Question

Which solution has the higher boiling point: 28.8 g C2H6O2 in 0.500 kg of H2O or 24.0 g NaCl in 0.500 kg of H2O?

Answer 1

elevation in boiling point ,
delta Tb = i*kb*mb

28.8 g C2H6O2 in 0.500 kg of H2O
moles of C2H6O2 = mass/molar mass
= 28.8 / (2*12 + 6*1 + 2*16)
=0.4645 mol
mb = 0.4645 mol / 0.500 kg = 0.929
i = 1
delta Tb = 0.929*kb

or 24.0 g NaCl in 0.500 kg of H2O
moles of NaCl = mass/molar mass
= 24.0 / (23 + 35.5)
=0.4103 mol
mb = 0.4103 mol / 0.500 kg = 0.8206
i = 2
delta Tb = 1.64*kb

clealt delta Tb is more for 2ndcase

so, higher boiling point for
24.0 g NaCl in 0.500 kg of H2O