Sebanyak 3,42 gram glukosa, C6H12O6 dilarutkan kedalam 150 mL air ( Ar C=12, H=1, O=16 ) dan massa jenis air 1 gram/mL

Jawaban:

Mr Glukosa = 180 gram/mol

Massa air (pelarut) = massa jenis × volume

Massa air (pelarut) = 1 gram / ml × 150 ml = 150 gram

m = (massa / Mr) × (1000 / massa pelarut)

m = (3,42 / 180) × (1000 / 150)

m = 0,1267 molal