Question

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 1

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

Answer 2

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

$n = \frac{mass}{molar mass}\\n = \frac{35.8*10^-^3}{598}\\ n = 0.00005986 moles$

Substitute the values and solve

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

The osmotic pressure of the solution is calculated as 0.0082 atm

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