Moles

1.	What is the percentage composition of iron in Fe2O3?
	A. 	30%
	B. 	70%
	C. 	35%
	D. 	20%

 

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		Answer: 	B
	111.6+48=159.6;  111.6÷159.6x100=70%
    
 


2.	What is the percentage composition of chlorine in PCl5?
	A. 	17%
	B. 	15%
	C. 	85%
	D. 	51%

 

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		Answer: 	C
	
	31.0 + 177.5 = 208.5;  177.5 ÷ 208.5 x 100 = 85%
	
	
    
 


3.	What is the percentage composition of carbon in C6H12O6?
	A. 	40.0%
	B. 	6.6%
	C. 	53.3%
	D. 	20.2%

 

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		Answer: 	A
	72 + 12 + 96 = 180;  72 ÷ 180 x 100 = 40%
    
 


4.	What is the percentage composition of hydrogen in (NH4)2HPO4?
	A. 	3.0 %
	B. 	6.1 %
	C. 	6.8 %
	D. 	3.8 %

 

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		Answer: 	C
	
	28 + 9 +  31 + 64 = 132;  9÷132x100 = 6.8%
	
	
    
 


5.	The mole is a unit which expresses a

	A. 	number
	B. 	volume
	C. 	mass
	D. 	formula

 

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		Answer: 	A
	
    
 


6.	The mole can be compared to a dozen in the sense that

	A. 	both represent the same number of things.
	B. 	both represent the mass of things.
	C. 	each represents a specific mass of things.
	D. 	each represents a specific number of things.

 

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		Answer: 	D
	
	
	
	
    
 



7.	The unit which does not belong to the list below, because it does not 
	represent a specific number of items in a group, is

	A. 	six pack
	B. 	gram
	C. 	mole
	D. 	dozen

 

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		Answer: 	B
	
	
	
	
    
 


8.	The unit which does not belong to the list below, because it does not 
	represent a specific number of items in a group, is

	A. 	pair
	B. 	mole
	C. 	litre
	D. 	dozen

 

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		Answer: 	C
	
	
	
	
    
 


9.	Which unit below is not measured directly?

	A. 	gram
	B. 	mole
	C. 	litre
	D. 	metre

 

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		Answer: 	B
	
	
	
	
    
 


10.	The unit which does not belong to the list below 
	(it is not measured directly) is

	A. 	mL
	B. 	mol
	C. 	m
	D. 	s

 

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		Answer: 	B
	
	
	
	
    
 


11.	The number of atoms in one mole of calcium is _______ the 
	number of atoms in one mole of iron.

	A. 	greater than
	B. 	fewer than
	C. 	the same as

 

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		Answer: 	C
	
	
	
	
    
 


12.	How many moles of copper are required to have the same number 
	of atoms as one mole of zinc?

	A. 	one
	B. 	two
	C. 	three 
	D. 	four

 

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		Answer: 	A
	
    
 


13.	The statement "equal volumes of gases at the same temperature 
	and pressure, contain an equal number of particles," represents
	A. 	Boyle's Law.
	B. 	Charles' Law.
	C. 	Avogadro's Hypothesis.
	D. 	Ideal Gas Equation.

 

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		Answer: 	C
	
	
	
	
    
 


14.	Avogadro's Hypothesis states that, at the same temperature and pressure

	A. 	equal volumes of two gases have equal number of molecules.
	B. 	equal volumes of two gases have equal masses.
	C. 	equal masses of two gases have equal volumes.
	D. 	equal masses of two gases have equal numbers of molecules.

 

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		Answer: 	A
	
    
 


15.	What is the mass of 3.1 x 1024 atoms of lithium?
 
	A. 	1.5 g
	B. 	4.4 g
	C. 	36 g
	D. 	89 g

 

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		Answer: 	C
	
	mass = 3.1 x 1024 atoms ÷  6.02 x 1023 atoms/mol  x  6.94 g/mol = 36 g
	
	
    
 


16.	What is the mass of 7.50 x 1024 atoms of chromium?
 
	A. 	6.48 g
	B. 	12.4 g
	C. 	124 g
	D. 	648 g

 

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		Answer: 	D
	
	mass = 7.50 x 1024 atoms ÷  6.02 x 1023 atoms/mol  x  51.994 g/mol = 648 g
	
	
    
 



17.	What is the mass of 1.8 x 1022 atoms of lead?
 
	A. 	0.62 g
	B. 	0.69 g
	C. 	6.2 g
	D. 	6.9 g

 

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		Answer: 	C
	
	mass = 1.8 x 1022 atoms ÷  6.02 x 1023 atoms/mol  x  207.2 g/mol = 6.2 g
	
	
    
 


18.	Which of the following has the least number of atoms?

	A. 	56.17 g of Si
	B. 	53.96 g of Al
	C. 	52.00 g of Cr
	D. 	61.94 g of P

 

