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Calculatethe freezingpoint of theone molaraqueoussolution(density 1.04 gL^(-1)) of KCl (k_(f) for wateer= 1.86 kg mol^(-1) , atomic mass of K = 39 ,Cl = 35.5) |
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Answer» Mass of solution` = 1000xx 1.04 = 1040 g` Mass of KCl= 74.5 g Mass of water= 1040-74.5 = 965.5 g . MOLALITYOF solution`= (1)/(965.5) XX 1000 = 1.0357` m SINCE KCl is strongelectrolyte, i = 2 `Delta T_(f) =i K_(f) xx m` `= 2 xx 1.86 xx 1.0357 = 3.852^(@)` Freezingpointof solution ` = 0 - 3.852 = - 3.852^(@)C` |
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