Sebanyak 18 gram glukosa C6 H12 O6 dimasukkan kedalam 500 gram air. Jika kalor beku molal air, Kf = 1,8`C/mol, Titik beku larutan tersebut adalah.....`C. (Ar : C = 12, H = 1, O = 16)
Muhammaddavid12
ΔTf = m. Kf = n. 1000/p . Kf = 18/180. 1000/500 . 1,8 = 0,1x2x1,8 =0,36
= n. 1000/p . Kf
= 18/180. 1000/500 . 1,8
= 0,1x2x1,8
=0,36
Tf = Tf° - ΔTf
= 0 - 0,36
= -0,36°C