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What is the osmotic pressure of a solution formed by dissolving 25 mg of aspirin in 0.25 L of water at 25 degrees Celsius?

1 Answer

5 votes

Answer:

Plug all values in equation (i)

π=1×(0.000568 mol/L)×(0.08206 Latm/molK)×298K

π=0.01389 atm

π=0.0139 atm

Therefore, the osmotic pressure of an aspirin solution is 0.0139 atm.

Step-by-step explanation:

The osmotic pressure of an aspirin solution formed by dissolving 25.0 mg of aspirin (C₉H₈O₄) in 0.250 l of water at 25°C is

The osmotic pressure is calculated as

π=i×M×R×T (i)

Where, i-van't hoff factor

M-Molarity

R=0.08206 Latm/molK

T=25°C

=25+273=298K

For a molecule C₉H₈O₄, the value of i is 1 as it is a non-electrolyte. We need to find out the molarity of the solution which is

Molarity=moles/volume in L

First calculate moles of C₉H₈O₄. The molar mass of C₉H₈O₄ is

Molar mass = [(9×12)+(8×1)+(4×16)] g/mol

=(108+8+64) g/mol

=176 g/mol

moles=25.0 mg×(1 g/1000 mg)×(1 mol/176 g)

moles=(25.0 mol/176×1000)

moles=(25.0 mol/176000)

moles=0.0001420 mol

The molarity of a solution is

Molarity(M)=0.0001420 mol/0.250 L

=0.000568 mol/L

answered
User Geoff Davids
by
8.8k points
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