Question

1. fraksi mol suatu larutan urea 1/6 maka kadar urea dlm larutan tsb adalah? (Mr Urea=60)

2. sebanyak2gram NaOH dan 18gram Glukosa (C6H12O6) dilarutkan kedalam 100ml air, (RoAir=1g/cm^3,Kfair=1,86). Titik Beku larutan tsb adalah?

tlg bantu yah pake caranya :3 thx

Answer 1

1. Gunakan rms fraksi mol terlarut
Xt = nt/(nt+np)
1/6=nt/(nt+np)
5nt=np
5.massa ter/Mr = massa pel/Mr
5.massa ter/60 = massa pel/18
massa ter/massa pel = 2/3

jadi % urea = 2/5 x 100% = 40%

2. ΔTf = ΔTf1 + ΔTf2
= m.kb.i + m.kb
= (2/40 x 1000/100 x 1,86 x 2) + (18/180 x 1000/100 x 1,86)
= 1,86 + 1,86
= 3,72
jadi Tf larutan = Tf pel - ΔTf = 0 -3,72 = -3,72