Question

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

Answer 1

Answer: (B)

Osmotic pressure, $\pi=C R T$
$=\frac{0.0020}{1000}\left( mol \,m ^{-3}\right) \times \frac{8.314}{101.266}$
$\left(L \,atm \,K ^{-1} \,mol ^{-1}\right) \times 293\, K$
$=4.87 \times 10^{-3} atm =4.87\, Pa$