A 1.43-g sample of an unknown gas has a volume of 804 mL and a pressure of 852 mmHg at 59.0 °C. Calculate the molar mass of this compound.

General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter5: The Gaseous State
Section: Chapter Questions
Problem 5.127QP: A 1.000-g sample of an unknown gas at 0C gives the following data: P(atm) V (L) 0.2500 3.1908 0.5000...
icon
Related questions
Question
A 1.43-g sample of an unknown gas has a volume of 804 mL and a pressure of 852 mmHg at 59.0 °C. Calculate the molar mass of this compound.
Approach
g/mol
The ideal gas law contains four variables (P. V. n, and T) and a constant (R = 0.082057 L atm/K mol):
PV = nRT
If you know three of the four variables, you can calculate the fourth. All properties must have units that match those in the gas constant: volume in liters, pressure in atmospheres, temperature in kelvins, and amount in moles. In this case you can solve for the amount (mol) of gas, n.
Once n is determined, the molar mass is calculated by dividing the mass by the amount (mol) of gas.
Transcribed Image Text:A 1.43-g sample of an unknown gas has a volume of 804 mL and a pressure of 852 mmHg at 59.0 °C. Calculate the molar mass of this compound. Approach g/mol The ideal gas law contains four variables (P. V. n, and T) and a constant (R = 0.082057 L atm/K mol): PV = nRT If you know three of the four variables, you can calculate the fourth. All properties must have units that match those in the gas constant: volume in liters, pressure in atmospheres, temperature in kelvins, and amount in moles. In this case you can solve for the amount (mol) of gas, n. Once n is determined, the molar mass is calculated by dividing the mass by the amount (mol) of gas.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
Mole Concept
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
General Chemistry - Standalone book (MindTap Cour…
General Chemistry - Standalone book (MindTap Cour…
Chemistry
ISBN:
9781305580343
Author:
Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:
Cengage Learning
Chemistry for Engineering Students
Chemistry for Engineering Students
Chemistry
ISBN:
9781337398909
Author:
Lawrence S. Brown, Tom Holme
Publisher:
Cengage Learning
Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781133611097
Author:
Steven S. Zumdahl
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781337399074
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning