#### Question

Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10^{−2} g of K_{2}SO_{4} in 2L of water at 25°C, assuming that it is completely dissociated.

(R = 0.0821 L atm K^{−1} mol^{−1}, Molar mass of K_{2}SO_{4} = 174 g mol^{−1})

#### Solution

w_{2} = 2.5 × 10^{−2} g (Mass of K_{2}SO_{4}) and M_{2} = 174 g mol^{−1} (Molar mass of K_{2}SO_{4})

V = 2L, R = 0.0821 L atm K^{−1} mol^{−1} and T = 25°C = 298 K

Osmotic pressure, ` pi=(w_2RT)/(M_2V)`

`pi=(2.5xx10^(-2)xx0.0821xx298)/(174xx2)=(61.1645xx10^(-2))/348=1.76xx10^(-3)atm`

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Solution Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10−2 g of K2SO4 in 2L of water at 25°C, assuming that it is completely dissociated. Concept: Colligative Properties and Determination of Molar Mass - Osmosis and Osmotic Pressure.