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

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(D) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.Given:$\pi = 2.57 	imes 10^{-3} \ bar$$V = 200 \ mL = 0.2 \ L$$w = 1.26

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$61039$

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(D) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.Given:$\pi = 2.57 	imes 10^{-3} \ bar$$V = 200 \ mL = 0.2 \ L$$w = 1.26 \ g$$T = 300 \ K$$R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$Substituting the values into the equation $\pi V = \frac{w}{M} RT$:$2.57 	imes 10^{-3} 	imes 0.2 = \frac{1.26}{M} 	imes 0.083 	imes 300$$M = \frac{1.26 	imes 0.083 	imes 300}{2.57 	imes 10^{-3} 	imes 0.2}$$M = \frac{31.374}{0.000514} \approx 61039 \ g \ mol^{-1}$.

$200 \ mL$ of an aqueous solution of a protein contains its $1.26 \ g$. The osmotic pressure of this solution at $300 \ K$ is found to be $2.57 \times 10^{-3} \ bar$. The molar mass of protein will be ......... $g \ mol^{-1}$.
$(R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1})$

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$200 \ mL$ of an aqueous solution of a protein contains its $1.26 \ g$. The osmotic pressure of this solution at $300 \ K$ is found to be $2.57 \times 10^{-3} \ bar$. The molar mass of protein will be ......... $g \ mol^{-1}$.$(R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1})$