0.05 M CuSO4 when treated with 0.01 M K2Cr2O7 gives green colour solution of Cu2Cr2O7. The two solutions are separated…

Strategy:\n> Osmosis is the movement of solvent from a side of low particle concentration (low osmotic pressure) to high particle concentration (high osmotic pressure). Determine the effective concentrations of both compartments.\n\nStep 1: Calculate effective concentrations ($i \cdot M$)\n- Side X ($\ce{K2Cr2O7}$, $M=0.01$): $\ce{K2Cr2O7 \o 2K^+ + Cr2O7^{2-}}$ ($i=3$). Eff = $0.03\\text{ M}$.\n- Side Y ($\ce{CuSO4}$, $M=0.05$): $\ce{CuSO4 \o Cu^{2+} + SO4^{2-}}$ ($i=2$). Eff = $0.10\\text{ M}$.\n\nStep 2: Determine solvent movement\nSolvent moves from Side X (0.03 M) to Side Y (0.10 M).\n\nStep 3: Evaluate local changes\n- Side X loses water $\implies$ its concentration increases.\n- Side Y gains water $\implies$ its concentration (molarity of $\ce{CuSO4}$) decreases/lowered.\n\nNote: No green colour forms because ions do not cross the SPM.\n\n$\boxed{\\text{Answer: (1)}}$