0.2 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 find the van't Hoff factor (i) for \( K_2SO_4 \) given that a 0.2 M solution of \( KCl \) is isotonic with a 0.2 M solution of \( K_2SO_4 \), we can follow these steps:

### Step 1: Understand 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 \( K_2SO_4 \) solution.

### Step 2: Write the Osmotic Pressure Formula
The osmotic pressure (\( \pi \)) can be expressed using the formula:
\[ \pi = iCRT \]
where:
- \( i \) = van't Hoff factor
- \( C \) = concentration of the solution (in molarity)
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 3: Set Up the Equation for Both Solutions
Let’s denote:
- \( C_1 = 0.2 \, M \) for \( KCl \)
- \( C_2 = 0.2 \, M \) for \( K_2SO_4 \)
- \( i_1 \) = van't Hoff factor for \( KCl \)
- \( i_2 \) = van't Hoff factor for \( K_2SO_4 \)

Since the solutions are isotonic:
\[ i_1 C_1 = i_2 C_2 \]

### Step 4: Determine the van't Hoff Factor for \( KCl \)
For \( KCl \):
- \( KCl \) dissociates into \( K^+ \) and \( Cl^- \), which gives us 2 ions.
- Therefore, \( i_1 = 2 \).

### Step 5: Substitute the Values into the Equation
Substituting the known values into the equation:
\[ 2 \times 0.2 = i_2 \times 0.2 \]

### Step 6: Solve for \( i_2 \)
To find \( i_2 \):
\[ 0.4 = i_2 \times 0.2 \]
\[ i_2 = \frac{0.4}{0.2} = 2 \]

### Conclusion
The van't Hoff factor (\( i \)) for \( K_2SO_4 \) is 2.