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Question
Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Solution
π = iCRT = `"i""n"/"V""RT"`
or, n = `(π xx "V")/("i" xx "R" xx "T")`
= `(0.75 "atm" xx 2.5 "L")/(2.47 xx 0.0821 "L atm K"^-1 "mol"^-1 xx 300 "K")`
= `1.875/60.836`
= 0.0308 mol
Molar mass of CaCl2 = 40 + 2 × 35.5 = 111 g mol−1
∴ Amount of CaCl2 dissolved = 0.0308 × 111 g = 3.42 g
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