Kedalam 200 gram air dimasukan 7,45 gram KCl. Jika K_f air = 1,86°C/molal K_b air = 0,52°C/molal . jika Ar K=39, Cl=35,5 tentukanlah : a. Titik didih larutan b

mol KCl = 7.45/74.5 = 0.1 mol
molal KCl = 0.1/0.2 = 0.5 molal
KCl --> K+ + Cl- 
n = 2
i = 1 + (n-1)alpa
i = 1 + (2-1)1
i = 2
a. delta Tb = m. kb. i = 0.5 x 0.52 x 2 = 0.52
maka titik didih larutan = To + delta Tb = 100 + 0.52 = 100.52
b. delta Tf = m . kf. i = 0.5 x 1.86 x 2 = 1.86
maka titik beku larutan = To - delta Tf = 0 - 1.86 = -1.86