Question

A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution freezes at 3.9°C. Can the chemist conclude that the compound is cocaine (C17H21N04)? What assumptions are made in the analysis? The freezing point of benzene is 5.5°C.

Answer 1

The given compound cannot be cocaine.

Step-by-step explanation:

The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.

Depression in freezing point of is related to molality as:

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant = 4.90°C/m

depression in freezing point = normal freezing point - freezing point of solution

depression in freezing point = 5.5-3.9 = 1.6°C

1.6°C = 4.90 X molality

`molality=(1.6)/(4.90) = 0.327 m`

we know that:

`molality=(moles)/(mass of solvent(kg))=(moles)/(0.008kg)`

therefore

moles = 0.327X0.008 = 0.00261 mol

`moles=(mass)/(molarmass)`

`molarmass=(mass)/(moles)=(0.50)/(0.00261)= 191.57g/mol`

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.