Question

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?

Answer 1

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

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

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

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