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 mL of water at 37 °C ?

Question

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 mL of water at 37 °C ? (A) 30.96 Pa (B) 34.36 Pa (C) 68.72 Pa (D) 48.25 Pa

Tardigrade Admin · Accepted Answer

No. of moles of polymer = (1/150,000) π = CRT = (n/V) RT π = (1/150,000) × (8.314 ×103×310/0.5) = 34.36 Pa (R = 8.314 × 103 Pa L K-1 mol-1)

Q. 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 $37^{°} C$ ?

$30.96 P a$

$34.36 P a$

$68.72 P a$

$48.25 P a$

No. of moles of polymer $= \frac{1}{150 , 000}$
$π = CRT = \frac{n}{V} RT$
$π = \frac{1}{150 , 000} \times \frac{8.314 \times 1 0 ^{3} \times 310}{0.5}$
$= 34.36 P a$
$(R = 8.314 \times 1 0^{3} P a L K^{- 1} m o l^{- 1})$

Q. 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 37°C ?

30.96Pa

34.36Pa

68.72Pa

48.25Pa