A solution containing `3.00gm` of calcium nitrate in `100 c.c.` of solution had an osmotic pressure of `11.2` atmosphere at `12^(@)C`. Calculate the degree of ionisation of calcium nitrate at this dilution and temperature.

Explanation

To solve the problem of calculating the degree of ionization of calcium nitrate in the given solution, we will follow these steps:

### Step 1: Calculate the Molar Mass of Calcium Nitrate
The formula for calcium nitrate is \( \text{Ca(NO}_3\text{)}_2 \).
- Molar mass of Calcium (Ca) = 40 g/mol
- Molar mass of Nitrogen (N) = 14 g/mol (2 Nitrogen atoms = 28 g/mol)
- Molar mass of Oxygen (O) = 16 g/mol (6 Oxygen atoms = 96 g/mol)

So, the molar mass of calcium nitrate is:
\[ \text{Molar mass} = 40 + 28 + 96 = 164 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Calcium Nitrate
Given mass of calcium nitrate = 3 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{164 \text{ g/mol}} \approx 0.0183 \text{ moles} \]

### Step 3: Calculate the Molarity of the Solution
The volume of the solution is given as 100 cm³, which is equivalent to 0.1 L.
Molarity (C) is given by:
\[ C = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.0183 \text{ moles}}{0.1 \text{ L}} = 0.183 \text{ mol/L} \]

### Step 4: Use the Osmotic Pressure Equation
The osmotic pressure (\( \Pi \)) is given by the formula:
\[ \Pi = iCRT \]
Where:
- \( \Pi = 11.2 \text{ atm} \)
- \( C = 0.183 \text{ mol/L} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 12^\circ C = 285 \text{ K} \)

Rearranging the formula to find the van't Hoff factor \( i \):
\[ i = \frac{\Pi}{CRT} \]
Substituting the values:
\[ i = \frac{11.2}{0.183 \times 0.0821 \times 285} \]

Calculating the denominator:
\[ 0.183 \times 0.0821 \times 285 \approx 4.29 \]
Thus,
\[ i \approx \frac{11.2}{4.29} \approx 2.61 \]

### Step 5: Determine the Degree of Ionization
Calcium nitrate dissociates in solution as follows:
\[ \text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_3^{-} \]
This means that for every 1 mole of calcium nitrate, we get 3 moles of ions (1 calcium ion and 2 nitrate ions).

Let \( \alpha \) be the degree of ionization. Then:
\[ i = 1 + 2\alpha \]
Substituting the value of \( i \):
\[ 2.61 = 1 + 2\alpha \]
Solving for \( \alpha \):
\[ 2\alpha = 2.61 - 1 = 1.61 \implies \alpha = \frac{1.61}{2} \approx 0.805 \]

### Final Result
The degree of ionization of calcium nitrate at this dilution and temperature is approximately \( 0.805 \) or \( 80.5\% \).

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