A chemist is given a white solid that is suspected of being pure cocaine (molar

Question

A chemist is given a white solid that is suspected of being pure cocaine (molar mass = 303.35 g/mol). When 1.22 g of the solid is dissolved in 15.60 g of benzene the freezing point is lowered by 1.32 degree C. The molar mass is calculated from these data to be 303 g. Assuming the following uncertainties, can the chemist be sure the substance is not codeine (molar mass 299.36 g/mol)? K_f for benzene is 5.12 degree C/mol. Uncertainties Mass of solid = plusminus 0.01 g Mass of benzene = plusminus 0.01 g Delta T (freezing point lowering) = plusminus 0.04 degree C K_f = plusminus 0.01 Support your answer with calculations.

Explanation / Answer

Tf = kf m
1.32 = 5.12 * m
molality of the unknown solid = 0.25781 mol / Kg benzene
(1.22 gm / molar mass of unknown) / 0.0156 Kg benzene = 0.25781 m
molar mass of the unknown = 303.344 gm/mol which is the molar mass of cocaine
Now lets calculate with uncertainities,
mass of the solid = 1.22 ± 0.01
mass of benzene = 15.60 ± 0.01
Tf = 1.32 ± 0.04
Kf = 5.12 ± 0.01
molality = (1.32 ± 0.04) / (5.12 ± 0.01)
calculate relative error in molality, dm
dm / (1.32/5.12) = (0.04/1.32) + (0.01/5.12)
dm / 0.25781 = 0.032256
dm = 8.3159*10-3
relative error of molality in % =  8.3159*10-3 *100 = 0.83159 %
This percent is very less and the change will not be huge. However the molar mass will be around
303.35 gm/mol because this uncertainity can influence the last 2 digits after the decimal if the error is rounded
off to 4 figures which is anyway insignificant in this case as molar mass of cocaine is rounded off 2 digits.
Hence chemist can be sure that it is not codeine.

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