JEE Main
Chemistry
Solutions
2026
JEE Main 2026 (04 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Question
At 27°C, 0.1 M, 1 L K_4[Fe(CN)_6] aqueous solution and 0.1 M, 1 L FeCl_3 aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct?
Options
- A. Blue color is formed on both sides.
- B. Ionic solutes in aqueous solution can pass through semi-permeable membrane.
- C. Solution on side 'y' is hypotonic.
- D. To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side 'x'.
Answer
C. Solution on side 'y' is hypotonic.
Step-by-step solution
For complete dissociation, the van 't Hoff factor i is equal to the number of ions produced per formula unit. For side 'x' containing K_4[Fe(CN)_6]: K_4[Fe(CN)_6] 4K^+ + [Fe(CN)_6]^ 4- Number of ions, i_x = 5 Effective concentration (osmolarity) = i_x C = 5 0.1 = 0.5 M For side 'y' containing FeCl_3: FeCl_3 Fe^ 3+ + 3Cl^- Number of ions, i_y = 4 Effective concentration (osmolarity) = i_y C = 4 0.1 = 0.4 M Since the effective concentration of side 'y' (0.4 M) is less than that of side 'x' (0.5 M), the solution on side 'y' has a lower osmotic pressure and is therefore hypotonic with respect to side 'x'. Evaluating the other options: A semi-permeable membrane (SPM) only allows solvent molecules to pass, not solute ions. Thus, Fe^ 3+ and [Fe(CN)_6]^ 4- cannot mix, and no blue color is formed. To cause reverse osmosis, the external pressure applied to the hypertonic side (side 'x') must be strictly greater than the osmotic pressure difference ( = _x - _y), not just any value. Answer: Solution on side 'y' is hypotonic.
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