Question

A 202 ml benzene solution containing 2.47 gm of organic polymer has an osmotic pressure of 8.63 mmHg at 294 k . Calculate the molecular mass of the polymer .

Answer 1

Answer:

p = 8.63 /760=0.0114 atm  

T = 21 + 273 =294 K  

Osmotic pressure = CRT  

0.0114 = C x 0.08206 x 294  

C = molar concentration of the solution = 0.000473 M  

moles polymer = 0.000473 M x 0.202 L=0.0000955  

molar mass = 2.47 g / 0.0000955 mol=2.59 x 10^4 g/mol

Answer 2

The molecular mass of the polymer is 25978.6 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 8.63 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polymer) = 2.47 g  

Volume of solution = 202 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $294K$

Putting values in above equation, we get:

$8.63mmHg=1\times \frac{2.47\times 1000}{\text{Molar mass of polymer}\times 202}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\\text{molar mass of polymer}=25978.6g/mol$

Hence, the molar mass of the polymer is 25978.6 g/mol

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