Question

sebanyak 36 gram glukosa (C6H12O6,M=180 g/mol) dilarutkan dalam 250 gram air.jika diketahui Kb air= 0,52° kg/mol.tentukan titik didih larutan tersebut

Answer 1

gr terlarut = 36 gr
Mr terlarut = 180
gr pelarut = 250 gr
Kb air = 0,52 °C kg/mol

Tb larutan = ........?
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ΔTb = Kb.m.i
ΔTb = Kb. (gr t / Mr t) . (1000/ gr p) .i
ΔTb = 0,52 x (36/180) x (1000/250) x 1
ΔTb = 0,416 °C

Tb = 100 + ΔTb
Tb = 100 + 0,416
Tb = 100,416 °C