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Calculatethe osmoticpressureof 0.2 Mglucosesolution at 300K. (R=8.314 J mol^(-1)K^(-1))

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Solution :Given :Concentrationof the solution`=c= 0.2 M`
`=0.2 "mol" dm^(-2)`
`=0.2 xx 10^(3) "mol" m^(-3)`
Temperature `= T= 300 K`
`R= 8.314 J mol^(-1) K^(-1)`
Theosmoticpressure`PI` is givenby
`pi` = CRT
`=0.2xx 10^(3)xx 8.314 xx 300`
`=4.988 xx 10^(5)Nm^(-2)`(or pa)`


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