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

Question

Anton [14]

3 years ago

15

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

Chemistry

1 answer:

likoan [24] · Answer 1 · 2022-05-12T11:08:03+03:00

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

Tb larutan = ........? 
--------------------------------------... 
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