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14 March, 08:20

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?

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  1. 14 March, 09:07
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    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
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