Osmotic Pressure Calculator

Q: What value of R is used?

R = 0.082057 L·atm·mol⁻¹·K⁻¹. We convert input pressures to atm before applying Π = iMRT.

Q: How do I choose the van’t Hoff factor i?

For nonelectrolytes like glucose use i≈1; NaCl≈2; CaCl₂≈3. Real solutions may have smaller effective i due to ion pairing or incomplete dissociation.

Q: Can I enter temperature in °C?

Yes. We convert to Kelvin internally (T_K = T_°C + 273.15).

Compute the osmotic pressure Π of a solution using the van’t Hoff equation. Supports common solutes (quick picks), temperature in °C or K, and reports Π in atm with handy conversions to kPa and mmHg. Includes step-by-step working for students.

Background

Osmotic pressure is a colligative property—it depends on the number of dissolved particles, not their identity. For dilute solutions, the van’t Hoff relation Π = i M R T approximates osmotic pressure well, where i is the van’t Hoff factor (effective particle count), M is molarity, R is the gas constant, and T is absolute temperature (K).

Enter values or use a quick pick:

Solve for:

Osmotic pressure Π

Molarity M

van’t Hoff factor i

Molarity (M, mol/L):

van’t Hoff factor (i):

Temperature:

Units: °C K

Osmotic pressure (Π):

Units: atm kPa mmHg

Quick picks:

Result:

No results yet. Choose what to solve for and enter values.

How to use this calculator

Solve Π: Enter i, M, and T (°C or K). We compute Π and convert to kPa & mmHg.
Solve M: Enter Π (atm/kPa/mmHg), i, and T. We compute M = Π/(iRT).
Solve i: Enter Π, M, and T. We compute i = Π/(MRT).
Typical i: glucose ≈ 1; NaCl ≈ 2; CaCl₂ ≈ 3 (real values can be lower due to non-ideality).

Assumes ideal dilute solutions and 25 °C reference for common approximations. For strong electrolytes, i may be slightly < theoretical due to ion pairing.

Formula & Equation Used

Van’t Hoff relation for osmotic pressure:

Π = i M R T

Where:

Π: osmotic pressure
i: van’t Hoff factor (effective particles per formula unit)
M: molarity (mol·L⁻¹)
R: 0.082057 L·atm·mol⁻¹·K⁻¹
T: absolute temperature (K)

Rearrangements:

M = \frac{Π}{i R T}

i = \frac{Π}{M R T}

Example Problems & Step-by-Step Solutions

Example 1 (Solve Π)

0.150 M NaCl (i ≈ 2) at 25 °C: Π = i M R T = 2 × 0.150 × 0.082057 × 298.15 ≈ 7.33 atm.

Example 2 (Solve M)

Π = 4.00 atm, i = 2, T = 25 °C ⇒ M = Π/(iRT) ≈ 4.00 / (2 × 0.082057 × 298.15) ≈ 0.0816 M.

Example 3 (Solve i)

Π = 2.50 atm, M = 0.100 M, T = 25 °C ⇒ i = Π/(MRT) ≈ 2.50 / (0.100 × 0.082057 × 298.15) ≈ 1.02.

Frequently Asked Questions

Q: What value of R is used?

We use R = 0.082057 L·atm·mol⁻¹·K⁻¹. For input Π in kPa or mmHg, we convert to atm before using the formula.

Q: How do I choose i?

For nonelectrolytes (glucose, urea), use i ≈ 1. For NaCl use ≈ 2; CaCl₂ ≈ 3. Real solutions may have smaller effective i due to ion pairing or incomplete dissociation.

Q: Can I enter temperature in °C?

Yes—enter °C and we convert to K internally (T_K = T_°C + 273.15).