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?

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

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = = ?

Depression in freezing point =

Depression in freezing point  is also given by formula:

= The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: = 40°C kg/mol

i = 1 (  non electrolyte)

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

you gotta add the following molecules so we can give u an answer

explanation: