Question

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?
[Report your answer as an integer, after multiplying with $100$.]

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
$4.98$ bar$ =\frac{36 g \times R \times T }{180 g mol ^{-1} \times 1 L }$
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
$\frac{1.52 \,bar }{4.98\, bar }=\frac{ W _{ B } \times R _{ R } \times T }{ M _{ B } \times V } \times \frac{180 \times 1\, L }{36 \times R \times T }$
$\frac{W_{B}}{M_{B} V}=\frac{1.52}{4.98 \times 5}=0.0610 \,mol\, L ^{-1}$
After multiplying with $100=6$