Question

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point taken from your sample is 165°C and b) the freezing point depression constant for camphor is 40°C kg/mol.

Answer 1

The molality of isoborneol in camphor is 0.53 mol/kg.

Step-by-step explanation:

Melting point of pure camphor= T =179°C

Melting point of sample =
`T_f` = 165°C

Depression in freezing point =
`\Delta T_f=?`

`\Delta T_f=T- T_f=179^oC-165^oC=14^oC`

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

i = van't Hoff factor

We have:
`K_f` = 40°C kg/mol

i = 1 ( organic compounds)

`\Delta T_f=14^oC`

`14^oC=1* 40^oC kg/mol* m`

`m=(14^oC)/(1* 40^oC kg/mol)=0.35 mol/kg`

The molality of isoborneol in camphor is 0.53 mol/kg.