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Step-by-Step Solution
Step 1: Understand the Problem
The problem compares the osmotic pressure of a dilute solution of an ionic compound XY with that of a 0.01 M solution of BaCl2. We need to find the concentration of XY if its osmotic pressure is four times that of the BaCl2 solution, assuming both salts dissociate completely.
Step 2: Recall the Formula for Osmotic Pressure
The osmotic pressure $ \pi $ of an electrolyte depends on:
$$
\pi = i M R T
$$
where:
$ i $ is the van 't Hoff factor (number of ions produced on complete dissociation)
$ M $ is the molar concentration of the solute
$ R $ is the gas constant
$ T $ is the temperature (in Kelvin)
Step 3: Determine the van 't Hoff Factors
For BaCl2, on complete dissociation:
$$
\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\,\text{Cl}^-
$$
The total number of ions = 3, hence $ i_{\mathrm{BaCl}_2} = 3 $.
For XY, on complete dissociation:
$$
\mathrm{XY} \rightarrow \mathrm{X}^+ + \mathrm{Y}^-
$$
The total number of ions = 2, hence $ i_{\mathrm{XY}} = 2 $.
Step 4: Express the Osmotic Pressures
Let the concentration of BaCl2 be $0.01\,\mathrm{M}$. Its osmotic pressure is:
$$
\pi_{\mathrm{BaCl}_2} = i_{\mathrm{BaCl}_2} \, M_{\mathrm{BaCl}_2} \, R \, T = 3 \times 0.01 \times R \, T = 0.03 \, R \, T.
$$
Let the concentration of XY be $ [\mathrm{XY}] $. Its osmotic pressure is:
$$
\pi_{\mathrm{XY}} = i_{\mathrm{XY}} \, [\mathrm{XY}] \, R \, T = 2 \times [\mathrm{XY}] \times R \, T.
$$
Step 5: Relate the Osmotic Pressures
We know from the problem that:
$$
\pi_{\mathrm{XY}} = 4 \times \pi_{\mathrm{BaCl}_2}.
$$
Substituting the expressions:
$$
2 \times [\mathrm{XY}] \times R \, T = 4 \times \left(0.03 \, R \, T\right).
$$
Since $ R \, T $ is a common factor, it cancels out on both sides, leaving:
$$
2 \times [\mathrm{XY}] = 4 \times 0.03.
$$
Step 6: Solve for the Concentration of XY
$$
2 \times [\mathrm{XY}] = 0.12 \quad \Longrightarrow \quad [\mathrm{XY}] = 0.06.
$$
Therefore,
$$
[\mathrm{XY}] = 6 \times 10^{-2} \,\mathrm{mol\,L}^{-1}.
$$
Final Answer
The concentration of XY is $6 \times 10^{-2}\,\mathrm{mol\,L}^{-1}$.
