The elevation in boilng point of a solution of 13.44 g of `CuCl_(2) ` 1 kg of water will be _____.
(Molecular mass of `CuCl_(2) = 134.4 and K_(b) =0.52 km^(-1))`

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 of a solution of 13.44 g of CuCl₂ in 1 kg of water, we will follow these steps:

### Step 1: Identify the given values
- Mass of solute (CuCl₂) = 13.44 g
- Molecular mass of CuCl₂ = 134.4 g/mol
- Mass of solvent (water) = 1 kg
- Boiling point elevation constant (K_b) = 0.52 °C kg/mol

### Step 2: Calculate the number of moles of the solute (CuCl₂)
To find the number of moles of CuCl₂, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molecular mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of CuCl₂} = \frac{13.44 \, \text{g}}{134.4 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ \text{Molality} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

### Step 4: Calculate the elevation in boiling point (ΔT_b)
The elevation in boiling point can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Substituting the values:
\[ \Delta T_b = 0.52 \, \text{°C kg/mol} \times 0.1 \, \text{mol/kg} = 0.052 \, \text{°C} \]

### Final Answer
The elevation in boiling point of the solution is **0.052 °C**.

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