Question

117. 200 c.c of an aqueous soloution contains 1.26
gms of a polymer. The osmotic pressure of such
solution at 300 K is found to be 2.57x10-3 bar.
Calculate the molar mass of the polymer
1) 61038 g/mole 2) 122076 g/mole
3) 610.38 g/mole 4) 122.076 g/mole​

Answer 1

Answer :

• Molar mass of polymer = 61038 g/mol.

Step-by-step explanation :

Given that,

• Volume of solution, V = 200 cm³ = 0.2 L.
• Mass of solute = 1.26 g.
• Absolute temperature, T = 300 K.
• Osmotic pressure, π = 2.57 × 10⁻³ bar.

Also,

• Gas constant, R = 0.083 L atm mol⁻¹ K⁻¹ .
• 1 bar = 0.986 atm ≈ 1 atm.
• Molar mass of solute (polymer), M = ?

$\hrulefill$

No. of moles of solute (polymer), n = $\rm{\dfrac{1.26\:g}{M}}$

Now, we know, $\rm{\pi\:V=nRT}$

Putting the values, we get,

$\implies\rm{2.57\times{}10^{-3}\:atm\times{}0.2\:L=\dfrac{1.26\:g}{M}\times{}0.083\:L\:atm\:mol^{-1}\:K^{-1}\times{}300\:K}$

$\implies\rm{M=\dfrac{1.26\:g\times{}0.083\:L\:atm\:mol^{-1}\:K^{-1}\times{}300\:K}{2.57\times{}10^{-3}\:atm\times{}0.2\:L}}$

Solving, we get,

$\implies\rm{M=\dfrac{61.038\:g\:mol^{-1}}{10^{-3}}=61.038\:\times{}10^3\:g\:mol^{-1}}$

$\implies\rm{M=}\:\bold{61038\:g\:mol^{-1}.}$