Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 11 - Gases - Exercises - Problems - Page 403: 71


The molar mass of the gas is $4.00$ grams/mol.

Work Step by Step

Using the ideal gas law $PV=nRT$ where "P" is the pressure the gas exerts, "V" is the volume the gas occupies, "n" represents the number of moles of the gas, "R" is the ideal gas constant and "T" represents the temperature, we can solve many gaseous questions that behave in ideal conditions. The question asks for the molar mass of the gas so we must isolate for the number of moles first from the ideal gas equation to obtain $n = \frac{PV}{RT}$. First we must convert the pressure in units of atm; $886 torr\times\frac{1 atm}{760 torr} =1.16579 atm$. We must also convert the temperature into Kelvin by simply adding $273.15$ to the Celcius value; $55^{\circ}C + 273.15 = 328.15 K$ $n = \frac{(1.16579 atm)(0.224 L)}{(0.08206 \frac{L\times atm}{mol\times K})(328.15 K)} = 9.697603486\times10^{-3} mols $. Lastly, we have to convert the mg to grams so $38.8 mg\div 1000 = 0.0388g$ Now that we have the number of moles, we can use the fact that the Molar Mass is the mass/number of moles. Thus the Molar mass is $0.0388 grams\div9.697603486\times10^{-3} mols = 4.00 g/mol$
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