1. Use the osmotic pressure relation

For a dilute solution, $\pi = i C R T$ where:

• $\pi$ = osmotic pressure
• $i$ = van’t Hoff factor
• $C$ = molar concentration
• $R,T$ are constant for both solutions

So, osmotic pressure is proportional to $iC$: $\pi \propto iC$

2. Find $i$ for $BaCl_2$

Assuming complete dissociation, $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ Thus, total number of ions produced = $3$. So, $i_{BaCl_2} = 3$

Given concentration of $BaCl_2$ is $0.01\,\text{M}$, hence $iC = 3 \times 0.01 = 0.03$

3. Find osmotic condition for $XY$

Given osmotic pressure of $XY$ solution is four times that of $0.01\,\text{M}\; BaCl_2$ solution: $\pi_{XY} = 4\pi_{BaCl_2}$

Therefore, $i_{XY} C_{XY} = 4 \times 0.03 = 0.12$

4. Find $i$ for ionic compound $XY$

Since $XY$ is an ionic compound and dissociates completely as $XY \rightarrow X^+ + Y^-$ or in general one cation + one anion, it produces $2$ ions. Hence, $i_{XY} = 2$

So, $2C_{XY} = 0.12$ $C_{XY} = \frac{0.12}{2} = 0.06\,\text{M}$

5. Match with options

$0.06 = 6 \times 10^{-2}\,\text{mol L}^{-1}$

So the correct option is B.

Final Answer

$\boxed{6 \times 10^{-2}\,\text{mol L}^{-1}}$