The Vapour pressure of solution containing `2g` of `NaCl` in `100g` of water at `100^(@)C` is `753.2mm` of `Hg`, then degree of dissociation of `NaCl`.

Options

A. `0.6`
B. `0.7`
C. `0.8`
D. `0.9`

Explanation

To solve the problem of finding the degree of dissociation of NaCl in a solution, we will follow these steps:

### Step 1: Calculate the Relative Lowering of Vapor Pressure
Using Raoult's Law, we know that the relative lowering of vapor pressure can be calculated as:

\[ \text{Relative lowering of vapor pressure} = \frac{P_0 - P_s}{P_0} \]

Where:
- \(P_0\) = Vapor pressure of pure solvent (water) at 100°C = 760 mm Hg
- \(P_s\) = Vapor pressure of the solution = 753.2 mm Hg

Substituting the values:

\[ \text{Relative lowering} = \frac{760 - 753.2}{760} = \frac{6.8}{760} \approx 0.008947 \]

### Step 2: Calculate the Mole Fraction of Solute
The mole fraction of solute (NaCl) can be approximated as the relative lowering of vapor pressure:

\[ \text{Mole fraction of solute} \approx \text{Relative lowering} \approx 0.008947 \]

### Step 3: Calculate the Moles of Solvent
The number of moles of solvent (water) can be calculated using its mass and molar mass:

\[ \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.56 \text{ moles} \]

### Step 4: Set Up the Equation for Mole Fraction
Using the mole fraction definition:

\[ \text{Mole fraction of solute} = \frac{\text{Moles of solute}}{\text{Moles of solute} + \text{Moles of solvent}} \]

Let \(n\) be the number of moles of NaCl. Since the solution is very dilute, we can neglect the moles of solute in the denominator:

\[ 0.008947 = \frac{n}{n + 5.56} \approx \frac{n}{5.56} \]

### Step 5: Calculate the Moles of Solute
Rearranging the equation gives:

\[ n \approx 0.008947 \times 5.56 \approx 0.0497 \text{ moles} \]

### Step 6: Calculate the Experimental Molar Mass of NaCl
The mass of NaCl used is 2 g, so we can find the experimental molar mass:

\[ \text{Molar mass of NaCl} = \frac{\text{mass}}{\text{moles}} = \frac{2 \text{ g}}{0.0497 \text{ moles}} \approx 40.24 \text{ g/mol} \]

### Step 7: Calculate the Degree of Dissociation
The theoretical molar mass of NaCl is 58.5 g/mol. The Van’t Hoff factor \(i\) can be calculated as:

\[ i = \frac{M_{\text{theoretical}}}{M_{\text{experimental}}} = \frac{58.5}{40.24} \approx 1.45 \]

For NaCl, which dissociates into two ions (Na\(^+\) and Cl\(^-\)), the relationship is:

\[ i = 1 + (n - 1) \alpha \]

Where \(n = 2\) (number of ions). Thus:

\[ 1.45 = 1 + (2 - 1) \alpha \implies 1.45 = 1 + \alpha \implies \alpha = 0.45 \]

### Final Answer
The degree of dissociation of NaCl is approximately **0.45**.