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 )

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

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

 

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 )

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

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

 

 

 

V1 / 44.0 seconds ( Mm 2 )

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

V1 / 80.0 seconds ( 28.0 g/mole )

 

 

 

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