At `0^@C` , the vapour pressure of pure water is 4.63 mm and an aqueous solution of 8.49 g of `NaNO_3` dissolved in 100g of water is 4.483 mm. What is the extent of ionisation of `NaNO_3` /

Answer

To solve the problem step by step, we will follow the reasoning and calculations as outlined in the video transcript.

### Step 1: Identify Given Values
- Vapor pressure of pure water (E₀) = 4.63 mm
- Vapor pressure of the solution (E) = 4.483 mm
- Mass of NaNO₃ (solute) = 8.49 g
- Mass of water (solvent) = 100 g

### Step 2: Calculate the Change in Vapor Pressure
Using the formula for the change in vapor pressure:
\[ \Delta E = E₀ - E = 4.63 \, \text{mm} - 4.483 \, \text{mm} = 0.147 \, \text{mm} \]

### Step 3: Calculate the Mole Fraction of the Solute
Using Raoult's Law, we can express the mole fraction of the solute (NaNO₃) in terms of the change in vapor pressure:
\[ \frac{\Delta E}{E₀} = \text{Mole Fraction of Solute} \times i \]
Where \(i\) is the van 't Hoff factor (number of particles the solute dissociates into).

Substituting the values:
\[ \frac{0.147}{4.63} = \text{Mole Fraction of Solute} \times i \]
Calculating the left side:
\[ \text{Mole Fraction of Solute} \times i = 0.0317 \]

### Step 4: Calculate Moles of Solute and Solvent
1. **Moles of NaNO₃**:
- Molar mass of NaNO₃ = 23 (Na) + 14 (N) + 48 (O) = 85 g/mol
- Moles of NaNO₃ = \(\frac{8.49 \, \text{g}}{85 \, \text{g/mol}} = 0.0998 \, \text{mol}\)

2. **Moles of Water**:
- Molar mass of water = 18 g/mol
- Moles of water = \(\frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}\)

### Step 5: Calculate Total Moles
Total moles = Moles of solute + Moles of solvent
\[ \text{Total moles} = 0.0998 + 5.56 \approx 5.66 \, \text{mol} \]

### Step 6: Calculate the Mole Fraction of Solute
\[ \text{Mole Fraction of Solute} = \frac{\text{Moles of Solute}}{\text{Total Moles}} = \frac{0.0998}{5.66} \approx 0.0176 \]

### Step 7: Substitute Back to Find \(i\)
From the equation:
\[ 0.0317 = 0.0176 \times i \]
Solving for \(i\):
\[ i = \frac{0.0317}{0.0176} \approx 1.80 \]

### Step 8: Determine the Extent of Ionization (α)
For NaNO₃, it dissociates into 2 ions (Na⁺ and NO₃⁻), so \(n = 2\). The extent of ionization (α) can be calculated using:
\[ \alpha = \frac{i - 1}{n - 1} \]
Substituting the values:
\[ \alpha = \frac{1.80 - 1}{2 - 1} = \frac{0.80}{1} = 0.80 \]

### Final Answer
The extent of ionization of NaNO₃ is approximately **0.80**.