Page 6 - 01 PG Bab 01 Sifat Koligatif.pmd

B. Uraiant! 4. ∆T = 101,3 – 100 = 1,3°C
b

∆T = × × K
b b
1. ∆T = K × ×
b b
1,3 = × × 0,52

= 0,52 × × M = 180
r
Rumus molekul = (CH O)
= 0,42°C 2 n
M = (12 + 2 × 1 + 16) × n
T larutan glukosa = ∆T air + ∆T r
b b b 180 = 30 × n
= (100 + 0,42)°C
n= 6
= 100,42°C
Rumus molekul senyawa tersebut C H O .
6 12 6

2. n = n = = 55
×

air A
2
p
P= x – P 0 5. n (mol pelarut, H O) =
− = 5,56 mol
A
$
= + & · P 0 n (mol terlarut formamid) = '

*<
−

$
t

17,37 = × 18
+ & = mol
'

*<

n = 2
B ∆P = P° – P = x × P°
n = n t
B zat X >°− >
x =
2= t >°

? −

?
M = 60
r =

?
3. ∆T = (100,65 – 100)°C = 0,65°C
b
Misal kadar gula dalam larutan = a% dalam = 1,9 × 10 –2
100 gram larutan:

– Gula = × 100 gram = a gram x =
t +

– Air = (100 – a) gram
Untuk larutan encer, harga n sangat kecil
t

∆T = × × K dibandingkan n . Oleh karena itu, harga n + n
b b p t p
p
dapat dianggap sama dengan n saja sehingga
0,65 = × × 0,52

−
x = .
t
0,4275 = −

42,75 – 0,4275a = a 1,9 × 10 = '

*<
–2
1,4275a = 42,75
a = 29,95

= 30 M formamid = × −
r
Jadi, kadar gula dalam larutan 30%.

=

= 45,45
–1
Berat molekul formamid = 45,45 mol .
Kimia Kelas XII 5