What is the osmotic pressure of a solution formed by dissolving 44.3 mg of aspirin (C9H8O4) in 0.358 L of water at 25 ∘C?

Osmotic pressure is 0.014 atm

Explanation:

Formula for osmotic pressure is π = M. R. T

where π is the osmotic pressure (atm)

M → Molarity (mol/L)

R → Universal constant of ideal gases (0.082 L. atm/mol. K)

T → Absolute temperature (T°C + 273)

With the data given, let's determine the molarity of aspirin in the solution

First of all, we convert the mass to g

44.3 mg. 1 g / 1000 mg = 0.0443 g

Afterwards we convert the mass (g) in moles

Aspirin molar mass = 216 g/mol

0.0443g. 1 mol / 216 g = 2.05*10⁻⁴ mol

Molarity is mol/L so → 2.05*10⁻⁴ mol / 0.358 L = 5.73*10⁻⁴M

T° → 25°C + 273 = 298K

Let's replace the data in the formula

π = 5.73*10⁻⁴M. 0.082 L. atm/mol. K. 298K

π = 0.014 atm