The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute aqueous solution, containing `7.0 g` of the salt per `100 g` of water at `100^(@)C` is `70%`. If the vapour pressure of water at `100^(@)C` is `760 mm`, calculate the vapour pressure of the solution.

Explanation

To solve the problem, we need to calculate the vapor pressure of a solution containing calcium nitrate, \( Ca(NO_3)_2 \), given its degree of dissociation and the vapor pressure of pure water at \( 100^\circ C \).

### Step 1: Determine the degree of dissociation
The degree of dissociation (\( \alpha \)) of \( Ca(NO_3)_2 \) is given as \( 70\% \) or \( 0.7 \).

### Step 2: Calculate the van 't Hoff factor (\( i \))
The dissociation of \( Ca(NO_3)_2 \) can be represented as:
\[ Ca(NO_3)_2 \rightarrow Ca^{2+} + 2 NO_3^{-} \]

From this equation, we can see that one formula unit of \( Ca(NO_3)_2 \) produces 3 ions (1 \( Ca^{2+} \) and 2 \( NO_3^{-} \)).

The van 't Hoff factor (\( i \)) is calculated as:
\[ i = 1 + 2\alpha \]
Substituting \( \alpha = 0.7 \):
\[ i = 1 + 2 \times 0.7 = 1 + 1.4 = 2.4 \]

### Step 3: Calculate the moles of solute
First, we need to find the molar mass of \( Ca(NO_3)_2 \):
- Calcium (Ca): \( 40 \, g/mol \)
- Nitrogen (N): \( 14 \, g/mol \) (2 Nitrogens in the formula)
- Oxygen (O): \( 16 \, g/mol \) (6 Oxygens in the formula)

Calculating the molar mass:
\[ M = 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164 \, g/mol \]

Now, we can calculate the moles of \( Ca(NO_3)_2 \) in 7 g:
\[ \text{Moles of } Ca(NO_3)_2 = \frac{7 \, g}{164 \, g/mol} = 0.04268 \, mol \]

### Step 4: Calculate the moles of solvent (water)
The mass of water is given as \( 100 \, g \). The molar mass of water (\( H_2O \)) is \( 18 \, g/mol \):
\[ \text{Moles of water} = \frac{100 \, g}{18 \, g/mol} = 5.5556 \, mol \]

### Step 5: Calculate the mole fraction of the solvent
The total number of moles in the solution is:
\[ \text{Total moles} = \text{Moles of } Ca(NO_3)_2 + \text{Moles of water} = 0.04268 + 5.5556 = 5.59828 \, mol \]

The mole fraction of the solvent (water) is:
\[ X_{solvent} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{5.5556}{5.59828} \approx 0.99236 \]

### Step 6: Calculate the vapor pressure of the solution
Using Raoult's Law, the vapor pressure of the solution (\( P_{solution} \)) can be calculated as:
\[ P_{solution} = P^0_{solvent} \times X_{solvent} \]
Where \( P^0_{solvent} \) is the vapor pressure of pure water at \( 100^\circ C \), which is \( 760 \, mmHg \).

Substituting the values:
\[ P_{solution} = 760 \, mmHg \times 0.99236 \approx 754.19 \, mmHg \]

### Final Answer
The vapor pressure of the solution is approximately \( 754.19 \, mmHg \).

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