Question

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 $2.57\times {10}^{-3}$ bar. Calculate the molar mass of the polymer?

• A. $61038$ g/mole
• B. $122076$ g/mole
• C. $610.38$ g/mole
• D. $122.076$ g/mole

Answer 1

The correct option is A $61038$ g/mole

$\sqcap =2.57\times {10}^{-3}bar$
and we know 1 bar $=0.986atm$
$\approx 1atm$
osmotic pressure $=C_{M}RT(\to molarity)$

$R=0.0821liter.atm/mole.K$

$2.57\times {10}^{-3}atm=\frac{01.26}{M.\omega t}\times \frac{1000}{200}\times 0.0821\times 300$

$M\omega t=\frac{1.26\times 1000\times 0.0821\times 30}{200\times 2.57\times {10}^{-3}atm}$

$\approx 61038g/mole$