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		Answer: 	C
	
	# of atoms = mass  ÷  molar mass  x  6.02 x 1023  
	
	
    
 


19.	How many atoms are present in 3.7 g of aluminum?

	A. 	1.2 x 10-23
	B. 	 8.2 x 1022
	C. 	0.14  
	D. 	7.3 x 1023
	E. 	4.1 x 1022

 

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		Answer: 	B
	
	# of atoms = 3.7 g  ÷  26.98 g/mol  x  6.02 x 1023 atoms/mole 
	=   8.2 x 1022 atoms
	
    
 


20.	The number of molecules in 332.9 g of SeO2 is
	A. 	1.806 x 1025 molecules.
	B. 	2.000 x 1027 molecules.
	C. 	2.000 x 1026 molecules.
	D. 	1.806 x 1024 molecules.
	E. 	1.812 x 1024 molecules.

 

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		Answer: 	D
	
	# of molecules = 332.9 g  ÷  110.96 g/mol
	  x  6.02 x 1023 molecules/mole
	 =   1.806 x 1024 molecules
    
 



21.	The number of atoms in a 34.5 gram sample of NO2 gas at STP is
	A. 	1.35 x 1024 atoms.
	B. 	9.27 x 1023 atoms.
	C. 	4.52 x 1023 atoms.
	D. 	9.03 x 1023 atoms.
	E. 	9.16 x 1023 atoms.

 

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		Answer: 	A
	# of molecules = 34.5 g  ÷  46.0 g/mol  
   	x  6.02 x 1023 x 3 atoms/molecule 
	=   1.35 x 1024 atoms.
    
 


22.	The mass of 2.71 x 1023 molecules of N2 is:
	A. 	0.450 g
	B. 	12.6 g
	C. 	6.30  g
	D. 	62.2 g
	E. 	12.9 g

 

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		Answer: 	B
	
	
	mass = 2.71 x 1023 molecules  
	 ÷ 6.02 x 1023 molecules/mol  x  28.0 g/mol = 12.6 g
    
 


23.	The mass of 2.11 x 1023 molecules of N2O is
	A. 	0.350 g
	B. 	7.95 mg
	C. 	12.2 g
	D. 	15.4 g
	E. 	0.740 g

 

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		Answer: 	D
	
	mass = 2.11 x 1023 molecules ÷  6.02 x 1023 
	molecules/mol  x  44.0 g/mol = 15.4 g
	
    
 



24.	The number of molecules in 496 g of SO3 is

	A. 	2.679 x 10-25 molecules.
	B. 	3.73 x 1024 molecules.
	C. 	2.99 x 1026 molecules.
	D. 	2.99 x 1027 molecules.
	E. 	3.01 x 1027 molecules.

 

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		Answer: 	B
	
	# of molecules = 496 g  ÷  80.0 g/mol  
	x  6.02 x 1023 molecules/mole 
	=   3.73 x 1024 molecules
	
    
 


25.	The number of molecules in a 10.08 g sample of nitrogen gas is

	A. 	4.331 x 1023 molecules.
	B. 	6.068 x 1024 molecules.
	C. 	1.702 x 1026 molecules.
	D. 	1.674 x 1024 molecules.
	E. 	2.167 x 1023 molecules.

 

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		Answer: 	E
	
	# of molecules = 10.08 g  ÷  28.0 g/mol 
	  x  6.02 x 1023 molecules/mole 
	= 2.167 x 1023 molecules.
    
 


26.	Given 20.0 g of each of the following elements, which contains 
	the least number of atoms?

	A. 	Si
	B. 	Al
	C. 	Fe
	D. 	V

 

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		Answer: 	C
	
	# of atoms = 20.0 g ÷ molar mass x 6.02 x 1023
	(largest molar mass contains the smallest number of moles
	and therefore the least atoms)
    
 


27.	Which of the following has the greatest number of atoms?

	A. 	183.85 g of W
	B. 	107.9 g of Ag
	C. 	47.9 g of Ti
	D. 	40.3 g of Ne

 

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		Answer: 	D
	
	# of atoms = mass ÷ molar mass x 6.02 x 1023 
	
	
    
 



28.	If 1.0 g of each of the following gases is taken at STP, which 
	would occupy the greatest volume?

	A. 	CO
	B. 	H2O
	C. 	CH4
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	C
	volume  =  1.0 g ÷ molar mass x 22.4 L/mol  
	(smallest molar mass gives the largest moles and 
	therefore the greatest volume at STP)
	
    
 


29.	If 1.0 g of each of the following gases is taken at STP, which 
	would occupy the greatest volume?

	A. 	CO
	B. 	H2O
	C. 	C2H4
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	B
	
	volume  =  1.0 g ÷ molar mass x 22.4 L/mol   
	(smallest molar mass gives the largest
	moles and therefore the greatest volume at STP)
    
 


