Calculate relative lowering of vapour pressure in an aqueous solution of `CaCI_(2)` haing 1 mole of `CaCI_(2)` dissolve3d in 324 g of water if degree of dissociain is 60%

Options

A. `(1)/(19)`
B. `(3)/(19)`
C. `(11)/(101)`
D. `(2.2)/(19)`

Answer

To calculate the relative lowering of vapor pressure in an aqueous solution of \( CaCl_2 \) with a degree of dissociation of 60%, we can follow these steps:

### Step 1: Determine the number of moles of water
Given that the mass of water is 324 g, we can convert this mass into moles. The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{324 \, \text{g}}{18 \, \text{g/mol}} = 18 \, \text{mol} \]

### Step 2: Calculate the effective number of particles from \( CaCl_2 \)
Since \( CaCl_2 \) dissociates into \( Ca^{2+} \) and 2 \( Cl^{-} \) ions, the total number of ions produced from 1 mole of \( CaCl_2 \) is:

\[ \text{Total ions} = 1 \, (Ca^{2+}) + 2 \, (Cl^{-}) = 3 \, \text{ions} \]

However, since the degree of dissociation is 60%, only 60% of the \( CaCl_2 \) will dissociate. Therefore, the effective number of moles of ions produced is:

\[ \text{Effective moles of ions} = 1 \, \text{mol} \times 0.6 \times 3 = 1.8 \, \text{mol} \]

### Step 3: Calculate the mole fraction of solute
The total number of moles in the solution is the sum of the moles of solute (ions) and the moles of solvent (water):

\[ \text{Total moles} = \text{Moles of ions} + \text{Moles of water} = 1.8 \, \text{mol} + 18 \, \text{mol} = 19.8 \, \text{mol} \]

Now, we can calculate the mole fraction of the solute (the ions):

\[ \text{Mole fraction of solute} (X_{solute}) = \frac{\text{Moles of ions}}{\text{Total moles}} = \frac{1.8}{19.8} \]

Calculating this gives:

\[ X_{solute} \approx 0.0909 \]

### Step 4: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure (\( \Delta P \)) can be calculated using Raoult's Law:

\[ \Delta P = X_{solute} \]

Thus, the relative lowering of vapor pressure is approximately:

\[ \Delta P \approx 0.0909 \]

### Final Answer
The relative lowering of vapor pressure in the aqueous solution of \( CaCl_2 \) is approximately 0.0909 or 9.09%.

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