The elevation in boiling point, when `13.44 g` of freshly prepared `CuCI_(2)` are added to one kilogram of water, is [Some useful data, `K_(b) (H_(2)O) = 0.52 kg K mol^(-1), "mol.wt of" CuCI_(2) = 134.4 gm]`

Options

A. 0.05
B. `0.10`
C. `0.16`
D. `0.20`

Answer

To solve the problem of finding the elevation in boiling point when 13.44 g of freshly prepared CuCl₂ is added to one kilogram of water, we will follow these steps:

### Step 1: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Find the number of moles of CuCl₂:**
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass of CuCl₂ = 13.44 g
- Molar mass of CuCl₂ = 134.4 g/mol

\[ \text{Number of moles of CuCl₂} = \frac{13.44 \, \text{g}}{134.4 \, \text{g/mol}} = 0.1 \, \text{mol} \]

2. **Calculate molality:**
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} \]
Given:
- Mass of solvent (water) = 1 kg

\[ \text{Molality} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

### Step 2: Determine the van 't Hoff factor (i)
The van 't Hoff factor (i) represents the number of particles the solute dissociates into in solution.

For CuCl₂:
- It dissociates into 1 Cu²⁺ ion and 2 Cl⁻ ions.
- Therefore, the total number of ions = 1 + 2 = 3.

Thus, \( i = 3 \).

### Step 3: Use the formula for elevation in boiling point
The elevation in boiling point (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( K_b \) for water = 0.52 kg K/mol
- \( m \) = 0.1 mol/kg
- \( i \) = 3

Substituting the values:
\[ \Delta T_b = 3 \cdot 0.52 \, \text{kg K/mol} \cdot 0.1 \, \text{mol/kg} \]
\[ \Delta T_b = 3 \cdot 0.052 = 0.156 \, \text{K} \]

### Step 4: Final result
The elevation in boiling point when 13.44 g of CuCl₂ is added to one kilogram of water is approximately \( 0.16 \, \text{K} \).

### Final Answer
The elevation in boiling point is \( 0.16 \, \text{K} \).

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