Ke dalam 300 gram air dilarutkan10gram glukosa(Mr 180).jika Kb=0,52°c/m. tentukan titik didih larutan benzenee

wp = 300 gram
wt = 10 gram
Mr = 180
Kb = 0,52
Tb = ...................?

ΔTb = wt/Mr x 1000/wp x Kb
= 10/180 x 1000/300 x 0,52
= 0,056 x 3,33 x 0,52
= 0,097°C

Tb = Tb° + ΔTb
= 100 + 0,097
= 100,097°C