   # In Exercise 96 in Chapter 8, the pressure of CO 2 in a bottle of sparkling wine was calculated assuming that the CO 2 was insoluble in water. This was a bad assumption. Redo this problem by assuming that CO 2 obeys Henry’ s law. Use the data given in that problem to calculate the partial pressure of CO 2 in the gas phase and the solubility of CO 2 in the wine at 25°C. The Henry’s law constant for CO 2 is 3.1 × w −2 mol/L · atm at 25°C with Henry’s law in the form C = kP, where C is the concentration of the gas in mol/L. ### 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 94AE
Textbook Problem
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## In Exercise 96 in Chapter 8, the pressure of CO2 in a bottle of sparkling wine was calculated assuming that the CO2 was insoluble in water. This was a bad assumption. Redo this problem by assuming that CO2 obeys Henry’ s law. Use the data given in that problem to calculate the partial pressure of CO2 in the gas phase and the solubility of CO2 in the wine at 25°C. The Henry’s law constant for CO2 is 3.1 × w−2 mol/L · atm at 25°C with Henry’s law in the form C = kP, where C is the concentration of the gas in mol/L.

Interpretation Introduction

Interpretation: The partial pressure and concentration of the gas has to be calculated.

Concept Introduction:

Henry’s law can be given as the quantity of gas dissolve in a solution is proportional to the pressure of the gas above the solution.

Henry’s law gives the relationship between the pressure of the gas and the concentration of gas dissolved. The equation can be given as,

C=kP

where, C= Concentration of the gas dissolved

k=constant

P=Partial pressure of the solute in gaseous solute

### Explanation of Solution

Record the given info

Volume of grape juice = 75.0mL

Density of juice = 1gcm-3

Volume of Ethanol = 825mL

Density of Ethanol = 0.79gcm-3

To calculate the moles of CO2

Molar Mass of Ethanol = 46.02g

Moles of CO2=750.0mLgrapejuice×12mLC2H5OH×0.79gC2H5OHmL×1molC2H5OH46.07g×2molCO22molC2H5OH=1.54mol

To calculate the partial pressure of Carbon dioxide

1.54 mol CO2 = total mol CO2 = mol CO2(g) + mol CO2(aq) = n+ naq

PCO2=ngRTV=ng(0.08206LatmmolK)(298K)75×10-3L=326ngPCO2=Ck=naq0

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