Question

At $300K$, $36g$ 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?
[Report your answer as an integer, after multiplying with $100$.]

Answer 1

Answer: 6

$\pi =CRT=\frac{W_B\times R\times T}{M_B\times V}$
For both solutions, $R,T$ and $V$ are constants.
For I solution
$4.98$ bar$=\frac{36g\times R\times T}{180gmo{l}^{-1}\times 1L}$
For Il solution
$1.52$ bar $=\frac{w_B\times R\times T}{M_B\times V}$
On dividing Eq. (ii) by Eq. (i), we get
$\frac{1.52bar}{4.98bar}=\frac{W_B\times R_R\times T}{M_B\times V}\times \frac{180\times 1L}{36\times R\times T}$
$\frac{W_B}{M_{B}V}=\frac{1.52}{4.98\times 5}=0.0610mol{L}^{-1}$
After multiplying with $100=6$