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A solution of sucrose (molar mass =324 g/mol) is prepared by dissolving 68.4 g in 1000 g of water. What is the freezing point and boiling point of the solution ? K_(f)= water is 1.86 K/mand K_(b) =0.52 K/m. |
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Answer» `K_(f)=1.86Km^(-1), K_(b)=0.52Km^(-1),T_(f)=?, T_(b)=?` `DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))""DeltaT_(f)=(K_(b)xxW_(B))/(M_(B)xxW_(A))` `=((186" K kg mol"^(-1))xx(68.4 g))/((342" g mol"^(-1))xx(1kg)),""DeltaT_(b)=((0.52" K kg mol"^(-1))xx(68.4g))/((342g)xx(1 kg))` `=0.372 K,"" =0.104 K` `T_(f)=T_(f)^(@)-DeltaT_(f)=273-0.372,""T_(b)=T_(b)^(@)+DeltaT_(b)=373+0.104` `=272.628 K""=373.104 K` |
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