sebanyak 6 gram urea (Mr=60) dilarutkan dalam 500 gram air (Kf air= 1,86 C /m). Tentukan penurunan titik bekunya

Dik :
m urea = 6g
Mr urea= 60g
p =500g
Kf air = 1,86 C/m

Dit : Tf?

Jwb :
cari dulu n urea : m/Mr = 6/60= 0,1 mol
delta
∆Tf = Kf (n × 1000/500)
∆Tf = 1,86 (0,1 ×1000/500)
∆Tf = 1,86 (0,2) = 0,372°C

Tf = 0° - 0,372° = -0,372°C