larutan asam florida (HCL) 18,25% massa mempunyai massa jenis (p/ro=1,1 gr/ml). Tentukan fraksi mol masing2 zat dlm larutan itu.

Asumsikan :
V larutan = 1000 mL

ro = 1,1 gr/mL
% HF = 18,25% m/m

m larutan = ro × V = 1,1 gr/mL × 1000 mL = 1100 gram

m HF = % HF × m larutan
= 18,25% × 1100 gram
= 200,75 gram

Mr HF = Ar H + Ar F = 1 + 19 = 20

n HF = m / Mr = 200,75 gr / 20 = 10,0375 mol

m air = m larutan - m HF = 1100 gr - 200,75 gr = 899,25 gr

Mr H2O = 2 Ar H + Ar O = 2(1) + 16 = 18

n H2O = m / Mr = 899,25 gr / 18 = 49,9583 mol

n tot = n HF + n air = 10,0375 mol + 49,9583 mol = 59,9958 mol

X HF = n HF / n tot = 10,0375 mol / 59,9958 mol = 0,1673

X air = n air / n tot = 49,9583 mol / 59,9958 mol = 0,8327