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

Answer:

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

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:

$\Delta T_f=i\times K_f\times m$

$K_f$ = 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)

$\Delta T_f=14^oC$

$14^oC=1\times 40^oC kg/mol\times m$

$m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg$

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