A solution containing 4 g of polymer in 4.0 | vedclass

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(B) The osmotic pressure formula is $\pi = CRT$, where $C$ is the molarity, $R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$, and $T$

Vedclass · Accepted Answer

$82000$

Vedclass · Answer

(B) The osmotic pressure formula is $\pi = CRT$, where $C$ is the molarity, $R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$, and $T$ is the temperature in Kelvin.Given: $\pi = 3.0 	imes 10^{-4} \ atm$, $T = 27 + 273 = 300 \ K$, $V = 4.0 \ L$, and mass of solute $w = 4 \ g$.$C = \frac{\pi}{RT} = \frac{3.0 	imes 10^{-4}}{0.0821 	imes 300} = 1.218 	imes 10^{-5} \ mol \ L^{-1}$.Since $C = \frac{w}{M 	imes V}$, where $M$ is the molar mass:$1.218 	imes 10^{-5} = \frac{4}{M 	imes 4.0}$.$M = \frac{4}{1.218 	imes 10^{-5} 	imes 4.0} = \frac{1}{1.218 	imes 10^{-5}} \approx 82101 \ g \ mol^{-1}$.Rounding to the nearest given option, the molar mass is $82000 \ g \ mol^{-1}$.

$A$ solution containing $4 \ g$ of polymer in $4.0 \ L$ solution at $27^{\circ} C$ shows an osmotic pressure of $3.0 \times 10^{-4} \ atm$. The molar mass of the polymer in $g \ mol^{-1}$ is

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