suatu larutan mengndung 18g glukosa (mr =180) dlm 500g air. tntukan pnurunan titik bwku larutan

Dik = massa C6H12O6(glukosa) = 18 gr
Mr = 180
massa H2O = 500 gr
kf H2O = 1,86 °C/m (ketetapan)
m = gr/Mr × 1000/p
= 18/180 × 1000/500
= 1/5
Dit = ∆Tf = ?
Jb:
∆Tf = m . kf
= 1/5 × 1,86
= 0,372 °C