30.	If 1.0 g of each of the following gases is taken at STP, which 
	would occupy the greatest volume?

	A. 	CO
	B. 	SO2
	C. 	C2H6
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	A
	volume  =  1.0 g ÷ molar mass x 22.4 L/mol  
	(smallest molar mass gives the largest moles and 
	therefore the greatest volume at STP)
    
 


31.	If 20.0 g of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO
	B. 	H2O
	C. 	CH4
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	D
	volume  =  20.0 g ÷ molar mass x 22.4 L/mol 
	(largest molar mass gives the smallest moles and
	therefore the smallest volume at STP)
	
    
 



32.	If 14.0 g of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO
	B. 	H2O
	C. 	C2H6
	D. 	NH3
	E. 	They would all occupy the same volume.

 

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		Answer: 	C
	volume  =  14.0 g ÷ molar mass x 22.4 L/mol  
	(largest molar mass gives the smallest moles and 
	therefore the smallest volume at STP)
	
    
 


33.	If 5.0 g of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO
	B. 	SO2
	C. 	C2H6
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	B
	volume  =  5.0 g ÷ molar mass x 22.4 L/mol  (largest molar mass gives the smallest  
	moles and therefore the smallest volume at STP)
	
	
    
 


34.	If 3.0 g of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO2
	B. 	O2
	C. 	C2H6
	D. 	NO
	E. 	They would all occupy the same volume.

 

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		Answer: 	A
	volume  =  3.0 g ÷ molar mass x 22.4 L/mol  
	(largest molar mass gives the smallest moles and 
	therefore the smallest volume at STP)
    
 


35.	The volume occupied by 12.6 g of nitrogen gas at STP is

	A. 	10.1 L.
	B. 	0.0990 L.
	C. 	19.8 L.
	D. 	4.95 L.
	E. 	22.4 L.

 

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		Answer: 	A
	volume = 12.6 g ÷ 28.0 g/mol x 22.4 L/mol = 10.1 L
    
 


36.	What is the mass of a sample of C3H8 gas if it occupies 100. mL at S.T.P.?
	A. 	0.196 g
	B. 	0.509 g
	C. 	5.09 g
	D. 	19.6 g
	E. 	196 g

 

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		Answer: 	A
	mass  = 0.100 L ÷ 22.4 L/mol x 44.0 g/mol = 0.196 g
    
 


37.	The mass of 1.456 L of CO2 at STP is:
	A. 	0.3497 g. 
	B. 	1435  g. 
	C. 	1.477 x 10-3 g.
	D. 	2.860 g. 
	E. 	1698 g. 

 

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		Answer: 	D
	
	mass CO2 = 1.456 L ÷ 22.4 L/mol x 44.0 g/mol = 2.860 g
	
	
    
 


38.	The volume occupied by 17.28 g of oxygen gas at STP is

	A. 	12.10 L.
	B. 	24.19 L.
	C. 	.02410 L.
	D. 	.04821 L.
	E. 	13.58 L.

 

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		Answer: 	A
	
	volume = 17.28 g ÷ 32.0 g/mol x 22.4 L/mol = 12.10 L
	
	
    
 


39.	The volume occupied by 8.40 g of NO at STP is

	A. 	0.159 L.
	B. 	80.0 L.
	C. 	6.27 L.
	D. 	11.2 L.
	E. 	188.2 L.

 

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		Answer: 	C
	
	volume = 8.40 g ÷ 30.0 g/mol x 22.4 L/mol = 10.1 L
	
	
    
 


40.	The mass of 55.552 L of SO2 at STP is
	A. 	6.290 x 10-3 g.
	B. 	158.72  g. 
	C. 	19.425 g. 
	D. 	119.04 g. 
	E. 	18.952 g. 

 

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		Answer: 	B
	
	mass SO2 = 55.552 L ÷ 22.4 L/mol x 64.0 g/mol = 158.72 g
	
	
    
 


41.	The number of moles in 8.0 g NaOH is
 
	A. 	0.20 mol
	B. 	0.33 mol
	C. 	5.0 mol
	D. 	320 mol

 

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		Answer: 	A
	moles NaOH = 8.0 g ÷ 40 g/mol = 0.20 mol
    
 


42.	The number of moles in 2.0 g NaOH is
 
	A. 	0.010 mol
	B. 	0.020 mol
	C. 	0.050 mol
	D. 	0.20 mol

 

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		Answer: 	C
	
	moles NaOH = 2.0 g ÷ 40 g/mol = 0.050 mol
	
	
    
 


43.	The number of moles in 40.6 g of FeCl3 is
 
	A. 	0.125mol
	B. 	0.250 mol
	C. 	4.00 mol
	D. 	8.00 mol

 

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		Answer: 	B
	
	moles  of FeCl3 = 40.6 g ÷ 162.3 g/mol = 0.250 mol
	
	
    
 


