The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl$$_2$$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L$$^{-1}$$) in solution is:

For osmotic pressure we always begin with van ’t Hoff’s expression

$$\pi = i\,C\,R\,T$$

where $$\pi$$ is the osmotic pressure, $$i$$ the van ’t Hoff factor (number of ions produced per formula unit on complete dissociation), $$C$$ the molar concentration, $$R$$ the gas constant and $$T$$ the absolute temperature. Because all the measurements are carried out at the same temperature and in the same solvent, the common factors $$R$$ and $$T$$ will cancel when we take ratios, so only the products $$iC$$ need be compared.

First we deal with the reference solution, $$\mathrm{BaCl_2}$$.

On complete dissociation

$$\mathrm{BaCl_2}\;\longrightarrow\;\mathrm{Ba^{2+}} + 2\,\mathrm{Cl^-}$$

This gives a total of $$1+2 = 3$$ ions, so

$$i_{\mathrm{BaCl_2}} = 3$$

The given concentration is $$C_{\mathrm{BaCl_2}} = 0.01\;\text{mol L}^{-1}$$. Hence

$$i_{\mathrm{BaCl_2}}\,C_{\mathrm{BaCl_2}} = 3 \times 0.01 = 0.03$$

Now we turn to the unknown electrolyte $$\mathrm{XY}$$. The formula $$\mathrm{XY}$$ tells us that it dissociates into one cation and one anion:

$$\mathrm{XY}\;\longrightarrow\;\mathrm{X^+} + \mathrm{Y^-}$$

or in whatever actual charges these ions carry, exactly two ions are produced. Therefore

$$i_{\mathrm{XY}} = 2$$

The statement of the problem says that the osmotic pressure of the $$\mathrm{XY}$$ solution is four times that of the $$\mathrm{BaCl_2}$$ solution. Translating this into the product $$iC$$ we write

$$i_{\mathrm{XY}}\,C_{\mathrm{XY}} = 4 \times (i_{\mathrm{BaCl_2}}\,C_{\mathrm{BaCl_2}})$$

Substituting all known values, we have

$$2\,C_{\mathrm{XY}} = 4 \times 0.03$$

$$2\,C_{\mathrm{XY}} = 0.12$$

Dividing both sides by 2 gives

$$C_{\mathrm{XY}} = \frac{0.12}{2} = 0.06$$

In scientific notation this is

$$C_{\mathrm{XY}} = 6 \times 10^{-2}\;\text{mol L}^{-1}$$

Hence, the correct answer is Option D.