1. Titik beku larutan 2 gram suatu zat non elektrolit dalam 500 gram air adalah -0,12 ° . Tentukan Mr zat non elekrtrolit, jika Kb air = 1,86 ° /mol. Jawab: Diketahui : Massa larutan = 2 gram Massa pelarut = 500 gram TbL = -0,12 ° Kb = 1,86 ° Ditanya : Mr zat = . . . .? Pembahasan : ∆Tb = TbP – TbP – TbL TbL = 0 – 0 – ( (- 0,12 ° ) = 0,12 ° ∆Tb = m × Kb 1000 × × Kb 2 1000 × × 1,86 0,12 0C = 500 7,44 .°/ 0
=
1,86 C =
°
Mr
7,44 °/ = 1,86 °
Mr
= 62
⁄
2. Titik beku larutan 0,1 M NaCl dalam air adalah -0,36° . Berapakah titik beku larutan Kalsium Klorida (CaCl2) 0,05 M dalam air? Jawab : Diketahui : TbL = -0,36 ° M NaCl = 0,1 M M CaCl2 = 0,05 M + NaCl → Na + Cli = 1 + (n - 1) α = 1 + (2 – (2 – 1) 1) 1 =2 CaCl2 → Ca2+ + 2Cli = 1 + (n - 1) α = 1 + (3 – (3 – 1) 1) 1 =3 Ditanya : TbL CaCl2 = . . . . ? Pembahasan : ∆Tb = TbP – TbP – TbL TbL ∆Tb = 0 – 0 – (-0,36 (-0,36° ) = 0,36 °
Titles you can't find anywhere else
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Titles you can't find anywhere else
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Kb NaCl
=
Kb CaCl2
∆Tb ×i
=
∆Tb ×i
0,36 ° 0,1 ×2
=
∆Tb 0,05 ×3
0,36 ° 0,2
=
∆Tb
0,054 ° 0,2
= =
∆Tb
∆Tb 0,15
0,27°
= TbP – TbP – TbL TbL
0,27° = 0 – TbL TbL TbL
= -0,27°
3. Salah satu MgCl2 dilarutkan dalam 500 gram air pada 25° . Tentukan tekanan osmotic larutan jika derajat ionisasi = 0,9! (Mg = 24, Cl = 35,5) Jawab : Diketahui : Massa MgCl2 = 1 gram Massa Pelarut = 500 gram T = 25 ° + 273 = 298 K α = 0,9 Ditanya : π = …. ? Pembahasan : n MgCl2
=
n MgCl2
=
n MgCl2
=
Molaritas MgCl2
1 95 /
0,011 mol = = =
MgCl2 → Mg2+ + 2Cl-
MgCl2 1000 × 1000 0,011 mol × 500
0,022 M (n = 2 + 1 = 3)
Titles you can't find anywhere else
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
=
1.147,6 mmHg
