   # 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? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 124CP
Textbook Problem
5 views

## A forensic chemist is given a white solid that is suspected of being pure cocaine (C17H21NO4, 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 (C18H21NO3, 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?

(a)

Interpretation Introduction

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,

Molarmass=MassofsubstanceAmountofsubstance

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

### Explanation of Solution

Record the given info

Molar mass of Cocaine = 303.35gmol-1

Mass of solid = 1.22±0.01g

Mass of Benzene = 15.60±0.01g

Freezing point = 1.32±0.04°C

To calculate the molality

m=ΔTfKf=1.32°C5.12°Ckg/mol=0.28molkg-1

To calculate the moles of the unknown solute

Molality = 0.28molkg-1

Moles of Unknown =  0.01560kg×0.258molunknownkg

=4

(b)

Interpretation Introduction

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,

Molarmass=MassofsubstanceAmountofsubstance

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

(c)

Interpretation Introduction

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,

Molarmass=MassofsubstanceAmountofsubstance

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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Which distance is farther, 100 km or 50 mi?

Introductory Chemistry: A Foundation

What is an atoll and how does it form?

Fundamentals of Physical Geography

For people with diabetes, the risk of heart disease, stroke, and dying on any particular day is cut in half. T ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What are neurotransmitters?

Biology (MindTap Course List)

50. Name each compound:

Chemistry In Focus

With the switch in the circuit of Figure 27.6a open, there is no current in R2. There is current in R1, however...

Physics for Scientists and Engineers, Technology Update (No access codes included)

How are Earths inner layers classified?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 