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What is the osmotic pressure of a $0.002 {mol dm}^{- 3}$ of sucrose $(C_{12} H_{22} O_{11})$ solution at $20 ° C$ . (Molar gas constant, $R = 8.314 J K^{-} {mol}^{-}, 1 {dm}^{3} = 0.001 m^{3}$ )

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4870 Pa

4.87 Pa

0.00487 Pa

0.33 Pa

answer is A.

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Detailed Solution

$\begin{array}{l} π = CRT \\ = 0.002 \times 1000 mol m^{- 3} \times 8.314 J K^{-} m o l^{-} \times (273 + 20) \\ = 4872.004 P a ≃ 4870 P a \end{array}$

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