At `300 K`, `36 g` of glucose present per litre in its solution has an osmotic pressure of `4.98 bar`. If the osmotic pressure of the solution is `1.52 bar` at the same temperature, what would be its concentration?

Explanation

`pi=CRT=(W_(B)xxRxxT)/(M_(B)xxV)`
For both the solution, R, T and V are constants
`(4.98 "bar")=((36g)xxRxxT)/((180 g mol^(-1))xxV)`
`(1.52"bar")=(W_(B)xxRxxT)/(M_(B)xxV)`
Dividing eqn. (ii) by eqn. (i),
`((1.52"bar"))/((4.98"bar"))=W_(B)/M_(B)xx(5mol)`
or `W_(B)/M_(B)(C)=(1.52)/(4.98)xx1/((5 mol))=0.0610 mol^(-1)`