At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar.

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.