Question

Calculate the boiling point of 20% (mass fraction) glucose solution  (K_b = 0.52).

 

Answer 1

n=m/Mr = 20/180 = 0,111111111 moles.

m=moles of solute/kg of solvent=0,111111111/0.08 = 1,38888889 Molality

∆Tb= Kb x m x i = 0.52 x 1,38888889 x 1 = 0,722222223.

T=100 + 0,722222223 = 100,722222223 °©.