berapa titik didih larutan 6,6 gram amonium sulfat dalam 250 gram larutan jika kb air 0,5 ? (ar amonium sulfat = 132)

∆Tb = Kb × m × i
∆Tb = Kb × ( m (NH4)2SO4 / Mr ) × ( 1000 / m lar ) × i
∆Tb = 0,5°C/molal × (6,6 gr / 132) × (1000/250 gr) × 3
∆Tb = 0,3°C

Tb = Tb° + ∆Tb
= 100°C + 0,3°C
= 100,3°C