1 gram MgCl2 dilarutkan dalam 500 gram air.tentukanlah: a.titik didih larutan b.titik beku larutan c.tekanan osmotik larutan pada 25*C.Jika derajat ionisasi 0,9

dik: gr= 1 gram
p=500gram.
1g/mL maka V= 500mL=0,5L
kb=0,2 C/m
kf=1,86 C/m
T=25+273=298K
mr=95
dit:a.Tb
b.Tf
c.π
jawab
a.delta Tb=kb×gr/mr×1000/p×i
=0,2×1/95×1000/500×3
=0,2×0,01×2×3
=0,012
Tb=100+0,012
=100,012 celcius
b.delta Tf=kf×gr/mr×1000/p×i
=1,86×1/95×1000/500×3
=1,86×0,01×2×3
=0,1116
Tf=0-0,1116
=-0,1116 celcius
c.π= M.R.T. i
=( (1/95)/0,5) x 0.082 x 298 x3
=0,02×24,436×3
=1,46616

a. MgCl2 –› Mg+ + 2Cl-
n = 3
i = 1 + (n-1)
i = 1 + (3-1) = 3
∆Tb = massa/mr . 1000/p . Kb . i
∆Tb = 1/95 . 1000/500 . 0,52 . 3
∆Tb = 0,033 °C
Tblar = Tbpel + ∆Tb
Tblar = (100 + 0,033) °C = 100,033 °C
b. ∆Tf = massa/mr . 1000/p . Kf . i
∆Tf = 1/95 . 1000/500 . 1,86 . 3
∆Tf = 0,117 °C
Tflar = Tfpel - ∆Tf
Tflar = (0 - 0,117) °C = -0,117 °C
c. i = 1 + (n-1)a
i = 1 + (3-1)0,9
i = 2,8
π = massa/mr . 1000/v . R . T . i
π = 1/95 . 1000/500 . 0,082 . 298 . 2,8
π = 1,44 atm