Jika ke dalam 180 gram air dilarutkan 30 gram asam cuka (CH3COOH),fraksi mol zat terlarut dan pelarut berturut-turut adalah….m(Ar:H=1,C=12,O=16)
Jawaban Terkonfirmasi
Nt (cuka) = w/Mr = 30/60 = 0,5 mol
np (air) = w/Mr = 180/18 = 10 mol
Xt (fraksi mol cuka) = nt/(nt+np) = 0,5/(0,5+10) = 0,5/10,5 = 0,05
Xp (fraksi mol air) = 1-Xt = 1-0,05 = 0,95
Dik.
mp = 180 gr
mt = 30 gr
mr air = 18
mr asam cuka = 60
np = gr/mr
= 180/18 =10
nt = gr/mr
= 30/60 = 0,5
Xt = nt/nt + np
= 0.5/ 0.5 + 10 = 0,0476190476 => 0.05
xt+xp= 1
0.0476190476 + xp = 1
xp = 1 – 0.0476190476
xp = 0.9523809524 => 0.95
jadi Xt = 0.05
Xp = 0.95