.52}{M 	imes 0.3} 	imes 0.08314 	imes 300$$M = \frac{2.52 	imes 0.08314 	imes 300}{5.04 	imes 10^{-3} 	imes 0.3}$$M = \frac{62.88384}{0.001512} \approx 41589 \ g \ mol^{-1} \approx 41.5 	imes 10^3 \ g \ mol^{-1}$.

$300 \ mL$ of an aqueous solution of a protein contains $2.52 \ g$ of the protein. If the osmotic pressure of such a solution at $300 \ K$ is $5.04 \times 10^{-3} \ bar$,the molar mass of the protein in $g \ mol^{-1}$ is:

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