A `1.2%` solution `(w/v)` of `NaCl` is isotonic with `7.2%` solution `(w/v)` of glucose. Calculate degree of ionization and Van't Hoff factor of `NaCl`.

Answer

To solve the problem of calculating the degree of ionization and Van't Hoff factor of NaCl, we will follow these steps:

### Step 1: Understand the Isotonic Condition
Given that a 1.2% (w/v) NaCl solution is isotonic with a 7.2% (w/v) glucose solution, we know that their osmotic pressures are equal.

### Step 2: Calculate the Concentration of NaCl and Glucose
1. **NaCl Concentration:**
- A 1.2% (w/v) solution means there are 1.2 grams of NaCl in 100 mL of solution.
- Molar mass of NaCl = 58.5 g/mol.
- Moles of NaCl = \( \frac{1.2 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0205 \text{ mol} \).
- Concentration of NaCl (C) = \( \frac{0.0205 \text{ mol}}{0.1 \text{ L}} = 0.205 \text{ mol/L} \).

2. **Glucose Concentration:**
- A 7.2% (w/v) solution means there are 7.2 grams of glucose in 100 mL of solution.
- Molar mass of glucose (C6H12O6) = 180 g/mol.
- Moles of glucose = \( \frac{7.2 \text{ g}}{180 \text{ g/mol}} = 0.04 \text{ mol} \).
- Concentration of glucose (C) = \( \frac{0.04 \text{ mol}}{0.1 \text{ L}} = 0.4 \text{ mol/L} \).

### Step 3: Set Up the Osmotic Pressure Equation
Since the osmotic pressures are equal, we can set up the following equation:
\[ \Pi_{\text{NaCl}} = \Pi_{\text{glucose}} \]
Using the formula for osmotic pressure:
\[ \Pi = i \cdot C \cdot R \cdot T \]
Where \( i \) is the Van't Hoff factor, \( C \) is the concentration, \( R \) is the gas constant, and \( T \) is the temperature. Since \( R \) and \( T \) are the same for both solutions, we can simplify to:
\[ i_{\text{NaCl}} \cdot C_{\text{NaCl}} = i_{\text{glucose}} \cdot C_{\text{glucose}} \]
Since glucose is a non-electrolyte, \( i_{\text{glucose}} = 1 \).

### Step 4: Substitute the Concentrations
Substituting the concentrations:
\[ i_{\text{NaCl}} \cdot 0.205 = 1 \cdot 0.4 \]
Solving for \( i_{\text{NaCl}} \):
\[ i_{\text{NaCl}} = \frac{0.4}{0.205} \approx 1.95 \]

### Step 5: Calculate the Degree of Ionization
The degree of ionization \( \alpha \) can be calculated from the Van't Hoff factor:
\[ i_{\text{NaCl}} = 1 + \alpha \]
Substituting the value of \( i_{\text{NaCl}} \):
\[ 1.95 = 1 + \alpha \]
Thus,
\[ \alpha = 1.95 - 1 = 0.95 \]

### Final Results
- The Van't Hoff factor \( i \) for NaCl is approximately **1.95**.
- The degree of ionization \( \alpha \) for NaCl is **0.95**.