Question

Consider the following setup separated by a semi–permeable membrane (SPM): Left compartment (x): \(K_4[Fe(CN)_6]\) solution \(0.1\,M\), \(100\,ml\) Right compartment (y): \(FeCl_3\) solution \(0.1\,M\), \(100\,ml\) Which of the following is correct for the above figure?

• A. \(y\) is hypotonic solution
• B. Both compartment will have blue colour
• C. To perform reverse osmosis, any pressure can be applied on x-side
• D. Solute can pass through semi-permeable membrane

Answer 1

The Correct Option is A

Concept: Osmotic pressure depends on the van’t Hoff factor \(i\). \[ \pi = iMRT \] Greater value of \(iM\) means higher osmotic pressure and the solution becomes hypertonic. The solution with smaller osmotic pressure is hypotonic.
Step 1: Calculate van’t Hoff factor for each solution. For \(K_4[Fe(CN)_6]\) \[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \] Total ions \(=5\) \[ i_x = 5 \] For \(FeCl_3\) \[ FeCl_3 \rightarrow Fe^{3+} + 3Cl^- \] Total ions \(=4\) \[ i_y = 4 \]
Step 2: Compare osmotic pressures. \[ \pi = iMRT \] Both solutions have same molarity. \[ \pi_x = 5MRT \] \[ \pi_y = 4MRT \] Thus, \[ \pi_x > \pi_y \] Hence, \[ x = \text{hypertonic}, \quad y = \text{hypotonic} \]
Step 3: Check other options. (B) Blue colour will not appear because ions cannot cross the SPM. (C) Reverse osmosis requires pressure greater than osmotic pressure, not any pressure. (D) Only solvent molecules pass through SPM. Therefore, the correct statement is: \[ \boxed{y \text{ is hypotonic solution}} \]