44.	The number of moles in 4.50 g of H2O  is
 
	A. 	0.132 mol
	B. 	0.250 mol
	C. 	4.00 mol
	D. 	81.0 mol

 

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		Answer: 	B
	
	moles  of H2O = 4.50 g ÷ 18 g/mol = 0.250 mol
	
	
    
 


45.	The number of moles in 50.0 g of H2O  is
 
	A. 	0.132 mol
	B. 	0.360 mol
	C. 	2.78 mol
	D. 	8.10 mol

 

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		Answer: 	C
	
	moles  of H2O = 50.0 g ÷ 18 g/mol = 2.78 mol
	
	
    
 


46.	The number of moles in 50.0 g of Cl2O  is
 
	A. 	0.132 mol
	B. 	0.575 mol
	C. 	1.74 mol
	D. 	2.78 mol

 

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		Answer: 	B
	
	moles  of Cl2O = 50.0 g ÷ 87 g/mol = 0.575 mol
	
	
    
 


47.	The number of moles in 25.0 g of Cl2O  is
 
	A. 	0.285 mol
	B. 	0.570 mol
	C. 	2.78 mol
	D. 	3.50 mol

 

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		Answer: 	A
	moles  of Cl2O = 25.0 g ÷ 87 g/mol = 0.285 mol
    
 


48.	If 0.250 mol of iron are required for an experiment, then the 
	mass of iron required is

	A. 	0.0717 g
	B. 	0.250 g
	C. 	14.0 g
	D. 	55.8 g

 

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		Answer: 	C
	
	
	mass of Fe = 0.250 mol x 55.8 g/mol = 14.0 g
	
    
 


49.	If 1.20 mol of carbon tetrachloride, CCl4, are required for 
	an experiment, then the mass of carbon tetrachloride required is

	A. 	184 g
	B. 	151 g
	C. 	128 g
	D. 	  47 g

 

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		Answer: 	A
	mass of CCl4 = 1.2 mol x 153.8 g/mol = 184 g
    
 


50.	If 0.250 mol of sulphur, S8, are required for an experiment, then the 
	mass of sulphur required is

	A. 	32.1 g
	B. 	128 g
	C. 	8.02 g
	D. 	64.1 g

 

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		Answer: 	D
	
	mass of S8 = 0.250 mol x 256.5 g/mol = 64.1 g
	
	
    
 



51.	If 4.00 mol of turpentine, C10H16, are required for an experiment, then 
	the mass of turpentine required is

	A. 	34.0 g
	B. 	136 g
	C. 	304 g
	D. 	544 g

 

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		Answer: 	D
	
	mass of C10H16 = 4.00 mol x 136 g/mol = 544 g
	
	
    
 



52.	The mass of 0.0202 mol of sodium hydroxide is

	A. 	0.0202 g
	B. 	0.808 g
	C. 	2.02 g
	D. 	40.0 g

 

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		Answer: 	B
	
	mass of NaOH = 0.0202 mol x 40.0 g/mol = 0.808 g
	
	
    
 


53.	The mass of 1.20 mol of calcium oxide (lime) is

	A. 	0.0214 g
	B. 	46.7 g
	C. 	56.1 g
	D. 	67.2 g

 

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		Answer: 	D
	
	mass of CaO = 1.20 mol x 56 g/mol = 67.2 g
	
	
    
 



54.	How many moles of calcium are in 120.24 g of calcium?

	A. 	one mole
	B. 	two moles
	C. 	three moles
	D. 	four moles

 

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		Answer: 	C
	
	moles of Ca = 120.24 g ÷ 40.06 g/mol = 3 moles
	
	
    
 


55.	How many grams of barium are required to make 2 moles of barium atoms?

	A. 	137.3 g
	B. 	274.6g
	C. 	411.9 g
	D. 	549.3 g

 

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		Answer: 	B
	
	mass of Ba = 2 moles x 137.3 g/mol = 274.6
	
	
    
 


56.	What is the mass of 2.50 moles of iodine gas?

	A. 	634.5 g
	B. 	126.9 g
	C. 	317.3 g
	D. 	253.8 g

 

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		Answer: 	A
	mass of I2 = 2.50 mol x 253.8 g/mol = 634.5 g
    
 


57.	What is the mass of two moles of  bromine?

	A. 	319.6 g
	B. 	159.8 g
	C. 	79.90 g
	D. 	639.2 g

 

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		Answer: 	A
	mass of Br2 = 2.0 mol x 159.8 g/mol = 319.6 g
    
 


58.	What is the mass of two moles of phosphorus?

	A. 	61.95 g
	B. 	123.9 g
	C. 	30.97 g
	D. 	247.8 g

 

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		Answer: 	D
	
	mass of P4 = 2.0 mol x 123.9 g/mol = 247.8 g
	
	
    
 



59.	The mass of 4.50 moles of Na2CO3 is
	A. 	106 g.
	B. 	477 g.
	C. 	23.6 g.
	D. 	0.042 g.
	E. 	465 g.

 

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		Answer: 	B
	
	
	mass of Na2CO3 = 4.50 mol x 106 g/mol = 477 g
	
    
 


60.	The number of moles in 7.2 x 10-2 g of Lead is
	A. 	15 moles. 
	B. 	2.8 x 103 moles.
	C. 	4.4 x 1022 moles.
	D. 	3.5 x 10-4 moles.
	E. 	3.0 x 10-4 moles.

