The van’t Hoff factor for KCl is i = 1.85. What is the boiling point of a 0.75 m solution of KCl in water? For water, Kb = 0.511°C kg/mol.
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Identify the formula for boiling point elevation: \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van’t Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.
Substitute the given values into the formula: \( i = 1.85 \), \( K_b = 0.511 \text{°C kg/mol} \), and \( m = 0.75 \text{ m} \).
Calculate \( \Delta T_b \) by multiplying the values: \( \Delta T_b = 1.85 \times 0.511 \times 0.75 \).
Determine the new boiling point of the solution by adding the boiling point elevation \( \Delta T_b \) to the normal boiling point of water, which is 100°C.
Conclude by stating that the boiling point of the solution is the sum of 100°C and \( \Delta T_b \).