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Question
Determine the osmotic pressure of a solution prepared by dissolving 2.32 × 10−2 g of K2SO4 in 2L of solution at 25°C assuming that K2SO4 is completely dissociated.
(R = 0.082 L atm K−1 mol, Molar mass K2SO4 = 174 g mol−1)
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Solution
Amount of K2SO4 dissolved = 2.32 × 10−2 g
Volume of solution = 2 L
Temperature = 25 + 273 = 298 K
Molar mass of K2SO4 = (2 × 39) + 32 + 4 × 16 = 174 g/mol
Since K2SO4 dissociates completely as
\[\ce{K2SO4 -> 2K^+ + SO^{2-}4}\]
Total ions produced after dissociation = 3
So, i = 3, π = iCRT
C = `4/"V"`
π = `(3 xx 2.32 xx 10^-2)/(174 xx 2) xx 0.082 xx 298`
= `85.14/(174 xx 100)`
= 0.0048
= 4.8 × 10−3 atm
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