fraksi mol larutan 36 gram glukosa (Mr=180) dalam 90 gram air (Mr=18)adalah
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Pertanyaan
fraksi mol larutan 36 gram glukosa (Mr=180) dalam 90 gram air (Mr=18)adalah
1 Jawaban
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1. Jawaban AiniLavoisier
wt = 36 (Mr = 180)
wp = 90 (Mr = 18)
Xt = ..........................?
Xp = ..........................?
nt = wt/Mr = 36/180 = 0,2 mol
np = wp/Mr = 90/18 = 5 mol
Xt = nt/(nt+np) = 0,2/(0,2+5) = 0,2/5,2 = 0,038
Xp = 1 - Xt = 1 - 0,038 = 0,962