Answer to Question #320104 in General Chemistry for Thel

Question

A mass glucose (C6H12O6) is 43.8g is dissolved into 359.9g of water. If the molal boiling point elevation of water is 0.51° C/m, what will be the change in the boiling point of water?

Answer 1

∆T = imK

∆T = change in temperature

i = van't Hoff factor = 1 for glucose because it does not ionize or dissociate

m = molality = moles of solute/mass of solvent in kg

Number of moles of glucose = mass/ MM = 43.8/120 = 0.243

m = 0.243mol / 0.3599kg = 0.676 m

K = boiling constant = 0.512

∆T = 1*0.676 * 0.512 = 0.35