 

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		Answer: 	D
	
	moles Pb = 7.2 x 10-2 g ÷ 207.2 g/mol = 3.5 x 10-4 moles
	
	
    
 



61.	The number of moles in 155 g of Titanium is

	A. 	3.24 moles.
	B. 	7428 moles.
	C. 	9.34x 1025 moles.
	D. 	0.309 moles.
	E. 	757 moles.

 

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		Answer: 	A
	moles Ti = 155 g ÷ 47.88 g/mol = 3.24 moles
    
 


62.	The number of moles in 158.9 g of Magnesium is

	A. 	13.24 moles.
	B. 	3862 moles.
	C. 	0.1529 moles.
	D. 	6.538 moles.
	E. 	13.16 moles.

 

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		Answer: 	D
	
	moles Mg = 158.9 g ÷ 24.305 g/mol = 6.538 moles
	
	
    
 


63.	How many moles of elemental chlorine is required to have the 
	same number of atoms as two moles of of elemental phosphorus?
	A. 	two moles
	B. 	three moles
	C. 	four moles
	D. 	one mole

 

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		Answer: 	C
	
	Chlorine is diatomic (Cl2) and phosphorus is tetraatomic 
	(P4) so twice as many.
	
    
 


64.	An unknown gas fills a 56.0 L container at STP.  How many moles of 
	gas are in the container?

	A. 	0.200 mol
	B. 	0.400 mol
	C. 	2.00 mol
	D. 	2.50 mol
	E. 	5.00 mol

 

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		Answer: 	D
	
	
	moles = 56.0 L ÷ 22.4 L/mol = 2.50 mol
	
    
 



65.	The number of moles in 195 L of carbon monoxide gas at STP is

	A. 	8.70 moles.
	B. 	0.115 moles.
	C. 	4368 moles.
	D. 	8.00 moles.
	E. 	2.75 moles.

 

    Depress mouse to view answer.
   	
		Answer: 	A
	moles = 195 L ÷ 22.4 L/mol = 8.70 mol
    
 


66.	The volume occupied by 42.39 moles of PH3(g) at STP is
	A. 	949.5 L. 
	B. 	1.053 mL. 
	C. 	1441 L. 
	D. 	6.940 x 10-4 L.
	E. 	995.4 L. 

 

    Depress mouse to view answer.
   	
		Answer: 	A
	
    
 


67.	The volume occupied by 0.0025 moles of NH3(g) at STP is
	A. 	1.1 x 10-4 L. 
	B. 	56 mL. 
	C. 	0.042 L. 
	D. 	0.95 L. 
	E. 	1.0 L. 

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	volume = 0.0025 mol x 22.4 L/mol = 0.056 L or 56 mL
	
	
    
 


68.	The number of moles in 1987 L of NO2(g)2(g) at STP is
	A. 	0.01127 moles.
	B. 	43.21 moles.
	C. 	967.8 moles.
	D. 	88.70 moles.
	E. 	966.3 moles.

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	moles = 1987 L ÷ 22.4 L/mol = 88.70 mol
	
	
    
 



69.	An unknown gas fills a 4.48 L container at STP.  How many moles of 
	gas are in the container?

	A. 	0.200 mol
	B. 	0.400 mol
	C. 	2.00 mol
	D. 	2.50 mol
	E. 	5.00 mol

 

    Depress mouse to view answer.
   	
		Answer: 	A
	
	moles = 4.48 L ÷ 22.4 L/mol = 0.200 mol
	
	
    
 


70.	If 4.0 moles of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO2
	B. 	O2
	C. 	C2H6
	D. 	NO
	E. 	They would all occupy the same volume.

 

    Depress mouse to view answer.
   	
		Answer: 	E
	Avogadro's hyposthesis: equal moles of different 
	gases at the same temperature and pressure will
	occupy the same volume.
	
    
 


71.	If 20.0 moles of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO
	B. 	H2O
	C. 	CH4
	D. 	NO
	E. 	They would all occupy the same volume.

 

    Depress mouse to view answer.
   	
		Answer: 	E
	Avogadro's hyposthesis: equal moles of different 
	gases at the same temperature and pressure will
	occupy the same volume.
	
    
 


72.	If 14.0 moles of each of the following gases is taken at STP, which 
	would occupy the smallest volume?

	A. 	CO
	B. 	H2O
	C. 	CH4
	D. 	NH3
	E. 	They would all occupy the same volume.

 

    Depress mouse to view answer.
   	
