What is the boiling point of a solution of ethanol that contains 4.9 grams of vanillin in 47.0 grams of ethanol?
Molar Mass(vanillin) = 152.14 g/mol; K_b(ethanol) = 1.22 ° C/ m ; bp(ethanol) = 78.5 ° C.

Question

Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔT_b = i * K_b * m, where i is the van't Hoff factor (1 for non-electrolytes like vanillin), K_b is the ebullioscopic constant, and m is the molality. First, calculate the molality: moles of vanillin = 4.9 g / 152.14 g/mol = 0.0322 mol; molality = 0.0322 mol / 0.047 kg = 0.685 m. Then, ΔT_b = 1 * 1.22 °C/m * 0.685 m = 0.8357 °C. The boiling point of the solution is 78.5 °C + 0.8357 °C = 79.3 °C.

What Is the Boiling Point of a Solution of Ethanol