The elevation in boiling point for 13.44 g of CuCl₂ dissolved in 1 kg of water as solvent will be (K_b = 0.52 kg/J, molar mass of CuCl₂ = 134.4 g/mol)

0.16

To calculate the elevation in boiling point caused by the addition of CuCl₂ to water, we can use the formula:

\(\Delta T_b = K_b \cdot m \cdot i\)

where:
\(\Delta T_b\) is the elevation in boiling point,
\(K_b\) is the ebullioscopic constant of the solvent (water),
\(m\) is the molality of the solute (CuCl₂), and
\(i\) is the Van't Hoff factor, which represents the number of particles formed when the solute dissolves.

First, we need to calculate the molality of CuCl₂. Molality (\(m\)) is defined as the moles of solute per kilogram of solvent. Since we have 13.44 g of CuCl₂ and 1 kg of water, we need to convert the mass of CuCl₂ to moles:

\(\text{moles of }CuCl_2 = \frac{{\text{mass of }CuCl_2}}{{\text{molar mass of }CuCl_2}}\)
\(\text{moles of }CuCl_2 = \frac{{13.44 \, \text{g}}}{{134.4 \, \text{g/mol}}} = 0.1 \, \text{mol}\)

Now we can calculate the molality:

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

The Van't Hoff factor (\(i\)) for CuCl₂ is 3 because it dissociates into three particles (Cu²⁺ and 2 Cl^-) when it dissolves.

Now we can calculate the elevation in boiling point:

\(\Delta T_b = K_b \cdot m \cdot i = 0.52 \, \text{kg/J} \cdot 0.1 \, \text{mol/kg} \cdot 3 = 0.156 \, \text{K}\)

The elevation in boiling point is 0.156 K, which is approximately equal to 0.16 K. Therefore, the answer is option (3).