		Answer: 	E
	Avogadro's hyposthesis: equal moles of different gases
	at the same temperature and pressure 
	 will occupy the same volume.
	
    
 


73.	100. litres of any gas at STP would contain how many moles?

	A. 	2 240 mol
	B. 	100. mol
	C. 	22.4 mol
	D. 	4.46 mol
	E. 	2.24 mol

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	moles = 100. L ÷ 22.4 L/mol = 4.46 mol
	
	
    
 



74.	10.0 litres of any gas at STP would contain how many moles?

	A. 	224 mol
	B. 	22.4 mol
	C. 	10.0mol
	D. 	4.46 mol
	E. 	0.446 mol

 

    Depress mouse to view answer.
   	
		Answer: 	E
	
	moles = 10.0 L ÷ 22.4 L/mol = 0.446 mol
	
	
    
 


75.	What is the volume, at STP,  of 3.06 x 1022 atoms of neon?
 
	A. 	1.14 litre
	B. 	5.10 litre
	C. 	35.1 litre
	D. 	114 litre

 

    Depress mouse to view answer.
   	
		Answer: 	A
	volume = 3.06 x 1022 atoms ÷ 6.02 x 1023 atoms/mol x 22.4 L/mol
   	 = 1.14 L
   	
    
 


76.	What is the volume, at STP,  of 3.06 x 1024 atoms of helium?
 
	A. 	1.14 litre
	B. 	5.10 litre
	C. 	35.0 litre
	D. 	114 litre

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	volume = 3.06 x 1024 atoms ÷ 6.02 x 1023 atoms/mol x 22.4 L/mol
	 = 114 L
	
    
 



77.	8.33 litres of neon gas at STP would contain how many atoms?

	A. 	2.24 x 1023 atoms
	B. 	2.68 x 1023 atoms
	C. 	2.24 x 1024 atoms
	D. 	2.68 x 1024 atoms
	E. 	2.24 x 1022 atoms

 

    Depress mouse to view answer.
   	
		Answer: 	A
	atoms = 8.33. L ÷ 22.4 L/mol  x 6.02 x 1023 atoms/mol
   	  = 2.24 x 1023 L
   	
    
 


78.	Upon analysis, a compound was found to contain 42.9% of carbon and 
	57.1% of oxygen.  The empirical formula is

	A. 	CO
	B. 	C2O
	C. 	CO2
	D. 	CO3
	E. 	C3O2

 

    Depress mouse to view answer.
   	
		Answer: 	A
	moles C = 42.9 g ÷ 12 g/mol =3.58 mol
	moles O = 57.1 g ÷ 16 g/mol = 3.57 mol 
	Ratio C to O is 1:1.
    
 


79.	Upon analysis, a compound was found to contain 27.3% of carbon and 
	72.7% of oxygen.  The empirical formula is

	A. 	CO
	B. 	C2O
	C. 	CO2
	D. 	CO3
	E. 	C3O2

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	moles C = 27.3 g ÷ 12 g/mol =2.28 mol
	moles O = 72.7 g ÷ 16 g/mol = 4.54 mol           
	Ratio O to C =4.54 ÷ 2.28 = 1.99 or 2  :1
    
 

81.	Upon analysis, a compound was found to contain 63.53% of iron and 
	36.47% of sulfur.  The empirical formula is

	A. 	FeS
	B. 	FeS2
	C. 	Fe2S
	D. 	Fe2S3
	E. 	Fe3S2

 

    Depress mouse to view answer.
   	
		Answer: 	A
	moles Fe = 63.53 g ÷ 55.8 g/mol =1.14 mol
	moles S = 36.47 g ÷ 32 g/mol =1.14 mol
	Ratio Fe to S = 1:1
    
 


82.	Upon analysis, a compound was found to contain 23.3% of magnesium, 
	30.7% of sulphur and 46.0% of oxygen.  The empirical formula is

	A. 	MgSO
	B. 	Mg2SO
	C. 	MgSO2
	D. 	MgSO3
	E. 	MgSO4     

 

    Depress mouse to view answer.
   	
		Answer: 	D
	moles Mg = 23.3 g ÷ 24.3 g/mol =0.959 mol
	moles S = 30.7 g ÷ 32 g/mol =0.959 mol
	moles O = 46.0 g ÷ 16 g/mol =2.88 mol
	Ratio Mg to S = 1:1   Ratio O to Mg = 2.88 ÷ 0.959 = 3.00
    
 



84.	After substantial heating, 0.194 g of uranium produced 0.246 g of a 
	compound with oxygen.  The empirical formula is

	A. 	UO2
	B. 	U5O2
	C. 	UO4
	D. 	U2O5
	E. 	UO3

 

    Depress mouse to view answer.
   	
