The elevation in boiling point of a solution of `13.44`g of `CuCl_(2)`(molecular weight =`134.4,k_(b)=0.52K "molality"^(-1))` in `1` kg water using the following information will be:

Options

A. 0.16
B. 0.05
C. 0.1
D. 0.2

Explanation

To find the elevation in boiling point of a solution of `CuCl2`, we can follow these steps:

### Step 1: Calculate the number of moles of `CuCl2`
We start with the mass of `CuCl2` given as `13.44 g`. The molecular weight of `CuCl2` is `134.4 g/mol`.

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

### Step 2: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, we have `1 kg` of water as the solvent.

\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

### Step 3: Determine the van 't Hoff factor (i)
`CuCl2` dissociates in solution into `1 Cu^2+` ion and `2 Cl^-` ions. Therefore, the total number of ions produced is:

\[ i = 1 + 2 = 3 \]

### Step 4: Use the formula for elevation in boiling point
The formula for the elevation in boiling point is given by:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \( K_b = 0.52 \, \text{K kg/mol} \)
- \( m = 0.1 \, \text{mol/kg} \)
- \( i = 3 \)

Substituting the values:

\[ \Delta T_b = 3 \cdot 0.52 \, \text{K kg/mol} \cdot 0.1 \, \text{mol/kg} \]

\[ \Delta T_b = 3 \cdot 0.052 = 0.156 \, \text{K} \]

### Step 5: Conclusion
The elevation in boiling point of the solution is `0.156 K`, which can be approximated to `0.16 K`.

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