Question

Sebanyak 4,16 gram HCL dilarutkan dalam 500 gram air jika Mr HCL 36,5 KB air = 0,52 derajat celcius/m nilai KF air = 1,86 derajat Celcius/m dan derajat ilimisasi 0, 9 Tentukan a.titik didih larutan b. titik beku larutan

Answer 1

Massa HCl = 4,16
Mr HCl = 36,5
massa air = 500 gr = 0,5 kg
Kb = 0,52
Kf = 1,86
α = 0,9
Tb, Tf = ?

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kita cari dulu faktor van't Hoff (i) dan molalitas (m) larutan.

HCl --> H+ + Cl-            (menghasilkan 2 ion, n = 2)

i = 1 + α(n - 1)
i = 1 + 0,9(2 - 1)
i = 1 + 0,9
i = 1,9

m HCl = (massa HCl / Mr) / massa air (kg)
m HCl = (4,16 gr / 36,5) / 0,5 kg
m HCl ≈ 0,228 m

(a) mencari Tb

ΔTb = m x Kb x i
ΔTb = 0,228 x 0,52 x 1,9
ΔTb = 0,225

Tb = Tb° + ΔTb
Tb = 100 + 0,225
Tb = 100,225°C

(b) mencari Tf
ΔTf = m x Kf x i
ΔTf = 0,228 x 1,86 x 1,9
ΔTf ≈ 0,806

Tf = Tf° - ΔTf
Tf = 0 - 0,806
Tf = -0,806°C