The osmotic pressure of solutions of PVC in c | vedclass

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(B) The osmotic pressure $\pi$ is given by the equation: $\pi = CRT = \left(\frac{W}{MV}\right)RT$,where $C$ is the concentration in $g \ L^{-1}$.Rear

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$41500$

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(B) The osmotic pressure $\pi$ is given by the equation: $\pi = CRT = \left(\frac{W}{MV}ight)RT$,where $C$ is the concentration in $g \ L^{-1}$.Rearranging the equation,we get: $\frac{\pi}{C} = \frac{RT}{M} + BC$,where $B$ is the second virial coefficient.For a dilute solution,the plot of $\frac{\pi}{C}$ versus $C$ is a straight line with slope $B$ and intercept $\frac{RT}{M}$.From the provided graph (assuming the slope is $6 	imes 10^{-4} \ L \ atm \ g^{-2}$ and intercept is $\frac{RT}{M}$),we can calculate $M$.Given: $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and intercept $\frac{RT}{M} = 0.6$ (based on standard problem data for this graph).Thus,$M = \frac{RT}{	ext{intercept}} = \frac{0.083 	imes 300}{0.6} = \frac{24.9}{0.6} = 41.5 	imes 10^3 = 41500 \ g \ mol^{-1}$.

The osmotic pressure of solutions of $PVC$ in cyclohexanone at $300 \ K$ are plotted on the graph. The molar mass of $PVC$ is $........ \ g \ mol^{-1}$ (Nearest integer) (Given : $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$ )

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