Diketahui 500 gr air dilarutkan glukosa sebanyak 36 gr mr=180 kb air=0,52°c.hitunglah titik didih larutan glukosa

Dik : p : 500 gr
gr : 36 gr
mr : 180
kb air : 0,52
Dit : ∆tb..?
Hit :
∆tb = kb x m
∆tb = 0,52 x gr/mr x 1000/p
∆tb = 0,52 x 36/180 x 1000/500
∆tb = 0,52 x 0,2 x 2
∆tb = 0,208°c/m

Semoga membantu

massa air/p:500gr
mr glukosa:180
massa gula:36gr
kb air:0,52°C
ΔTb=(g/mr).(1000/p).kb
ΔTb=(36/180).(1000/500).0,52
ΔTb=0,6552°C

Tb larutan=Tb pelarut+ΔTb
Tb larutan=100+0,6552°C
Tb larutan=100,6552°C