1. Use osmotic pressure to determine direction of osmosis
   For dilute solutions, $\pi = iMRT$ So we compare $iM$ for the two sides.

• For $\mathrm{CuSO_4}$: $\mathrm{CuSO_4 \rightarrow Cu^{2+} + SO_4^{2-}}$ Hence, ideally $i=2$. Given molarity $=0.05\,\mathrm{M}$, so $iM = 2 \times 0.05 = 0.10$

• For $\mathrm{K_2Cr_2O_7}$: $\mathrm{K_2Cr_2O_7 \rightarrow 2K^+ + Cr_2O_7^{2-}}$ Hence, ideally $i=3$. Given molarity $=0.01\,\mathrm{M}$, so $iM = 3 \times 0.01 = 0.03$

Thus, $\pi_{\mathrm{CuSO_4}} > \pi_{\mathrm{K_2Cr_2O_7}}$ So solvent flows from the $\mathrm{K_2Cr_2O_7}$ side to the $\mathrm{CuSO_4}$ side.

2. Effect on molarity
   Since solvent enters the $\mathrm{CuSO_4}$ solution, that solution gets diluted. Therefore its molarity decreases.

So,

• molarity of $\mathrm{CuSO_4}$ decreases,
• molarity of $\mathrm{K_2Cr_2O_7}$ would increase (not decrease), because solvent leaves that side.

Hence:

• A is false
• C is true

3. Where will green colour appear?
   The green colour is due to formation of $\mathrm{Cu_2Cr_2O_7}$ when the two solutions come into contact. But a semi-permeable membrane allows only solvent molecules to pass, not solute ions. Therefore the solutes do not mix across the membrane.

So green colour due to reaction product will not be observed merely because of osmosis on either side X or Y.

Hence:

• B is false
• D is false

4. Final answer
   The correct option is: $\boxed{\text{C}}$