Question

A 2.0 g/L solution of a particular preparation of a polymer (polystyrene) in toluene was measured to give an osmotic pressure of 58.25 Pascal at room temperature (298K). Estimate the molecular weight of the polystyrene

Answer 1

solution) Π = iMRT or Π = i(n/V)RT

Π = osmotic pressure

i = van't hoff factor

M = molarity of solution

R = gas constant

T = temperature

osmotic pressure = 58.25 pascal (given), we have to chnage this into atm

1atm = 1.013 * 10⁵Pa

Therefore, osmotic pressure = 5.74 * 10^-4 atm

Temperature = 298 K(given)

R = 0.08206 Latm/mol.K

i = 1 ( i will be assumed to be 1 in this question since polystyrene will not dissociate in toluene)

Now we will put the values of Π , R , i , T in equation Π = iMRT and calculate the molarity.

Π = iMRT

M = Π/ iRT

M = (5.74 * 10^-4 atm) / (1) (0.08206 Latm/mol.K) (298 K)

M = 0.24* 10^-4 mol / L

The molecular weight of the polystyrene can be determined by the following method :

Molarity = moles of solute / volume of solution(L)

moles of solute = (mass of solute / molecular weight of solute)

Molarity = (mass of solute / molecular weight of solute) / volume of solution (in L)

Molarity = (mass of solute) / molecular weight of solute volume of solution (in L)

molecular weight of solute =  (mass of solute) / Molarity   volume of solution (in L)

In question it is given 2.0 g/L solution which means that 2gm of solute is dissolved in 1 L of solution

Therefore, mass of solute = 2gm

Volume of solution =1 L

Molarity = 0.24* 10^-4 mol / L

By putting the above values in equation:

molecular weight of solute =  (mass of solute) / Molarity   volume of solution (in L)

molecular weight of solute = 2g / 0.24* 10^-4 mol / L 1 L

molecular weight of solute = 8.33 * 10⁴ g/mol

molecular weight of the polystyrene is 8.33 * 10⁴ g/mol