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(D) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.

Vedclass · Accepted Answer

$415$

Vedclass · Answer

(D) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.Given:Mass of protein $(w)$ = $2.0 \ g$Molar mass of protein $(M)$ = $60 \ kg \ mol^{-1} = 60000 \ g \ mol^{-1}$Volume of solution $(V)$ = $200 \ mL = 0.2 \ L$Temperature $(T)$ = $27^{\circ} C = 300 \ K$$R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$Calculating molarity $(C)$:$C = \frac{w}{M 	imes V} = \frac{2.0 \ g}{60000 \ g \ mol^{-1} 	imes 0.2 \ L} = \frac{2}{12000} \ mol \ L^{-1} = \frac{1}{6000} \ mol \ L^{-1}$Calculating osmotic pressure $(\pi)$:$\pi = C 	imes R 	imes T = \frac{1}{6000} 	imes 0.083 	imes 300 = \frac{0.083}{20} = 0.00415 \ bar$Converting to Pascals $(Pa)$:Since $1 \ bar = 10^{5} \ Pa$,$\pi = 0.00415 	imes 10^{5} \ Pa = 415 \ Pa$.

The osmotic pressure exerted by a solution prepared by dissolving $2.0 \ g$ of protein of molar mass $60 \ kg \ mol^{-1}$ in $200 \ mL$ of water at $27^{\circ} C$ is $.....Pa$. [integer value]
(use $R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$)

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The osmotic pressure exerted by a solution prepared by dissolving $2.0 \ g$ of protein of molar mass $60 \ kg \ mol^{-1}$ in $200 \ mL$ of water at $27^{\circ} C$ is $.....Pa$. [integer value](use $R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$)