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

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

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

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

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