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?

Answer

`"10.98 g L"^(-1)`

Explanation

`pi_(1)=C_(1)RT," "pi_(2)=C_(2)RT" "therefore" "pi_(1)//pi_(2)=C_(1)//C_(2)`
`"4.98bar = 1.52 bar"=(36//"180 mol L"^(-1))//C_(2)" or "C_(2)=0.061" mol L"^(-1) = 0.061xx180gL^(-1)=10.98gL^(-1)`