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Determine the amount of CaCl_(2)dissolved in 2.5L at 27^(@)C such that its osmotic pressure is 0.75 atm at 27^(@)C. (i for CaCl_(2) = 2.47) |
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Answer» Solution :For `CaCl_(2) , "" i= 2.47` `pi = ICRT` `=i(n_(B))/(V) XX RT` `0.75 = (2.47 xx n_(B) xx 0.082 xx 300)/(2.5)` `n_(B) = (0.75 xx 2.5)/(2.47 xx 0.082 xx 300)` `n_(B) = 0.0308` MOL Amount = `0.0308` mol `xx` 111 g `mol^(-1)` `= 3.148` g |
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