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Chem 101: Help with Unit 5 Assessment Question #9

chem101

#1
  1. What is the pressure in atmospheres of a gas mixture that consists of 8.80 grams of nitrogen and 8.80 grams of carbon dioxide in a 2.01 liter container at 27°C?
    Select one:
    a. 6.30 atm Correct
    b. 3.85 atm
    c. 2.45 atm
    d. 0.971 atm

My calculations give an answer of 10.1 atm. Thanks in advance for any guidance correcting my calculations.

P=nRT/V ; solve for pressure

n=0.828 mol ; convert to mol 8.80g N * 1 mol/14.007g N = 0.628 mol N + 8.80 g CO2 * 1 mol/43.999 g CO2 = 0.200 mol CO2

R = 0.0821 Latm/molK

T = 300 K ; convert celsius to kelvin 27 + 273.15

V = 2.01 L

P = (0.828 mol * 300 K * 0.0821 L * atm)/(2.01 L * mol * K)

P = 10.1 atm ; rounded to 3 sig. digits


#2

I get the same, but I’m no chemist. I can get the given answer, 6.30 atm, if I use 4.40 grams nitrogen instead of 8.80 – seems like maybe that’s what we meant.

@jazinheira care to dust off your chemistry?
cc: @devrit

FWIW, I used these:
https://www.convertunits.com/from/grams+Nitrogen/to/moles
https://www.convertunits.com/from/grams+Carbon+Dioxide/to/moles
https://www.chemicool.com/cgi-bin/gaslaws.pl


#3

Yep. I got the same answer. 10.14 atm.

You can either calculate the pressure of each individual gas and add them or figure out the number of mols for both gases combined. It looks like that’s what the question is trying to test.


#4

@jazinheira so will it work to change the question to 4.40 grams of nitrogen rather than 8.80 grams, so that our given answer of 6.30 is correct?


#5

Sorry @sean! Thought I had replied to this earlier. Yeah I agree; it looks like 8.80 grams of Nitrogen is a typo. If I use 4.40 grams of Nitrogen, we get the ‘correct’ answer of 6.30 atm. Also, the partial pressure of just the Nitrogen alone comes out conveniently to 3.85 atm (answer b).

@devrit


#6

Thanks all. I’ve updated this assessment question in the course.