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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08/01/2022 5:52 pm
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Determine the amount of CaCl_{2} (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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08/01/2022 5:59 pm
We know that,
π = i \(\frac{n}{V}\)RT
⇒ π = i \(\frac{w}{MV}\)RT
⇒ w = i \(\frac{\pi MV}{iRT}\)
π = 0.75 atm
V = 2.5 L
i = 2.47
T = (27 + 273)K = 300K
Here,
R = 0.0821 L atm K^{1}mol^{1}
M = 1 × 40 + 2 × 35.5
= 111g mol^{1}
Therefore, w = \(\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}\)
= 3.42 g
Hence, the required amount of CaCl_{2} is 3.42 g.
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