What is the osmotic pressure of a 0.0020 mol d m-3 sucrose (C12 H22 O11) solution at 20° C ? (Molar gas constant, R=8.314 J K-1 mol -1 ; 1 dm 3=0.001 m 3 )

Question

What is the osmotic pressure of a 0.0020 mol d m-3 sucrose (C12 H22 O11) solution at 20° C ? (Molar gas constant, R=8.314 J K-1 mol -1 ; 1 dm 3=0.001 m 3 ) (A) 4870 Pa (B) 4.87 Pa (C) 0.00487 Pa (D) 0.33 Pa

Tardigrade Admin · Accepted Answer

Osmotic pressure, π=C R T =(0.0020/1000)( mol m -3) × (8.314/101.266) (L atm K -1 mol -1) × 293 K =4.87 × 10-3 atm =4.87 Pa

Q. What is the osmotic pressure of a $0.0020 m o l d m^{- 3}$ sucrose $(C_{12} H_{22} O_{11})$ solution at $2 0^{\circ} C$ ?
(Molar gas constant, $R = 8.314 J K^{- 1} m o l^{- 1}; 1 d m^{3} = 0.001 m^{3}$ )

4870 Pa

4.87 Pa

0.00487 Pa

0.33 Pa

Osmotic pressure, $π = CRT$
$= \frac{0.0020}{1000} (m o l m^{- 3}) \times \frac{8.314}{101.266}$
$(L a t m K^{- 1} m o l^{- 1}) \times 293 K$
$= 4.87 \times 1 0^{- 3} a t m = 4.87 P a$

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