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?

Answer 1

In the first case, 

π₁ = C₁RT

i.e., 4.98 bar = \(\frac{36g\times R\times300K}{180g\,mol^{-1}}\) ——– (1)

In second case, 

π₂ = C₂RT

i.e., 1.52 bar = C₂R × 300 K ———- (2)

\(\frac{Eqn.(2)}{Eqn.(1)}\Rightarrow \frac{1.52\,bar}{4.98\,bar}\)\(=\frac{C_2\times R\times 300K\times 180\,g\,mol^{-1}}{36\,gL^{-1}\times R\,\times\,300K}\)

∴ \(C_2=\frac{1.52\,bar\times\,36\,g\,L^{-1}}{4.98\,bar\times 180\,g\,mol^{-1}}\)

= 0.061 mol L^-1