Didalam 400 ml larutan terlarut 3,4 gram NH3(Ar N=14 , H=1) hitunglah molaritas larutan dan fraksi mol

Diketahui =

massa pelarut = 400 ml = 400 gram 
massa terlarut  = NH3 = 3,4 gram 
Mr NH3 = 17 

penyelesaian 

fraksi mol 

nt = 3,4 / 17 = 0,2 mol
np = 400/17 = 23,5 mol

xt = 0,2/0,2+23,5 =0,008
xp = 23,5/ 0,2+23,5 =0,991 

1. Molaritas
M = gr/Mr * 1000/400
Μ = 3,4/17 * 1000/400
Molaritas = 0,5 M

2. Fraksi mol
mol :
NH3 = nt = 3,4/17 = 0,2
H2O = np = (400-3,4)/18 = 396,6/18 = 22,033

Xt = nt/(nt+np)
→ 0,2/(0,2+22,033)
→ 0,2/22,233
→ 0,899 = 0,9