   Chapter 13, Problem 90IL

Chapter
Section
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,

R=0.082057L.atm K1mol1T=250C=(25+273.2)K=298.2K

Molar mass of the heroin is 369.41g/mol

Molar mass of the lactose is 342.30g/mol

Total mass of the mixture is 1.00g

Osmotic pressure,π=539mmHg

The osmotic pressure of the solution is,

π=cRT

Hence,

The concentration is calculated by,

Concentration,c=πRT=(539mmHg)(1atm760mmHg)(0.082057L.atmK1mol1)(298.2K)=2.898×102mol/L

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

2.898×102mol/L1L(0.100L)=2.898×103mol

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

Numberofmole=GivenmassofthesubstanceMolarmass

Total mass of the mixture is 1.00g

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