The boiling point of a 5.00 m solution of C₆H₄Cl₂ in CCl₄ is 101.6°C. Assume that C₆H₄Cl₂ is not volatile and K_b= 4.95°C/m.

Question

Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔTb = i * Kb * m, where ΔTb is the boiling point elevation, i is the van't Hoff factor (which is 1 for non-electrolytes like the given compound), Kb is the ebullioscopic constant of the solvent (4.95°C/m for water), and m is the molality of the solution (5.00 m). Thus, ΔTb = 1 * 4.95°C/m * 5.00 m = 24.75°C. Since the normal boiling point of water is 100°C, the boiling point of the solution would be 100°C + 24.75°C = 124.75°C, indicating an error in the provided statement.

The Boiling Point of a 5