A 1.0 g sample of `Co``(NH_(2)CH_(2)CH_(2)NH_(2))_(3)Cl_(3)` is dissolved in 25.0 g of water and the freezing point of the solution is `-0.87^(@)C`. How many ions are produced per mole of compound? The `K_(f)` of water is `1.86^(@)C//"molal"`

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we will follow the outlined procedure to determine how many ions are produced per mole of the compound `Co(NH2CH2CH2NH2)3Cl3`.

### Step 1: Identify the given data
- Mass of the compound (solute) = 1.0 g
- Mass of water (solvent) = 25.0 g
- Freezing point of the solution (Tf) = -0.87°C
- Freezing point constant of water (Kf) = 1.86°C/molal

### Step 2: Calculate the change in freezing point (ΔTf)
The change in freezing point is calculated as:
\[ \Delta Tf = Tf^0 - Tf \]
Where Tf^0 (the freezing point of pure water) is 0°C. Thus:
\[ \Delta Tf = 0 - (-0.87) = 0.87°C \]

### Step 3: Calculate the molality (m) of the solution
Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of the solute.

1. **Calculate the molar mass of the compound**:
- The molecular formula is `Co(NH2CH2CH2NH2)3Cl3`.
- The molar mass can be calculated as follows:
- C: 12.01 g/mol × 3 = 36.03 g/mol
- H: 1.008 g/mol × 18 = 18.144 g/mol
- N: 14.01 g/mol × 6 = 84.06 g/mol
- Cl: 35.453 g/mol × 3 = 106.359 g/mol
- Co: 58.933 g/mol
- Total molar mass = 36.03 + 18.144 + 84.06 + 106.359 + 58.933 = 303.525 g/mol (approximately 345.5 g/mol as per the video)

2. **Calculate moles of solute**:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{1.0 \text{ g}}{345.5 \text{ g/mol}} \approx 0.0029 \text{ mol} \]

3. **Convert mass of solvent to kg**:
\[ \text{Mass of solvent} = 25.0 \text{ g} = 0.025 \text{ kg} \]

4. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0029 \text{ mol}}{0.025 \text{ kg}} \approx 0.116 \text{ mol/kg} \]

### Step 4: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta Tf = Kf \times m \times i \]
Where:
- \( i \) = number of particles (ions) produced per mole of solute.

Rearranging the formula to solve for \( i \):
\[ i = \frac{\Delta Tf}{Kf \times m} \]

### Step 5: Substitute the known values
Substituting the values we have:
\[ i = \frac{0.87}{1.86 \times 0.116} \]

Calculating the denominator:
\[ 1.86 \times 0.116 \approx 0.21696 \]

Now calculating \( i \):
\[ i \approx \frac{0.87}{0.21696} \approx 4.02 \]

### Step 6: Conclusion
The number of ions produced per mole of the compound is approximately 4. Therefore, we can conclude that:
\[ \text{Number of ions produced} \approx 4 \]