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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.
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Solution
Given: Osmotic pressure (π) = 0.75 atm
Volume (V) = 2.5 L
i = 2.47
R = 0.0821 L atm K−1 mol−1
Temperature (T) = 27°C = 27 + 273 = 300 K
π = iCRT
or, π = `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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