Question

Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that the melting point of pure camphor is 179°C and its freezing point depression constant is 40°C·kg/mol?

Answer 1

168°C is the melting point of your impure sample.

Step-by-step explanation:

Melting point of pure camphor= T =179°C

Melting point of sample =
`T_f` = ?

Depression in freezing point =
`\Delta T_f`

Depression in freezing point is also given by formula:

`\Delta T_f=i* K_f* m`

`K_f` = The freezing point depression constant

m = molality of the sample = 0.275 mol/kg

i = van't Hoff factor

We have:
`K_f` = 40°C kg/mol

i = 1 ( non electrolyte)

`\Delta T_f=1* 40^oC kg/mol* 0.275 mol/kg`

`\Delta T_f=11^oC`

`\Delta T_f=T- T_f`

`T_f=T- \Delta T_f=179^oC-11^oC=168^oC`

168°C is the melting point of your impure sample.