A forensic chemist is given a white solid that is suspected of being pure cocaine (C 17 H 21 NO 4 , molar mass = 303.35 g/mol). She dissolves 1.22 ± 0.01 g of the solid in 15.60 ± 0.01 g benzene. The freezing point is lowered by 1.32 ± 0.04°C. a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calculate the uncertainty in the molar mass. b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine (C 18 H 21 NO 3 , molar mass = 299.36 g/mol)? c. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?

Interpretation: The molar mass and improvisation in the results has to be calculated.

Concept Introduction:

Molar mass of a compound can be given by mass of a substance to the amount of the substance. It can be given by expression,

Molar mass = Mass of substanceAmount of substance

The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.

The depression in freezing point can be given by the equation,

ΔT=Kfmsolute

 Where, ΔT=change in freezing point depression

                         Kf= molal freezing point depression constant

                                                msolute = molality of solute

Record the given info

Molar mass of Cocaine = 303.35 g mol-1

Mass of solid = 1.22±0.01 g

Mass of Benzene = 15.60±0.01 g

Freezing point = 1.32±0.04°C

To calculate the molality

m=ΔTfKf    =1.32°C5.12°C kg/mol     =0.28 mol kg-1

To calculate the moles of the unknown solute

Molality = 0.28 mol kg-1

Moles of Unknown =  0.01560 kg×0.258 mol unknownkg

                              =4

Interpretation: The molar mass and improvisation in the results has to be calculated.

Concept Introduction:

Molar mass of a compound can be given by mass of a substance to the amount of the substance. It can be given by expression,

Molar mass = Mass of substanceAmount of substance

The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.

The depression in freezing point can be given by the equation,

ΔT=Kfmsolute

 Where, ΔT=change in freezing point depression

                         Kf= molal freezing point depression constant

                                                msolute = molality of solute

Interpretation: The molar mass and improvisation in the results has to be calculated.

Concept Introduction:

Molar mass of a compound can be given by mass of a substance to the amount of the substance. It can be given by expression,

Molar mass = Mass of substanceAmount of substance

The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.

The depression in freezing point can be given by the equation,

ΔT=Kfmsolute

 Where, ΔT=change in freezing point depression

                         Kf= molal freezing point depression constant

                                                msolute = molality of solute