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

Answer:

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

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:

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

$K_f$ = 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)

$\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg$

$\Delta T_f=11^oC$

$\Delta T_f=T- T_f$

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

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