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 bars at the same temperature, what would be its concentration?
Solution:
Here,
T = 300 K
π = 1.52 bar
R = 0.083 bar L K−1 mol−1
Applying the relation,
π = CRT
$\Rightarrow C=\frac{\pi}{\mathrm{R} T}$
$=\frac{1.52 \mathrm{bar}}{0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}$
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.