A 1% (w/v) aqueous solution of potassium chloride has extent of ionisation 0.82 in water it `18^@C` . Calculate the osmatic pressure.

Answer

To calculate the osmotic pressure of a 1% (w/v) aqueous solution of potassium chloride (KCl) with an extent of ionization of 0.82 at 18°C, we can follow these steps:

### Step 1: Determine the Ionization Factor (i)
Potassium chloride dissociates in water as follows:
\[ \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \]

From one KCl molecule, we get two ions (K⁺ and Cl⁻). Therefore, the value of \( x \) (the number of particles produced from one formula unit) is 2.

The degree of ionization \( \alpha \) is given as 0.82. The relationship between the van 't Hoff factor \( i \), the degree of ionization \( \alpha \), and \( x \) is:
\[ \alpha = \frac{i - 1}{x - 1} \]

Substituting the known values:
\[ 0.82 = \frac{i - 1}{2 - 1} \]

Solving for \( i \):
\[ 0.82 = i - 1 \]
\[ i = 1 + 0.82 = 1.82 \]

### Step 2: Calculate the Molarity of the Solution
A 1% (w/v) solution means there are 1 gram of solute (KCl) in 100 mL of solution. Therefore, in 1000 mL (1 L), there are:
\[ 10 \text{ grams of KCl} \]

Next, we need to calculate the molar mass of KCl:
- Potassium (K) = 39 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Now, calculate the number of moles of KCl in 10 grams:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{74.5 \text{ g/mol}} \approx 0.134 \text{ moles} \]

Since this is in 1 L of solution, the molarity \( C \) is:
\[ C = 0.134 \text{ M} \]

### Step 3: Convert Temperature to Kelvin
The temperature given is 18°C. To convert to Kelvin:
\[ T = 18 + 273 = 291 \text{ K} \]

### Step 4: Calculate the Osmotic Pressure
The formula for osmotic pressure \( \pi \) is given by:
\[ \pi = iCRT \]
Where:
- \( i = 1.82 \)
- \( C = 0.134 \text{ M} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 291 \text{ K} \)

Substituting the values:
\[ \pi = 1.82 \times 0.134 \times 0.0821 \times 291 \]

Calculating this gives:
\[ \pi \approx 5.82 \text{ atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **5.82 atm**.

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