Calculate the boiling point of a 4.0 m solution of KI in water? Remember that KI is an electrolyte.The normal boiling point of pure water is 100.0$\degree$C.K_b(water) = 0.52$\degree$C/m A)101.0 $\degree$C B)104.2$\degree$C C)96.0$\degree$C D)102.1$\degree$C E)4.16$\degree$C

Question

Quizplus · Accepted Answer

The boiling point elevation equation is ΔTb = iKb*m, where ΔTb is the change in boiling point, i is the van't Hoff factor (number of particles into which the solute dissociates), Kb is the molal boiling point elevation constant, and m is the molality of the solution.

For KI, i = 2 because it dissociates into K+ and I- ions. Kb for water is 0.52 °C/m. We need to convert 4.0 m to molality by dividing by the molar mass of KI (166 g/mol) to get 0.024 mol/kg.

ΔTb = 2 * 0.52 °C/m * 0.024 mol/kg = 0.025 °C

Therefore, the boiling point of the solution is:

Boiling point = 100 °C + ΔTb = 100.025 °C

When converted to Fahrenheit, the answer is approximately 104.2 °F. Therefore, the best choice is B.

Calculate the Boiling Point of a 4 °\degree° CKb (water) = 0

Calculate the Boiling Point of a 4 $\degree$ CK_b(water) = 0