larutan NaOH 4% (Mr=40) akan membeku pada suhu... C. Kf air = 1,86 C/molal

Larutan NaOH 4% (Mr = 40)
Kf air = 1,86°C/m

ditanya = Tf larutan ... ?

Larutan NaOH 4%. Misal massa larutan 100 gram, maka massa NaOH-nya adalah 4 gram. Sedangkan massa pelarutnya (air) = 100 - 4 = 96 gram.

*molalitas larutan
m = massa NaOH/Mr NaOH . 1000/massa pelarut
m = 4/40 . 1000/96
m = 1,0416 molal

* NaOH => Na^+ + OH^-
n = 2

*NaOH terionisasi sempurna, jadi α = 1.

*Penurunan titik beku
∆Tf = Kf . m . i
∆Tf = 1,86 . 1,0416 . (1 + (n-1) α)
∆Tf = 1,86 . 1,0416 . (1 + (2-1) 1)
∆Tf = 1,86 . 1,0416 . (1 + 1)
∆Tf = 1,86 . 1,0416 . 2
∆Tf = 3,88°C

*Titik beku larutan
∆Tf = Tf pelarut - Tf larutan
Tf larutan = Tf pelarut - ∆Tf
Tf larutan = 0 - 3,88
Tf larutan = -3,88°C