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
Answer = 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
Answer =
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