1 g of polymer having molar mass 1,60,000 g is dissolved in 800 mL water. Calculate the osmotic pressure in pascal at 27°C. (R = 8.314 J K-1mole-1)

Question

1 g of polymer having molar mass 1,60,000 g is dissolved in 800 mL water. Calculate the osmotic pressure in pascal at 27°C. (R = 8.314 J K-1mole-1) (A) 19.4 (B) 0.90 (C) 0.50 (D) 1.20

Tardigrade Admin · Accepted Answer

π=CRT π=(1/1.6×105)×(8.314×300/800×10-6)=19.4 Pa

Q. 1 g of polymer having molar mass 1,60,000 g is dissolved in 800 mL water. Calculate the osmotic pressure in pascal at 27 $^{\circ}$ C.
$(R = 8.314 J K^{- 1} m o l e^{- 1})$

19.4

0.90

0.50

1.20

$π =$ CRT
$π = \frac{1}{1.6 \times 1 0 ^{5}} \times \frac{8.314 \times 300}{800 \times 1 0 ^{- 6}} = 19.4 P a$

Q. 1 g of polymer having molar mass 1,60,000 g is dissolved in 800 mL water. Calculate the osmotic pressure in pascal at 27∘C. (R=8.314JK−1mole−1)