Question

What would be the boiling point of 500 mL of chloroform if 10.50 grams of caffeine (molar mass = 194.19) was added? The normal boiling point of chloroform is 61.3oC, the density of chloroform is 1.49g/mL, and the Kb of chloroform is 3.63 oC/m.

Answer 1

∆T=iKbm.

n=10.5/194.19=0.054 mol.

∆T= 1 x 3.63 x 0.054/0.7555= 0.26.

61.3+0.26 = 61.56 °C.