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Question
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 liter of water at 25°C, assuming that it is completely dissociated.
Solution
When K2SO4 is dissolved in water, K+ and \[\ce{SO^{2-}4}\] ions are produced.
\[\ce{K2SO4 -> 2K^+ + SO^{2-}4}\]
Total number of ions produced = 3
∴ i = 3
Given,
w = 25 mg = 0.025 g
V = 2 L
T = 25°C = (25 + 273) K = 298 K
Also, we know that:
R = 0.0821 L atm K-1 mol-1
M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol-1
Applying the following relation,
`pi = "i" "n"/"V""RT"`
= `"i" xx "w"/"M" xx 1/"V""RT"`
= `3xx0.025/174 xx 1/2xx0.0821xx298`
= 5.27 × 10−3 atm
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