CHM 2045C Module Six Sample Exam ANSWERS

Module Six Part H-1 Gas Density Problem                   3 points

Calculate the gas density of octane gas C8H18 vapor at STP?

First: find the volume of gas:  1.00 mole C8H18  = 22.4 L@STP by definition

Second: what is the Mass of 1 Mole of C8H18 :

8 carbons @ 12.0 =    96.0 g carbon atoms

18 hydrogens@ 1.0 = 18.0 g hydrogen atoms

-----------------------------------------------------

1 mole C8H18  =    114.0 g C8H18

Third: Find the density of octane:

Density @STP = mass/volume  = 114.0 g C8H18 / 22.4 L@STP = 5.09 g/L@STP

Module Six Part H-2 Molecular Mass Determination Problem  2 points

Calculate the molecular weight of an unknown liquid that when vaporized at 99 oC and 755 torre , gave 125.0 mL of vapor with a mass of 0.673 grams .

PV = nRT

P=755/760 atm = 0.993 atm

T= 99+273=372K

V= 120.0 mL/1000mL/L = 0.120L

n= PV/RT   = ( 0.993 ) (0.120) / (372) (0.08205) mole = 0.00390 moles

Molar Mass = Mass/Mole = 0.673 g/0.00390 mol = 172 g/mol

Module Six Part I Effusion of Gasses Problem                   5 points

A sample of Nitrogen gas escapes through a tiny hole in 44.0 seconds.  An unknown amount of a gas escapes under the same conditions in 80.0 seconds.  Calculate the molecular mass of the unknown.

Rate of effusion of gas 1           ( molar mass of gas 2 ) ½

----------------------------------    =       ----------------------------

Rate of effusion of gas 2            ( molar mass of gas 1 ) ½

Rate1 = Volume escaped1/Time elapsed1        R1 = V1/ T1    =    V1 /  44.0 seconds

Rate2 = Volume escaped2/Time elapsed2        R2 = V2/ T2   =    V2 /  80.0 seconds

Volume escaped1 = Volume escaped2    V1  =   V2  therefore by substitution:

R2 =  V1 / 80.0

Molar mass N2 = 28.0 g/mole

Molar Mass Gas2  =  ?  g/mole

Substitute the values in the equation:

Rate of effusion of gas 1           ( molar mass of gas 2 ) ½

----------------------------------    =       ----------------------------

Rate of effusion of gas 2            ( molar mass of gas 1 ) ½

V1 /  44.0 seconds                           ( Mm 2 ) ½

------------------------         =            -----------------

V1 / 80.0 seconds                            ( 28.0 g/mole ) ½

Square both sides to remove the square root on the right, Cancel V1 on left

( 80.0 / 44.0 )2  =  Mm 2 /  28.0 g/mole

Cross multiple and divide 80.0 by 44.0

( 1.82 )2  28.0 g/mole =  Mm 2

Square the 1.82 and multiply by 28.0

Answer:  92.7 g/mole =  Mm 2