kedalam 250 gram air dilarutkan 17,4 gram K2SO4b .Jika Kb air=0,52 c/m maka kenaikan titik didih larutan tersebut adalah ... (Ar K=39 ,S=32 ,O=16)

Diketahui :
P = 250gr
m = 17,4gr
Kb air = 0,52
Mr K2SO4 = (39×2) + 32 + (16×4) = 174


∆Tb = Kb × molal
∆Tb = Kb × (gr/Mr) × (1000/P)
∆Tb = 0,52 × (17,4/174) × (1000/250)
∆Tb = (52/100) × (1/10) × 4
∆Tb = (52×4) / (100×10)
∆Tb = 208/1000
∆Tb = 0,208°C