What will be the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 g$ of polymer of molar mass $150, 000$ in $500 m L$ of water at $37^{o} C$ ?

30.96

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34.36

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68.72

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48.25

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Solution

The correct option is D $34.36$ Pa
No of moles of polymer $= \frac{1}{150000}$

$π = C R T = \frac{n}{V} R T$

$π = \frac{1}{150000} \times \frac{8.314 \times 10^{3} \times 310}{0.5} = 34.36 P a$

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What will be the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 150,000 in 500mL of water at 37oC?

The correct option is D 34.36 PaNo of moles of polymer =1150000π=CRT=nVRTπ=1150000×8.314×103×3100.5=34.36Pa

What will be the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 g$ of polymer of molar mass $150, 000$ in $500 m L$ of water at $37^{o} C$ ?

The correct option is D $34.36$ Pa
No of moles of polymer $= \frac{1}{150000}$

$π = C R T = \frac{n}{V} R T$

$π = \frac{1}{150000} \times \frac{8.314 \times 10^{3} \times 310}{0.5} = 34.36 P a$