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Osmotic Pressure

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Question

Calculate the osmotic pressure of a solution prepared by dissolving $0.025 g$ of $K_{2} S O_{4}$ in $2.0$ litres of water at $25^{o} C$ assuming that $K_{2} S O_{4}$ is: (completely dissociated)
(mol. wt. of $K_{2} S O_{4}$ = $174 g m o l^{- 1}$ )

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Solution

No. of moles of $K_{2} S O_{4} = \frac{0.025 g}{174}$
Concentration of $K_{2} S O_{4} = \frac{0.025}{174 \times 2} m o l e / L$
Osmotic pressure $= π = C R T = \frac{0.025}{174 \times 2} m o l e / L \times 0.0821 a t m - L m o l^{- 1} K^{- 1} = 1.75 \times 10^{- 3} a t m$

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2.5 \times 10^{- 2} g

K_{2} S O_{4}

2 L

25^{\circ} C

(T = 0.0821 L a t m K^{- 1} m o l^{- 1}

K_{2} S O_{4} = 174 g m o l^{- 1})

2.5 \times 10^{- 2} g

K_{2} S O_{4}

2 L

25^{\circ} C

(R = 0.0821 L a t m K^{- 1} m o l^{- 1}

K_{2} S O_{4} = 174 g m o l^{- 1})

2.5 \times 10^{- 2} g

K_{2} S O_{4}

25^{0} C

R = 0.0821 L a t m

K^{- 1} m o l^{- 1}

K_{2} S O_{4} = 174 g m o l^{- 1}

25 m g

K_{2} S O_{4}

2

25^{\circ} C

25 m g

K_{2} S O_{4}

2

25^{\circ} C

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