220. At 300 K, 36 g of glucose (C₆H₁₂O₆) present per litre in its aqueous solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another solution of glucose is 1.52 bar at the same temperature, what would be its concentration? 

Answer:

$$\begin{gathered} {\mathrm{\pi }}_1 = {\mathrm{C}}_1\mathrm{RT} ....................1 \\ {\mathrm{\pi }}_2 = {\mathrm{C}}_2\mathrm{RT} .................... 2 \\ \mathrm{therfore} \mathrm{on} \mathrm{dividing} \mathrm{the} 1 \mathrm{by} 2 \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{\pi }}_1}{{\mathrm{\pi }}_2} = \frac{{\mathrm{C}}_1}{{\mathrm{C}}_2} \end{gathered}$$

Let us calculate the concentration of the first solution with osmatic pressure 4.98bar

mass of glucose = 36g
molar mass of gulcose = 180g/ mol
therefore number of moles of gulcose =  36/180
                                                         = 0.2 moles

volume of the solution  = 1L
  molarity  = No. of moles of glucose / vol.of solution

molarity  =0.2/1L

C_{1 =} 0.2 moles/L
$$\begin{gathered} {\mathrm{C}}_2 = \frac{{\mathrm{\pi }}_2{\mathrm{C}}_1}{{\mathrm{\pi }}_2} \\ {\mathrm{C}}_2 = \frac{1.52\times 0.2}{4.98} = 0.61\mathrm{M} \end{gathered}$$