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. The solution contains 1 mole of NaCl and 3 moles of water.

### Step-by-Step Solution:

1. **Understand the Concept of Relative Decrease in Vapor Pressure**:
The relative decrease in vapor pressure (RL) is given by the formula:
\[ RL = \frac{P^0 - P}{P^0} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \cdot i \]
where \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into), \(n_{solute}\) is the number of moles of solute, and \(n_{solvent}\) is the number of moles of solvent.

2. **Identify Given Values**:
- Relative decrease in vapor pressure, \(RL = 0.4\)
- Moles of NaCl (solute), \(n_{solute} = 1\)
- Moles of water (solvent), \(n_{solvent} = 3\)

3. **Calculate Total Moles**:
\[ n_{total} = n_{solute} + n_{solvent} = 1 + 3 = 4 \]

4. **Calculate Mole Fraction of Solute**:
\[ \text{Mole fraction of NaCl} = \frac{n_{solute}}{n_{total}} = \frac{1}{4} = 0.25 \]

5. **Set Up the Equation for Relative Decrease in Vapor Pressure**:
\[ 0.4 = i \cdot \text{Mole fraction of NaCl} \]
Substituting the mole fraction:
\[ 0.4 = i \cdot 0.25 \]

6. **Solve for the van 't Hoff Factor (i)**:
\[ i = \frac{0.4}{0.25} = 1.6 \]

7. **Determine the Percentage Ionization**:
NaCl dissociates into two ions: Na\(^+\) and Cl\(^-\), so \(N = 2\).
The formula relating ionization (\(\alpha\)) to the van 't Hoff factor is:
\[ \alpha = \frac{i - 1}{N - 1} \]
Substituting the values:
\[ \alpha = \frac{1.6 - 1}{2 - 1} = \frac{0.6}{1} = 0.6 \]

8. **Convert to Percentage**:
\[ \text{Percentage ionization} = \alpha \times 100 = 0.6 \times 100 = 60\% \]

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