Question

What is the osmotic pressure of a 0.0020 mol $dm^{- 3}$ sucrose $\left(\right.C_{12}H_{22}O_{11}\left.\right)$ solution at $20 \,{}^{o}C$ ?

(Molar gas constant, $R=8.314 \, JK^{- 1}mol^{- 1}$ ; $1 \, dm^{3}=0.001 \, m^{3}$ )

• A. 4872 Pa
• B. 4.87 Pa
• C. 0.00487 Pa
• D. 0.33 Pa

Answer 1

Answer: (A)

$\pi =$ CRT

$C=0.002 \, mol \, dm^{- 3}=2 \, mol \, m^{- 3}$

$R=8.314 \, J \, K^{- 1}mol^{- 1},T=293 \, K$

$\pi =CRT=2\times 8.314\times 293=4872.004 \, Pa$