0.2 M aq. Solution of KCl is istonic with 0.2 M `K_(2)SO_(4)` at same temperature. What is the van't Hoff fector of `K_(2)SO_(4)` ?

Answer

To determine the van't Hoff factor (i) of K₂SO₄, we can follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. This means that the osmotic pressure of the KCl solution is equal to that of the K₂SO₄ solution.

### Step 2: Write the formula for osmotic pressure
The osmotic pressure (π) can be expressed using the formula:
\[ \pi = CRTi \]
where:
- \(C\) = concentration of the solution (in molarity)
- \(R\) = gas constant
- \(T\) = temperature (in Kelvin)
- \(i\) = van't Hoff factor (number of particles the solute dissociates into)

### Step 3: Set up the equation for isotonic solutions
Since the two solutions are isotonic, we can set their osmotic pressures equal to each other:
\[ \pi_{KCl} = \pi_{K_2SO_4} \]
Substituting the formula for osmotic pressure, we get:
\[ C_{KCl}RTi_{KCl} = C_{K_2SO_4}RTi_{K_2SO_4} \]

### Step 4: Cancel out common terms
Since both solutions are at the same temperature and the gas constant is the same for both, we can cancel out \(R\) and \(T\) from both sides:
\[ C_{KCl}i_{KCl} = C_{K_2SO_4}i_{K_2SO_4} \]

### Step 5: Substitute known values
We know:
- \(C_{KCl} = 0.2 \, M\)
- \(i_{KCl} = 2\) (because KCl dissociates into K⁺ and Cl⁻)
- \(C_{K_2SO_4} = 0.2 \, M\)

Substituting these values into the equation:
\[ 0.2 \times 2 = 0.2 \times i_{K_2SO_4} \]

### Step 6: Solve for the van't Hoff factor of K₂SO₄
Now, we can simplify the equation:
\[ 0.4 = 0.2 \times i_{K_2SO_4} \]
Dividing both sides by 0.2:
\[ i_{K_2SO_4} = \frac{0.4}{0.2} = 2 \]

### Conclusion
The van't Hoff factor \(i\) for K₂SO₄ is 2.

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