tentukan fraksi mol zat terlarut dan pelarut jika kedalam 90 gram air dilarurkan 15 gram asam cuka (CH3COOH)!

Penyelesaian :Mr CH3COOH = 60a g/molMr H2O = 18 g/mol
nCH3COOH = 15 g / 60 g/mol = 0,25 mol = ntnH2O = 90 g / 18 g/mol = 5 mol = npXCH3COOH  = nt/(nt  + np) = 0,25/(0,25 + 5) = 0,05XH2O ­= 1 – 0,05 = 0,95