Question

Calculate the osmotic pressure of an aqueous solution of FeCl3 that is 28.8% by mass with a solution density of 1.280g/mL, at 25 degrees Celsius. Please show all work.

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.28 g/mL

we have below equation to be used:

mass = density * volume

= 1.28 g/mL *1*10^3 mL

= 1280.0 g

This is mass of solution

mass of FeCl3 = 28.8 % of mass of solution

= 28.8*1280.0/100

= 368.64 g

Molar mass of FeCl3 = 1*MM(Fe) + 3*MM(Cl)

= 1*55.85 + 3*35.45

= 162.2 g/mol

mass of FeCl3 = 368.64 g

we have below equation to be used:

number of mol of FeCl3,

n = mass of FeCl3/molar mass of FeCl3

=(368.64 g)/(162.2 g/mol)

= 2.273 mol

volume , V = 1 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.273/1

= 2.273 M

This is concentration of FeCl3 solution

FeCl3 dissociates into 4 particles. So, i=4

T= 25.0 oC

= (25.0+273) K

= 298 K

we have below equation to be used:

P = i*C*R*T

P = 4.0*2.273*0.0821*298.0

P = 2.273 atm

Answer: 2.273 atm