What would be the boiling point of a 0.52 m solution of KBr in water (K_b= 0.52 $\circ$ C)? A)99.7 $\circ$ C B)100.0 $\circ$ C C)100.3 $\circ$ C D)100.5 $\circ$ C E)101.1 $\circ$ C

Question

Quizplus · Accepted Answer

The boiling point elevation formula is ΔT = i * Kb * m. For KBr, i = 2 (since it dissociates into K+ and Br-). ΔT = 2 * 0.52 * 0.52 = 0.54°C. The boiling point of pure water is 100°C, so the solution's boiling point is 100 + 0.54 = 100.54°C, closest to 100.5°C.

What Would Be the Boiling Point of a 0 ∘\circ∘ C)? A)99

What Would Be the Boiling Point of a 0 $\circ$ C)?
A)99