		Answer: 	C
	mass of oxygen = 0.246 - 0.194 = 0.052 g
	moles U = 0.194 g ÷ 238 g/mol = 0.000815 mol
	moles O = 0.052 g ÷ 16 g/mol = 0.00325 mol  
	Ratio O to U =0.00325  ÷ 0.000815 = 3.99
    
 


86.	A sample of a compound contains 1.36 moles of nitrogen atoms and 
	2.72 moles of oxygen atoms.  What is the empirical formula?

	A. 	NO
	B. 	N2O
	C. 	NO2
	D. 	N2O3
	E. 	N3O2

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	moles of O ÷ moles of N = 2.72 ÷ 1.36 = 2.00
	
	
    
 


87.	A sample of a compound contains 1.36 moles of nitrogen atoms and 
	0.68 moles of oxygen atoms.  What is the empirical formula?

	A. 	NO
	B. 	N2O
	C. 	NO2
	D. 	N2O3
	E. 	N3O2

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	moles of N ÷ moles of O = 1.36 ÷ 0.68 = 2.00
	
	
    
 


88.	A sample of a compound contains 1.81 moles of nitrogen atoms and 
	2.72 moles of oxygen atoms.  What is the empirical formula?

	A. 	NO
	B. 	N2O
	C. 	NO2
	D. 	N2O3
	E. 	N3O2

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	moles of O ÷ moles of N = 2.72 ÷ 1.81 = 1.50 or 3:2
	
	
    
 



89.	A sample of a compound contains 4.08 moles of nitrogen atoms and 
	2.72 moles of oxygen atoms.  What is the empirical formula?

	A. 	NO
	B. 	N2O
	C. 	NO2
	D. 	N2O3
	E. 	N3O2

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	moles of N ÷ moles of O = 4.08 ÷ 2.72 = 1.50 or 3:2
	
	
    
 


90.	At S.T.P. 22.4 L of a gas has a mass of 32 grams.  Identify the gas.

	A. 	oxygen
	B. 	nitrogen
	C. 	fluorine
	D. 	chlorine
	E. 	argon

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	moles of gas = 22.4 ÷ 22.4 = 1.00 mole
	molar mass = 28 g ÷ 1.00 mole = 28 g/mole
	N2 or Nitrogen
    
 


91.	What is the mass of one mole of a gas which has a density of 
	2.86 g/L at STP?

	A. 	0.130 g
	B. 	7.80 g
	C. 	8.60 g
	D. 	64.1 g

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	moles of gas = 1.00 L ÷ 22.4 = 0.0446 mole
	molar mass = 2.86 g ÷ 0.0446 mole = 64.1 g/mole
	
    
 



92.	An ideal gas has a density of 1.20 g/L at S.T.P.  The molar
	mass of the gas is

	A. 	18.7 g.
	B. 	25.4 g.
	C. 	26.9 g.
	D. 	44.2 g.
	E. 	35.6 g.

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	moles of gas = 1.00 L ÷ 22.4 = 0.0446 mole
	molar mass = 1.20 g ÷ 0.0446 mole = 26.9 g/mole
	
    
 


93.	1.0 mol is represented by
	A. 	78 g of potassium sulfide.
	B. 	164 g of calcium nitrate.
	C. 	188 g of chromiun (III) oxide.
	D. 	92 g of carbon tetrabromide.

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	
	Ca(NO3)2 = 40 + (2 x 14) + (6 x 16) = 164 g/mol
	
    
 


94.	1.0 mol is represented by
	A. 	78 g of potassium sulfide.
	B. 	102 g of calcium nitrate.
	C. 	152 g of chromiun (III) oxide.
	D. 	92 g of carbon tetrabromide.

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	Cr2O3 = (2 x 52) + (3 x 16) = 152 g/mol
	
	
    
 


95.	1.0 mol is represented by
	A. 	78 g of potassium sulfide.
	B. 	102 g of calcium nitrate.
	C. 	188 g of chromiun (III) oxide.
	D. 	332 g of carbon tetrabromide.

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	CBr4 = 12 + (4 x 80) = 332 g/mol
	
	
    
 



96.	What is the molar mass of Mg(NO3)2?
	A. 	148.3 g
	B. 	86.3 g
	C. 	134.3 g
	D. 	172.6 g

 

    Depress mouse to view answer.
   	
		Answer: 	A
	Mg(NO3)2 = 24.3 + (2 x 14) + (6 x 16) = 148.3 g/mol
	
	
    
 


97.	What is the molar mass of SrBr2?
	A. 	167.5 g
	B. 	87.62 g
	C. 	247.4 g
	D. 	255.1 g

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	
	SrBr2 = 87.6 + (2 x 79.9)  = 247.4 g/mol
	
    
 


98.	What is the molar mass of Co(CH3COO)2?
	A. 	117.9 g
	B. 	120.9 g
	C. 	146.0 g
	D. 	152.9 g
	E. 	238.7 g

 

    Depress mouse to view answer.
   	
