The osmotic pressure of 0.2 M KCl solution at 310 K is

Options

A. 10.17 atm
B. 5.084 atm
C. 8.36 atm
D. 12.2 atm

Answer

To calculate the osmotic pressure of a 0.2 M KCl solution at 310 K, we can use the formula for osmotic pressure:

\[ \pi = i \times C \times R \times T \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = concentration of the solution (in molarity)
- \(R\) = ideal gas constant
- \(T\) = temperature (in Kelvin)

### Step 1: Determine the van 't Hoff factor (i)
KCl dissociates into two ions:
- KCl → K⁺ + Cl⁻

Thus, the van 't Hoff factor \(i\) for KCl is:
\[ i = 2 \]

### Step 2: Identify the concentration (C)
The concentration of the KCl solution is given as:
\[ C = 0.2 \, \text{M} \]

### Step 3: Use the ideal gas constant (R)
The ideal gas constant \(R\) is:
\[ R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \]

### Step 4: Use the temperature (T)
The temperature is given as:
\[ T = 310 \, \text{K} \]

### Step 5: Substitute the values into the osmotic pressure formula
Now, we can substitute \(i\), \(C\), \(R\), and \(T\) into the osmotic pressure formula:
\[ \pi = 2 \times 0.2 \, \text{M} \times 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \times 310 \, \text{K} \]

### Step 6: Calculate the osmotic pressure
Calculating the above expression:
\[ \pi = 2 \times 0.2 \times 0.0821 \times 310 \]
\[ \pi = 2 \times 0.2 \times 25.453 \]
\[ \pi = 2 \times 5.0906 \]
\[ \pi = 10.1812 \, \text{atm} \]

Thus, the osmotic pressure of the 0.2 M KCl solution at 310 K is approximately:
\[ \pi \approx 10.18 \, \text{atm} \]

### Final Answer
The osmotic pressure of the 0.2 M KCl solution at 310 K is approximately **10.18 atm**.