3 mL of solution was made by dissolving 20 | vedclass

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(C) The osmotic pressure formula is given by $\pi = \frac{W_B 	imes R 	imes T}{V 	imes M_B}$.Given values: $W_B = 20 \ mg = 0.02 \ g$, $V = 3 \ mL

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$3 	imes 10^4$

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(C) The osmotic pressure formula is given by $\pi = \frac{W_B 	imes R 	imes T}{V 	imes M_B}$.Given values: $W_B = 20 \ mg = 0.02 \ g$, $V = 3 \ mL = 0.003 \ L$, $T = 0^{\circ} C = 273 \ K$, $\pi = 3.8 \ torr = \frac{3.8}{760} \ atm = 0.005 \ atm$.Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$, we rearrange for $M_B$:$M_B = \frac{W_B 	imes R 	imes T}{V 	imes \pi} = \frac{0.02 	imes 0.0821 	imes 273}{0.003 	imes 0.005}$.$M_B = \frac{0.448266}{0.000015} \approx 29884 \ g \ mol^{-1} \approx 3 	imes 10^4 \ g \ mol^{-1}$.

$3 \ mL$ of solution was made by dissolving $20 \ mg$ of protein at $0^{\circ} C$. The osmotic pressure of the resulting solution is $3.8 \ torr$. The molecular weight of the protein is approximately $(g/mol)$

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