Page 13 - 01 PG Bab 01 Sifat Koligatif.pmd

2. ∆T = m · K
f f Z − Z
0,372 = m · 1,86 0,416 = {( ) + ( )} × × 0,52
m = 0,2 molal
Z +
Z
−
∆T = m · K 0,416 = × 2,08
b b
= 0,2 · 0,52
74,88 = –4,16x + 168,48
= 0,104°C
4,16x = 93,6
Titik didih urea = 100 + 0,104 = 100,104°C
x = 22,5 gram
3. Massa larutan = 100 ml × 1,04 g/ml Massa glukosa = x = 22,5 gram
= 104 gram Massa urea = (27 – x)
= 27 – 22,5
∆T = · · K
f f = 4,5 gram
Z Jadi, perbandingan massa glukosa : urea
0 – (–3,44°C) = · · 1,86°C
− Z = 22,5 : 4,5 = 5 : 1.
= 3,44°C
7.
1.860x = 21.456,6 – 206,4x
2.066,4x = 21.465,6 ∆T = m · K b
b
x = 10,4 gram
= · · K
b

% massa CO(NH ) = × 100% = 10%
2 2 (100,26 – 100) =
· · 0,52

4. ∆T = · · K
]
b b 0,26 =

]
1= · · K b
M = = 60
r
K = 1,22 °C/m Jadi, massa molekul relatif zat itu = 60.
b
5. massa sukrosa = 6,84 gram

M sukrosa = 342 8. a. 0,300 g urea = −
r
–1 –1
R = 0,082 L atm mol K = 5 · 10 mol
–3
V = 2 L −
larutan ]
Molalitas larutan = −
T = 25 + 273 = 298 K ]
larutan
π = M × R × T = 0,5 mol kg –1
∆T = m · K
b b
= × × R × T –1 –1
# = 0,5 mol kg · 0,512°C kg mol

= 0,256°C

= × × 0,082 × 298 Titik didih larutan

= titik didih pelarut murni (H O) + ∆T
= 0,24 atm 2 b
= 100,00°C + 0,256°C
Jadi, tekanan osmotik larutan sukrosa sebesar
0,24 atm. = 100,256°C
b. ∆T = m · K f
f
6. ∆T = (100,416 – 100)°C = 0,416°C –1 –1
b = 0,5 mol kg · 1,86°C kg mol
Misal: massa glukosa = x gram = 0,93°C
massa urea = (27 – x) gram
Titik beku larutan

∆T = { × × K } = titik beku pelarut murni (H O) – ∆T
b b 2 f

= 0,00°C – 0,93°C

= –0,93°C
+ { × × K }
b

12 Sifat Koligatif Larutan