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Osmotic Pressure

The osmotic p...

Question

The osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 g$ polymer of molar mass $185, 000$ in $450 m L$ of water at $37^{o} C$ is :

10.2 P a

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30.9 P a

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20.4 P a

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40.6 P a

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Solution

The correct option is B $30.9 P a$
Given= Mass of polymer= $1.0 g$

Molar mass= $185, 000$

Volume of water= $V = 450 m l = 0.45 L$

Temperature= $(37 + 273) K = 310 K$

Osmotic pressure is given by,

$π = \frac{n}{v} R T$
$= \frac{1}{185000} \times \frac{1}{0.45} \times 8.314 \times 10^{3} \times 310 K$

$π = 30.95 P a$

Option B is correct.

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The osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g polymer of molar mass 185,000 in 450mL of water at 37oC is :

The correct option is B 30.9PaGiven= Mass of polymer= 1.0gMolar mass= 185,000Volume of water= V=450ml=0.45LTemperature= (37+273)K=310KOsmotic pressure is given by,π=nvRT =1185000×10.45×8.314×103×310Kπ=30.95PaOption B is correct.

The osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 g$ polymer of molar mass $185, 000$ in $450 m L$ of water at $37^{o} C$ is :