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An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 "g mol"^(-1)) in 600 g of water. Calculate the freezing point of the solution. (K_(f) for water =1.86" K kg mol"^(-1)) |
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Answer» SOLUTION :Given : `w_(1)=600" g,"w_(2)=31" g,"M_(2)=62" g mol"^(-1)` Molality of the solution `=(w_(2)//M_(2))/(w_(1)//1000)=(31//62)/(600//1000)=(31)/(62)xx(1000)/(600)=5/6m` Apply the RELATION : `DeltaT_(f)=K_(f)m` `=1.86" K kg mol"^(-1)xx5/6" mol kg"^(-1)` `=1.86xx5/6=1.55` FREEZING point of the solution `=-1.55^(@)C` or `271.45K` |
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