QUESTION 2.22

At 300 K, 36 g of glucose present in a litre of 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?

$\pi =CRT=\frac{W_B\times R\times T}{M_B\times V}$

For both solutions, R, T and V are constants.

For I solution

$\left(4.98bar\right)=\frac{\left(36g\right)\times R\times T}{\left(180gmo{l}^{-1}\right)\times V}$

For II solution

$\left(1.52bar\right)=\frac{W_B\times R\times T}{M_B\times V}$

On dividing Eq. (ii) by Eq. (i), we get

$\frac{\left(1.52bar\right)}{\left(4.98bar\right)}=\frac{W_B\times R\times T}{M_B\times V}\times \frac{180\times V}{36\times R\times T}$

$\frac{W_B}{M_B}=\frac{1.52}{4.98\times 5}=0.0610mol{L}^{-1}$