A solution of `XY` (`100%` ionised) has osmotic pressure equal to four times the osmotic pressure of `0.01 M BaCl_(2)` (`100%` ionised). Find the molarity of `XY`

Options

A. `6xx10^(-2)M`
B. `3xx10^(-2)M`
C. `4xx10^(-2)M`
D. `12xx10^(-2)M`

Answer

To solve the problem, we need to find the molarity of the solution `XY` given that its osmotic pressure is four times that of a `0.01 M` solution of `BaCl2`, which is fully ionized.

### Step-by-Step Solution:

1. **Understand Osmotic Pressure**:
The osmotic pressure (\( \pi \)) of a solution can be calculated using the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \( i \) = Van't Hoff factor (number of particles the solute breaks into)
- \( C \) = molarity of the solution
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Identify Van't Hoff Factor for BaCl2**:
For `BaCl2`, it dissociates into one barium ion (`Ba^2+`) and two chloride ions (`2Cl^-`), giving a total of three ions. Thus, the Van't Hoff factor \( i \) for `BaCl2` is:
\[ i_{BaCl2} = 1 + 2 = 3 \]

3. **Calculate Osmotic Pressure for BaCl2**:
The osmotic pressure for the `0.01 M` solution of `BaCl2` can be expressed as:
\[ \pi_{BaCl2} = i_{BaCl2} \cdot C_{BaCl2} = 3 \cdot 0.01 = 0.03 \text{ atm} \]

4. **Osmotic Pressure for XY**:
According to the problem, the osmotic pressure of the `XY` solution is four times that of the `BaCl2` solution:
\[ \pi_{XY} = 4 \cdot \pi_{BaCl2} = 4 \cdot 0.03 = 0.12 \text{ atm} \]

5. **Identify Van't Hoff Factor for XY**:
Since `XY` is fully ionized into two particles (X and Y), the Van't Hoff factor \( i \) for `XY` is:
\[ i_{XY} = 2 \]

6. **Set Up the Equation for XY**:
Using the osmotic pressure formula for `XY`, we have:
\[ \pi_{XY} = i_{XY} \cdot C_{XY} \]
Substituting the known values:
\[ 0.12 = 2 \cdot C_{XY} \]

7. **Solve for Molarity of XY**:
Rearranging the equation gives:
\[ C_{XY} = \frac{0.12}{2} = 0.06 \text{ M} \]

### Final Answer:
The molarity of the `XY` solution is:
\[ \boxed{0.06 \text{ M}} \]