1 . 46   g of a biopolymer dissolved in a 100   mL water at 300   K exerted an osmotic…

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42×10-3bar 

The molar mass of the biopolymer is -×104 g mol-1. (Round off to the Nearest Integer)
[Use: R=0.083 L bar mol-1 K-1

Solution

π=CRT; π= osmotic pressure
C= molarity
T= Temperature of solution let the molar mass be

Molar mass of polymer = M gm/mol 

2.42×10-3 bar==1.46 gM gm/mol0.1×0.083.barmol.K×300 K

 M=15.02×104 g/mol

Asked in: JEE Main 2021 (27 Jul Shift 1)

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