Boiling Point Chemistry

Question

20.0 g of NaOH was dissolved in 200.0 g of water. Calculate the boiling point of the solution. K_b for water is 0.51 ^oC/m

Answer 1

∆T = imK

∆T = change in boiling point temperature

i = van't Hoff factor = 2 for NaOH since it dissociates into 2 particles (Na+ and OH-)

m = molality = moles NaOH/kg solvent = 20.0 g x 1 mo/40 g = 0.5 mol/0.200 kg = 2.5 m

K = boiling point constant for water = 0.52º/m

∆T = (2)(2.5)(0.52) = 2.6ºC

Boiling point of new solution = 100º + 2.6º = 102.6ºC

Answer 2

ΔT_b = m x K_b

Molar Mass NaOH = 1.008amu + 23amu + 16amu = 40.01 g/mol

20g NaOH x mol NaOH/40.01 g NaOH = .5 mol NaOH

m(molality) = moles solute / kg solvent

m = .5 mol NaOH/ .2kg = 2.5m

T_b = 2.5m x .51ºC/m = 1.3ºC

New boiling point is 101.3ºC