1% aqueous solution of ` Ca(NO_(3))_(2)` has freezing point

Options

A. `0^(@)C`
B. less than ` 0^(@)C`
C. `1^(@)C`
D. `2^(@)C`

Answer

To solve the problem of determining the freezing point of a 1% aqueous solution of calcium nitrate (Ca(NO₃)₂), we will follow these steps:

### Step 1: Understand the Concept of Freezing Point Depression
Freezing point depression is a colligative property that depends on the number of solute particles in a solution. The formula to calculate freezing point depression is:
\[ \Delta T_f = -i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) = freezing point depression,
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into),
- \(K_f\) = molal freezing point depression constant,
- \(m\) = molality of the solution.

### Step 2: Calculate the Molecular Weight of Calcium Nitrate
The molecular weight of calcium nitrate (Ca(NO₃)₂) is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Nitrogen (N) = 14 g/mol (2 Nitrogen atoms = 2 × 14 = 28 g/mol)
- Oxygen (O) = 16 g/mol (6 Oxygen atoms = 6 × 16 = 96 g/mol)

Thus, the total molecular weight is:
\[ \text{Molecular weight of Ca(NO}_3\text{)}_2 = 40 + 28 + 96 = 164 \text{ g/mol} \]

### Step 3: Determine the Mass of Solute in the Solution
Given a 1% mass/volume solution, it means there is 1 gram of calcium nitrate in 100 mL of solution. Therefore, in 1 liter (1000 mL) of solution, there will be:
\[ \text{Mass of Ca(NO}_3\text{)}_2 = 10 \text{ g} \]

### Step 4: Calculate the Number of Moles of Calcium Nitrate
Using the molecular weight calculated earlier:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{10 \text{ g}}{164 \text{ g/mol}} \approx 0.061 \text{ moles} \]

### Step 5: Calculate the Molality of the Solution
Since we are dealing with 1 liter of solution (which is approximately 1 kg of water for dilute solutions), the molality (m) can be calculated as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.061 \text{ moles}}{1 \text{ kg}} = 0.061 \text{ mol/kg} \]

### Step 6: Determine the van 't Hoff Factor (i)
Calcium nitrate dissociates in solution as follows:
\[ \text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2 \text{NO}_3^{-} \]
This means it produces 3 particles (1 calcium ion and 2 nitrate ions), so:
\[ i = 3 \]

### Step 7: Use the Freezing Point Depression Formula
Now substituting the values into the freezing point depression formula:
\[ \Delta T_f = -i \cdot K_f \cdot m = -3 \cdot 1.86 \cdot 0.061 \]
Calculating this gives:
\[ \Delta T_f \approx -0.34 \text{ °C} \]

### Step 8: Determine the Freezing Point of the Solution
The normal freezing point of water is 0 °C. Therefore, the freezing point of the solution is:
\[ \text{Freezing point} = 0 \text{ °C} + \Delta T_f \approx 0 \text{ °C} - 0.34 \text{ °C} \approx -0.34 \text{ °C} \]

### Conclusion
The freezing point of a 1% aqueous solution of calcium nitrate is less than 0 °C.

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