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?

(A) The osmotic pressure $\pi$ is given by the formula $\pi = C R T$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Given:
$T = 300 \, K$
$\pi = 1.52 \, bar$
$R = 0.083 \, bar \, L \, K^{-1} \, mol^{-1}$
Rearranging the formula for concentration $C$:
$C = \frac{\pi}{R T}$
Substituting the values:
$C = \frac{1.52 \, bar}{0.083 \, bar \, L \, K^{-1} \, mol^{-1} \times 300 \, K}$
$C = \frac{1.52}{24.9} \, mol \, L^{-1}$
$C \approx 0.061 \, mol \, L^{-1}$
Thus,the concentration of the solution is $0.061 \, M$.