ke dalam 1 liter air dilarutkan 60 g urea (Mr=60); 11,7 g NaCl (Mr=111).tentukan titik beku larutan itu.Kf=1,86 c/m

Diket : massa urea : 60
Mr urea : 60
P : 1 Liter = 1000 mL
massa Nacl : 11.7 gr
Mr Nacl = 111
Kf : 1,86
dit : Tf larutan?
jawab :
∆Tf : m. kf
∆Tf = gr/mr × 1000/p × kf
= (60 + 11.7)/60 + 111 × 1000/1000 .1.86
= 0.77
∆Tf = Tf pelarut - Tf larutan
0.77 = 0° - Tf laruran
Tf larutan = - 0.77 °c