What is the osmotic pressure of a 0.0020 mol dm- 3 sucrose (.C12H22O11.) solution at 20 oC ? (Molar gas constant, R=8.314 JK- 1mol- 1 ; 1 dm3=0.001 m3 )

Question

What is the osmotic pressure of a 0.0020 mol dm- 3 sucrose (.C12H22O11.) solution at 20 oC ? (Molar gas constant, R=8.314 JK- 1mol- 1 ; 1 dm3=0.001 m3 ) (A) 4872 Pa (B) 4.87 Pa (C) 0.00487 Pa (D) 0.33 Pa

Tardigrade Admin · Accepted Answer

π = CRT C=0.002 mol dm- 3=2 mol m- 3 R=8.314 J K- 1mol- 1,T=293 K π =CRT=2× 8.314× 293=4872.004 Pa

Q. What is the osmotic pressure of a 0.0020 mol $d m^{- 3}$ sucrose $(C_{12} H_{22} O_{11})$ solution at $20^{o} C$ ?

(Molar gas constant, $R = 8.314 J K^{- 1} m o l^{- 1}$ ; $1 d m^{3} = 0.001 m^{3}$ )

4872 Pa

4.87 Pa

0.00487 Pa

0.33 Pa

$π =$ CRT

$C = 0.002 m o l d m^{- 3} = 2 m o l m^{- 3}$

$R = 8.314 J K^{- 1} m o l^{- 1}, T = 293 K$

$π = CRT = 2 \times 8.314 \times 293 = 4872.004 P a$

Q. What is the osmotic pressure of a 0.0020 mol dm−3 sucrose (C12​H22​O11​) solution at 20oC ? (Molar gas constant, R=8.314JK−1mol−1 ; 1dm3=0.001m3 )