Suatu larutan glukosa (Mr =180) mengandung 20% glukosa ,berapahkah kenaikan titik didih larutan tersebut?(Ar H=1; O=16; C=12; KB air =0,52 c/molal?

dik :
20 % glukosa arutan glukosa = 20 g glukosa / 100 ml air
mr glukosa = 180
kb = 0,52 °c / molal
p = V air × massa jenis air = 100 ml × 1 gram/ ml = 100 gram
 
dit :
ΔTB ?

jawab :

ΔTb = kb × g/mr × 1000/p
= 0,52 °c/molal × 20 g/180 × 1000/100 g
=0,58 °c