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

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

No is the answer to the question

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