Question

An aqueous solution containing 12.48 gm of barium chloride in 1.0 kg of water, boils at 373.0832 K. Calculate the degree of dissociation of barium chloride. (\( K_b = 0.52 \) kg mol\(^{-1}\), molar mass of BaCl\(_2\) = 208.34)

Hint:

The van't Hoff factor (\( i \)) accounts for ionization in colligative properties and is used to determine dissociation in solutions.

Answer 1

Step 1: The elevation in boiling point is given by: \[ \Delta T_b = K_b \times m \times i \] where \( m \) is the molality and \( i \) is the van't Hoff factor. 

Step 2: Calculate molality: \[ m = \frac{\text{Mass of solute (g)}}{\text{Molar mass (g/mol)} \times \text{Mass of solvent (kg)}} \] \[ m = \frac{12.48}{208.34 \times 1} = 0.0599 \text{ mol/kg} \] 

Step 3: Calculate van't Hoff factor \( i \): \[ \Delta T_b = 373.0832 - 373 = 0.0832 \] \[ i = \frac{\Delta T_b}{K_b \times m} = \frac{0.0832}{0.52 \times 0.0599} = 2.67 \] 

Step 4: Degree of dissociation \( \alpha \) is calculated as: \[ i = 1 + \alpha(n - 1) \] For BaCl\(_2\), \( n = 3 \): \[ 2.67 = 1 + \alpha(3 - 1) \] \[ \alpha = \frac{2.67 - 1}{2} = 0.835 \] Thus, the degree of dissociation is **0.835**.