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 bar at the same temperature, what would be its concentration?

[Report your answer upto two decimal places.]

Answer 1

Answer: 0.06

$\pi =\text{CRT}=\frac{{\text{W}}_{\text{B}}\times \text{R}\times \text{T}}{{\text{M}}_{\text{B}}\times \text{V}}$

For both solutions, R, T and V are constants.

For I solution

$(4.98\mathrm{bar})=\frac{(36\mathrm{g})\times \mathrm{R}\times \mathrm{T}}{\left(180\mathrm{g}{\mathrm{mol}}^{-1}\right)\times \mathrm{V}}$ ...(i)

For II solution

(1.52 bar) $=\frac{{\mathrm{W}}_{\mathrm{B}}\times \mathrm{R}\times \mathrm{T}}{{\mathrm{M}}_{\mathrm{B}}\times \mathrm{V}}$...(ii)

On dividing Eq. (ii) by Eq. (i), we get

$\frac{\left(\text{1.52 bar}\right)}{\left(\text{4.98 bar}\right)}=\frac{{\left(\text{W}\right)}_{\text{B}}\times \text{R}\times \text{T}}{{\left(\text{M}\right)}_{\text{B}}\times \text{V}}\times \frac{\text{180}\times \text{V}}{\text{36}\times \text{R}\times \text{T}}$

$\frac{{\text{W}}_{\text{B}}}{{\text{M}}_{\text{B}}}=\frac{\text{1.52}}{\text{4.98}\times 5}={\text{0.0610 mol L}}^{-1}$