Question
Given below are two statements:
| Chamber 1 | Semipermeable membrane | Chamber 2 |
|---|---|---|
| 18 g glucose in 100 mL aqueous solution | 30 g glucose in 250 mL aqueous solution |
Statement I : $H_{2}O$ molecules move from the chamber 1 to chamber 2.
Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass $= 174\text{ }g/mol$ ) in 2 L of water (at $27^{\circ}C$ ) is 0.0107 bar. (Given : $R = 0.083{dm}^{3}$ bar $K^{- 1}{\text{ }mol}^{- 1}$ and assume complete dissociation of electrolyte)
In the light of the above statements, choose the correct answer from the options given below :
Options
Solution
Sol. [Key Concept: Compute the two molar concentrations ($1$ M vs $\frac{2}{3}$ M): water flows toward the more concentrated side, so Statement I is false, while the $K_{2}{SO}_{4}$ osmotic pressure $\pi = iCRT$ with $i = 3$ gives $0.0107$ bar, making Statement II true.]
On side (I) $C_{1} = \frac{18/180}{0.1} = 1M$
On side (II) $C_{2} = \frac{30/180}{250/1000} = \frac{2}{3}M\ ;\ \left\{ \pi_{1} > \pi_{2} \right\}$
Hence water will flow from chamber-II to Chamber-I.
$\pi = iCRT\ ;i = 3$ for $K_{2}{SO}_{4}$
And $\left( \pi = 3 \times \frac{50 \times 10^{- 3}/174}{2} \right) \times 0.083 \times 300$
$= 0.0107$ bar.
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