Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 7.3
Find the osmolarity and osmotic pressure of a 0.5% by weight solution of glucose at room
temperature.
Solution
A 0.5% solution of glucose equals
g
mL
¼
5
100
0
:
5
g
L
. Since the molecular weight of glucose is
5
L
180
M
¼
g
mol
, we have a molarity of
0278
M
180
L
. Since glucose does not dissociate into separate
ions in solution, the osmolarity is equal to the molarity of glucose, giving 27
0
:
8
mOsm
L
:
. The
osmotic pressure is therefore
p ¼
cRT
¼
0
:
0278
62
:
367
310
¼
537
:
5 mmHg
:
EXAMPLE PROBLEM 7.4
Consider a cell with an internal osmolarity of 300
mOsm
L
and volume of 2 nL, in a 30 nL
solution of 300
mOsm
L
. A 5 nL, 2%
by weight solution is added to the extracellular space.
NaCl
Assuming that
is impermeable and that the moles inside the cell do not change, describe
the events that take place until steady state is achieved. What is the volume of the cell at steady
state?
NaCl
Solution
The first step in the solution is to determine the number of osmoles inside and outside the cell
0
. Inside the cell, we have 300
mOsm
L
10
9
mOsm. Outside the cell, we have
at
t
¼
2nL
¼
600
300
mOsm
L
10
9
mOsm
:
Next, determine the number of osmoles outside the cell at
30 nL
¼
9, 000
0
þ
. The 5 nL, 2% solution of
t
¼
NaCl
2
100
g
mL
¼
20
g
g
M
,
added to the solution outside the cell has
L
, and with a molecular weight of 58
:
5
20
L
58
342
M
10
9
10
9
moles of
has a molarity of
5
M
¼
0
:
L
:
In 5 nL, there are 0.342
5
¼
1.71
NaCl
.
:
10
9
mOsm are added to the solution outside the cell,
giving a total of 12,420 mOsm. The outside osmolarity is then
12420
35
Since
NaCl
separates into two ions, 3,420
355
mOsm
L
¼
. This difference
in osmolarity causes an osmotic pressure of p ¼
D
cRT
¼
(355
300)
62.367
310
¼
1,063 mmHg,
driving water out of the cell at room temperature.
At steady state, the total inside and outside cell volume is 37 nL, and total inside and outside
10
9
mOsm. The total osmolarity then is
13420
37
352
mOsm
L
osmoles equal 13,420
¼
, which is the
osmolarity inside and outside the cell. The final volume of the cell is therefore
600
352
¼
1
:
7nL
:
