Chemistry Reference
In-Depth Information
Solution
By examining the Huggins equation, one finds that only two flow times corresponding to two
concentrations,
C
1
and
C
2
, are needed to obtain [
η
]:
1
(
x
1
, y
1
)
1
C
t
0
— 1
(
x
1/2
, y
1/2
)
(
x
0
,
y
0
)
C
1
= 2
C
1/2
[η]
C
1/2
C
1
C
By collinearity,
y
1
2y
0
2x
0
5
y
1
2y
0
2
x
1
x
1
2x
0
2
2 ½
t
1
1
C
1
t
1
t
0
21
1
C
1
2
t
0
21
2 ½
2
i:e:;
5
C
1
20
C
1
20
2
2C
1
C
1
C
t
2C
1
½5
C
t
1
t
0
21
1
1
2
t
0
21
2
½
2
C
1
2
1
2
2
C
t
2 t
0
t
0
1
t
1
2 t
0
t
0
1
½5
2
2
2
C
1
1
2
2
½5
2 t
1
14t
1
23t
0
2
C
1
t
0
Taking the average flow times from experimental data,
t
0
573:4s;
t
1
5229:1 s;
t
1
5137:3 s
2
½η 5
2
229
:
1
1
4
ð
137
:
3
Þ 2
3
ð
73
:
4
Þ
73
:
4
3
0
:
7g
=
100
mL
5
194 mL
=
g
From
Table 3.1
, a50.73 and K50.017 at 25
C.
i.e.,
0:73
194
5
0
:
017
ðM
v
Þ
0
@
1
A
5
194
0:017
ln
0
:
73ln
ðM
v
Þ
M
v
D
361000
