If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol of `H_(2)O`, then % ionization NaCl is

Options

A. 0.6
B. 0.8
C. 0.4
D. 1

Answer

To solve the problem, we need to determine the percentage ionization of NaCl in a solution where the relative decrease in vapor pressure is given as 0.4.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Relative decrease in vapor pressure (ΔP/P₀) = 0.4
- Moles of NaCl (solute) = 1
- Moles of water (solvent) = 3

2. **Using the Formula for Relative Decrease in Vapor Pressure**:
The formula for relative decrease in vapor pressure is given by:
\[ \frac{\Delta P}{P_0} = i \cdot \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
where:
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(n_{solute}\) = moles of solute
- \(n_{solvent}\) = moles of solvent

3. **Substituting the Known Values**:
Substituting the values into the formula:
\[ 0.4 = i \cdot \frac{1}{1 + 3} \]
This simplifies to:
\[ 0.4 = i \cdot \frac{1}{4} \]

4. **Solving for \(i\)**:
Rearranging the equation to find \(i\):
\[ i = 0.4 \cdot 4 = 1.6 \]

5. **Understanding the Degree of Ionization**:
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
where \(n\) is the number of particles the solute dissociates into. For NaCl, it dissociates into Na⁺ and Cl⁻, so \(n = 2\).

6. **Substituting Values to Find Degree of Ionization**:
Substituting the values into the degree of ionization formula:
\[ \alpha = \frac{1.6 - 1}{2 - 1} = \frac{0.6}{1} = 0.6 \]

7. **Calculating Percentage Ionization**:
To find the percentage ionization, we multiply the degree of ionization by 100:
\[ \text{Percentage Ionization} = 0.6 \times 100 = 60\% \]

### Final Answer:
The percentage ionization of NaCl in the solution is **60%**.