Chapter 13, Problem 43PS

### 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

# Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Interpretation Introduction

Interpretation: The molar mass of benzyl acetate 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

Change in boiling point is calculated by using the equation,

ΔTbp=Kbpmsolute

where,

Kbp is the molal boiling point elevation constant.

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

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

The molar mass of benzyl acetate is calculated.

Given,

â€‚Â Î”Tâ€‰=â€‰(61.820Câ€‰âˆ’â€‰61.700Câ€‰)â€‰=â€‰0.120C

â€‚Â Mass of benzyl acetate is 0.125â€‰g

â€‚Â Mass of chloroform is 25â€‰gâ€‰=â€‰0.025â€‰kg

â€‚Â Molal boiling point elevation constant of chloroform is 3.630C/m

Change in boiling point is calculated by using the equation,

â€‚Â Î”Tbpâ€‰=â€‰Kbpâ€‰msolute

Hence,

The concentration is calculated by,

â€‚Â Concentration,msolute=â€‰Î”TbpKbp=â€‰(â€‰0.120C)(3.630C/m)=â€‰0

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