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)`.

Explanation

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`