		Answer: 	D
	
	Co(CH3COO)2 = 58.9 + (2 x 12) + (4 x 16) + (6 x 1.0) = 152.9 g/mol
	
	
    
 


99.	The mass of 0.25 mole of a compound with the molecular 
	formula A3B is 17 g. If the molar mass of element A is 11 g/mol, 
	the molar mass of element B, in g/mol, is

	A. 	57
	B. 	19
	C. 	35
	D. 	33

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	molar mass compound = 17 g ÷ 0.25 mol = 68 g/mol
	molar mass of element B = 68 - 3(11)  = 35 g/mol
	
    
 


100.The mass of 0.25 mole of a compound with the molecular formula 
	AB is 17 g. If the molar mass of element A is 11 g/mol, the molar 
	mass of element B, in g/mol, is

	A. 	57
	B. 	19
	C. 	35
	D. 	6

 

    Depress mouse to view answer.
   	
		Answer: 	A
	molar mass compound = 17 g ÷ 0.25 mol = 68 g/mol
	molar mass of element B = 68 - 11 = 57 g/mol
    
 


101.	What is the mass of one mole of elemental sulphur?

	A. 	32.06 g
	B. 	256.48 g
	C. 	64.12 g
	D. 	128.24 g

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	molar mass of S8 = 8 (32.06) = 256.48 g/mol
	
	
    
 


102.	What is the mass of one mole of elemental phosphorus?

	A. 	31 g
	B. 	62 g
	C. 	124 g
	D. 	248 g

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	molar mass of P4 = 4 (31) = 124 g/mol
	
	
    
 


103.	What is the mass of one mole of elemental fluorine?

	A. 	19 g
	B. 	38 g
	C. 	76 g
	D. 	152 g

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	molar mass of F2 = 2 (19) = 38 g/mol
	
	
    
 


104.	What is the mass of one mole of elemental nitrogen?

	A. 	14 g
	B. 	28 g
	C. 	56 g
	D. 	112 g

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	molar mass of N2 = 2 (14) = 28 g/mol
	
	
    
 


105.	What is the mass of one mole of elemental iodine?

	A. 	127 g
	B. 	254 g
	C. 	508 g
	D. 	1016 g

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	molar mass of I2 = 2 (127) = 254 g/mol
	
	
    
 


106.	The molar volume of Carbon Dioxide gas, at S.T.P., is

	A. 	1.00 L.
	B. 	6.02 L.
	C. 	22.4 L.
	D. 	273 L.
	E. 	24.5 L.

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	
	Any gas at STP occupies 22.4 L of space!
	
    
 


107.All gases with equal temperature, pressure and number of moles, 
	are also equal in

	A. 	mass.
	B. 	volume.
	C. 	speed.
	D. 	diffusion rate.

 

    Depress mouse to view answer.
   	
		Answer: 	B
	
	Avogadro's Hypothesis
	
	
    
 


108.What is the molecular formula for a compound that is 85.8 % C 
	and 14.2% H and has a molar mass of 112 g/mole?

	A. 	C6H6
	B. 	C8H16
	C. 	C4H6
	D. 	C5H12

 

    Depress mouse to view answer.
   	
		Answer: 	B
	moles of C = 85.8 g ÷ 12 g/mol = 7.15 mol
	moles of H = 14.2 g ÷ 1.0 g/mol = 14.2 mol
	ratio of H to C = 14.2 ÷ 7.15 = 1.99 or CH2 with a molar mass of 12 + 2(1) = 14 g/mol
	112 g/mol ÷ 14 g/mol = 8  so 8 times CH2 is C8H16.
    
 


109.What is the molecular formula for a compound that is 92.3% C 
	and 7.7% H and has a molar mass of 78.0 g/mole?

	A. 	C6H6
	B. 	C8H16
	C. 	C4H6
	D. 	C5H12

 

    Depress mouse to view answer.
   	
		Answer: 	A
	moles of C = 85.8 g ÷ 12 g/mol = 7.15 mol
	moles of H = 14.2 g ÷ 1.0 g/mol = 14.2 mol
	ratio of H to C = 14.2 ÷ 7.15 = 1.99 or CH2 l
	with a molar mass of 12 + 2(1) = 14 g/mo
	112 g/mol ÷ 14 g/mol = 8  so 8 times CH2 is C8H16.
	
    
 


110.The molar mass of a compound is 58.  What is the molecular formula 
	of the compound if the empirical formula of the compound is C2H5?
	A. 	C2H5
	B. 	C6H15
	C. 	C4H10
	D. 	C8H20

 

    Depress mouse to view answer.
   	
		Answer: 	C
	
	molar mass of C2H5 is 29 g/mol.
	ratio = 58 ÷ 29 = 2.......so moleclar formula = 2(C2H5) or C4H10