Chapter 13, Problem 90IL

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# In a police forensics lab, you examine a package that may contain heroin. However, you find the white powder is not pure heroin but a mixture of heroin (C12H23O5N) and lactose (C12H22O11). To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volumetric flask. You find that the solution has an osmotic pressure of 539 mm Hg at 25 °C. What is the composition of the mixture?

Interpretation Introduction

Interpretation: The composition of heroin and lactose in the mixture has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Osmotic pressure: The pressure created by the column of solution for the system at equilibrium is a measure of the osmotic pressure and is calculated by using the equation,

π=cRT

where,

c is the molar concentration

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

Given,

â€‚Â Râ€‰=â€‰0.082057â€‰L.atmÂ Kâˆ’1molâˆ’1Tâ€‰=â€‰250Câ€‰=â€‰(25â€‰+â€‰273.2)â€‰Kâ€‰=â€‰298.2â€‰K

â€‚Â Molar mass of the heroin is 369.41â€‰g/mol

â€‚Â Molar mass of the lactose is 342.30â€‰g/mol

â€‚Â Total mass of the mixture is 1.00â€‰g

â€‚Â Osmotic pressure,Ï€â€‰=â€‰539â€‰mmâ€‰Hg

The osmotic pressure of the solution is,

â€‚Â Ï€â€‰=â€‰cRT

Hence,

The concentration is calculated by,

â€‚Â Concentration,câ€‰=â€‰Ï€RT=â€‰(539â€‰mmâ€‰Hg)(1â€‰atm760â€‰mmâ€‰Hg)(0.082057â€‰â€‰L.atmâ€‰Kâˆ’1â€‰molâˆ’1)(298.2â€‰K)=â€‰2.898â€‰Ã—â€‰10âˆ’2â€‰mol/L

The amount of starch dissolved in one liter of water is,

â€‚Â 2.898â€‰Ã—â€‰10âˆ’2â€‰mol/L1â€‰Lâ€‰(0.100â€‰L)=â€‰2.898â€‰Ã—â€‰10âˆ’3â€‰mol

The number of moles of any substance can be determined using the equation

â€‚Â Numberâ€‰ofâ€‰moleâ€‰=Givenâ€‰massâ€‰ofâ€‰theâ€‰substanceMolarâ€‰massâ€‰

Total mass of the mixture is 1.00â€‰g

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