SME
Cell & Molecular Biology · Problem type
Osmotic pressure
Use this for “What osmotic pressure does 0.15 M NaCl have at body temperature?”
Diagram
Solute 0 M outside, 0.3 M inside: the outside is hypotonic, so water moves into the cell through the aquaporin, toward more solute. i × C = 2 × 0.15 M = 0.3 M of particles.
One dot ≈ 0.025 M.
This is an example. Type your own values to start a new problem.
π (atm)
=
i
×
C (M)
×
R
×
T (K)
Formulas
π
=
i
×
C
×
0.08206
×
T
7.632 = 2 × 0.15 × 0.08206 × 310
Assumptions
•
π = iCRT with R = 0.08206 L·atm/(mol·K); body temperature is 310 K (37 °C).
•
i counts the particles each unit dissolves into: glucose 1, NaCl 2, CaCl₂ 3.
•
The membrane passes water but not the solute; π is the pressure that stops water flowing in.
•
The solution is dilute enough to behave ideally.
Step-by-step
WE KNOW
C = 0.15 M, i = 2, T = 310 K
FIND
osmotic pressure (π)
Step 1 · Find osmotic pressure (π)
Use π = i × C × 0.08206 × T
Dissolved particles pull water across: iC is the concentration of particles, times RT.
π
=
2
×
0.15
×
0.08206
×
310
π
=
0.3
×
0.08206
×
310
π
=
0.02462
×
310
π
=
7.632
atm
Check
Put the numbers back in. Both sides should match.
7.632 = 2 × 0.15 × 0.08206 × 310 ✓
Topic check
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