1.46 , g of a biopolymer dissolved in a 100 , | vedclass

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(D) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.Given: $\pi = 2.42

Vedclass · Accepted Answer

$15$

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(D) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.Given: $\pi = 2.42 	imes 10^{-3} \, bar$,$V = 100 \, mL = 0.1 \, L$,$T = 300 \, K$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,and mass $w = 1.46 \, g$.$C = \frac{n}{V} = \frac{w}{M 	imes V} = \frac{1.46}{M 	imes 0.1} \, mol \, L^{-1}$.Substituting these values into the equation: $2.42 	imes 10^{-3} = \left(\frac{1.46}{M 	imes 0.1}ight) 	imes 0.083 	imes 300$.$M = \frac{1.46 	imes 0.083 	imes 300}{2.42 	imes 10^{-3} 	imes 0.1} = \frac{36.354}{0.000242} \approx 150223 \, g \, mol^{-1}$.$M \approx 15.02 	imes 10^{4} \, g \, mol^{-1}$.Rounding to the nearest integer,we get $15$.

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$1.46 \, g$ of a biopolymer dissolved in a $100 \, mL$ water at $300 \, K$ exerted an osmotic pressure of $2.42 \times 10^{-3} \, bar$.The molar mass of the biopolymer is $..... \times 10^{4} \, g \, mol^{-1}$. (Round off to the Nearest Integer)[Use : $R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$ ]