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A compound is found to have a molar mass of 598 g/mol. If 35.8 mg of the compound is dissolved in enough water to make 175 ml of solution at 25°C, what is the osmotic pressure of the resulting solution?

Answer :

The osmotic pressure of the solution is 0.0082atm

Data;

  • Molar mass = 598 g/mol
  • mass = 35.8mg
  • v = 175mL
  • T = 25°C = 298.15K
  • R = 0.0821 J/ mol K

Osmotic Pressure

The osmotic pressure of a solution can be calculated as

πv = nRT

Let's calculate the number of moles

[tex]n = \frac{mass}{molar mass}\\n = \frac{35.8*10^-^3}{598}\\ n = 0.00005986 moles[/tex]

Substitute the values and solve

[tex]\pi v = nRT\\\pi = \frac{nRT}{V}\\ \pi = \frac{0.00005986*0.0821*298.15}{175*10^-^3}\\ \pi = 0.0082atm[/tex]

The osmotic pressure of the solution is calculated as 0.0082 atm

Learn more on osmotic pressure here;

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Rewritten by : Barada

The computation for molarity is:
(x) (0.175 L) = 0.0358 g / 598 g/mol
x = 0.000342093 M
Whereas the osmotic pressure calculation:
pi = iMRT
pi = (1) (0.000342093 mol/L) (0.08206 L atm / mol K) (298 K)
pi = 0.0083655 atm
Converting the answer to torr, will give us:
0.0083655 atm times (760 torr/atm) = 6.35778 torr
which rounds off to 6.36 torr