Chapter 13, Problem 70GQ

### 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

# Cigars are best stored in a “humidor” at 18 °C and 55% relative humidity. This means the pressure of water vapor should be 55% of the vapor pressure of pure water at the same temperature. The proper humidity can be maintained by placing a solution of glycerol [C3H5(OH)3] and water in the humidor. Calculate the percent by mass of glycerol that will lower the vapor pressure of water to the desired value. (The vapor pressure of glycerol is negligible.)

Interpretation Introduction

Interpretation:

Mass percentage of glycerol should be determined under the given condition of temperature and pressure.

Concept introduction:

• Raoult’s law: In a solution, vapour pressure of solvent is proportional to its mole fraction.

Psolvent=XsolventP0solvent

Where,

P0solvent is the vapour pressure of pure solvent.

• Mass percentage is one way to represent the concentration of an element in a compound or a component in a mixture. Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture ( nA +  nB +  n...).
• Mole fraction of A (χA)= nA nA +  nB +  n...
• Massofthesubstance=Numberofmoles×Molarmassofthesubstance
Explanation

Given,

Pressure of water vapor should be 55% of vapor pressure of pure water at the same temperature.

The mole fraction of water can be determined by using Raoultâ€™s law. And here glycerol is a nonvolatile substance and thus its vapor pressure is negligible.

Then,

According with Raoultâ€™s law,

Psolutionâ€‰â€‰=â€‰â€‰0.55â€‰â€‰Pwater0Â

The number of mole of water = 0.55â€‰â€‰mol

The total mole fraction of all components in solution or mixture of substance is 1.

Therefore the number of moles of glycerol, nglycerolâ€‰=â€‰â€‰1.00â€‰âˆ’â€‰0.55â€‰â€‰=â€‰â€‰0.45â€‰â€‰molâ€‰Â

Masses of water and glycerol can be calculated by using the equation given below,

Massâ€‰â€‰ofâ€‰â€‰theâ€‰â€‰substanceâ€‰â€‰=â€‰â€‰Numberâ€‰â€‰ofâ€‰molesâ€‰â€‰Ã—â€‰â€‰Molarâ€‰â€‰massâ€‰â€‰ofâ€‰â€‰theâ€‰â€‰substanceâ€‰

Massâ€‰â€‰ofâ€‰â€‰waterâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰0.55â€‰molâ€‰â€‰Ã—18.02â€‰â€‰g/molâ€‰â€‰â€‰=â€‰â€‰9

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