Chapter 20, Problem 6PS

### 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

# Suppose you find the CO concentration in your home is 10. ppm by volume at 1.00 atm pressure and 25 °C. What is the CO concentration in mg/L and in ppm by mass. (The average molar mass for dry air is 28.96 g/mol at 1.00 atm pressure and 25 °C.)

Interpretation Introduction

Interpretation:

The concentration of CO in mg/L and by mass should be determined.

Concept introduction:

• Parts per million (ppm): It is one of the important concentration expressing terms, and which describes grams of solute per million grams of total solutions.

ppm=MassofsoluteMassofsolution×1061ppm=1mg/L

• Mass of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Massofsubstance

Explanation

Given,

Carbon monoxide concentration is 10â€‰ppm.

1â€‰ppmâ€‰â€‰=â€‰â€‰â€‰1â€‰mg/L10â€‰â€‰ppmâ€‰â€‰=â€‰â€‰10â€‰â€‰mg/L

Therefore the concentration of carbon monoxide in mg/L from the given value of ppm is,

10â€‰mg/L

â€‚Â 10 ppm of CO in mg/L is 10â€‰mg/L

The given value of concentration of carbon monoxide in ppm can also expressed in mass as follows

The molar mass of dry air is 28.96â€‰g/mol.

10â€‰mg/L of CO means in 1 L there are 0.001â€‰molâ€‰â€‰CO present

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