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Dennis_Churaev [7]
3 years ago
5

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.
Chemistry
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

The given compound cannot be cocaine.

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=\frac{1.6}{4.90} = 0.327 m

we know that:

molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}

therefore

moles = 0.327X0.008 = 0.00261 mol

moles=\frac{mass}{molarmass}

molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.

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