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InterviewSolution
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Freezing Point Depression: when `15.0 grams` of ethyl alcohol `(C_(2)H_(5)OH)` is dissolved in `750 grams` of formic acid, the freezing point of the solution is `7.20^(@)C`. The freezing point of pure formic acid is `8.40^(@)C` Evalulate `K_(f)` for formic acid. Strategy: First calculate the molality and the depression of the freezing point. Then solve the equation `DeltaT_(f)=K_(f)m` for `K_(f)` by substituting values for `m` and `DeltaT_(f)`. |
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Answer» Step 1. Calculate depression of the freezing point `DeltaT_(f)` is the depression of freezing point. It is defined as `DeltaT_(f)=T_(f("solvent"))-T_(f("solution"))` So it is always positive. Here `DeltaT_(f)=T_(f) ("formic acid")-T_(f)("solution")` `=8.40^(@)C-7.20^(@)C` `=1.20^(@)C` (depression) Step 2. Calculate moles of solute `n_(solute)=(mass (solute))/(molar mass (solute))` `=(15.0 g C_(2)H_(5)OH)/(46.0 g C_(2)H_(5)OH//mol C_(2)H_(5)OH)` `=0.326 mol C_(2)H_(5)OH` Step 3. Calculate molality of solute `Molality (m)=n_(solute)/g_(solvent)xx(1000 g)/(kg)` `=(0.326 mol)/(750 g)xx(1000 g)/(kg)` `=0.435 mol Kg^(-1)` Setp 4. Calculate the molal freezing point depression constant, `K_(f)` According to Equation (2.68), we have `K_(f)=(DeltaT_(f))/m` Thus `K_(f)=(1.20^(@)C)/(0.435 m)` `=2.76^(@)C//m` |
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