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Mole Concepts Problems with Solutions





Given below are the Mole Concepts Questions with Detailed Solutions
a. Concepts questions
b. Calculation problems
c. percentage composition
d. Mole fraction & molarity

Known Values
Mole Concepts Numericals with Detailed Solutions

Question 1
How many molecules of water are there in 54 g of H2O ?

Answer

Molar Mass of H2O = 2 + 16 = 18 g/moles
So ,number of moles of H2O = Mass/Molar Mass = 54/18 =3 moles
Now 1 moles = 6.022×1023 molecule
So 3 moles will have 18.066×1023 molecule


Question 2
Calculate the mass of 6.022×1023 molecule of NH4Cl ?

Answer

Molar mass (Molecular mass in gram) of NH4Cl= 14+4+35.5= 53.5g
No. of moles of NH4Cl
=6.022×10236.022×1023
= 1 mole
Mass of NH4Cl
=No. of moles×molar mass
=1×53.5g=53.5g.


Question 3
Calculate the mass of 12.044×1023 Oxygen atoms.

Answer

No. of moles of Oxygen atoms
=12.044×10236.022×1023
= 2 mole
Mass of Oxygen atoms
=No. of moles×atomic mass
=2×16
= 32 g


Question 4
How many atoms of hydrogen are there in 36 g of NH4?

Answer

Molar mass (Molecular mass in gram) of NH4= 14+4= 18 g
No. of moles of NH4?
= 36/18 = 4 moles
Now Total Moles of Hydrogen Atoms
= 16 moles
=16x6.022×1023=96.352×1023


Question 5
Calculate the number of hydrogen atoms in 1 mole of H2.

Answer

1 molecule of H2= 2 hydrogen atoms
So, 1 mole of H2= 2 mole hydrogen atoms
=2×6.022×1023=12.044×1023 hydrogen atoms.


Question 6
Calculate the number of Cu atoms in 0.3175 g of Cu.

Answer

No. of moles of Cu
=Mass of CuAtomic mass
= 0.3175/63.5
=0.005 mole
No. of Cu atoms
=No. of moles×Avogadro constant
=0.005×6.022×1023
=30.11×1020 Cu atoms.



Question 7
Calculate the number of molecules in 22.4 litres of CH4 gas at NTP.

Answer

1 mole of CH4= 22.4 L (at NTP)
Therefore
22.4 L of CH4gas= 1 mole CH4gas= 6.022×1023 CH4 gas molecules.


Question 8
a. Find the volume of 1 g of H2gas in litres at N.T.P
b.Find the volume of 20g H2at NTP.
c.What is the volume occupied by 6.022×1023 molecules of any gas at NTP?

Answer

a. Molar Mass of H2gas = 2 gm
Number of moles=MassMolar Mass=12=.5
1 mole of H2= 22.4 L (at NTP)
Therefore
.5 mole of H2= .5 x 22.4 = 11.2 litre
b.
No. of moles of H2= 20/2 =10
1 mole of H2= 22.4 L (at NTP)
Therefore
10 moles = 10×22.4
=224 L
c.
6.022×1023 molecules = 1 mole molecules, and
1 mole molecules of any ideal gas occupies 22.4 L at NTP.



Question 9
An atom of some element Y weighs 6.644×1023 g. Calculate the number of gram-atoms in 40 kg of it.

Answer

Mass of 1 mole Y atoms
=mass of 1 atom×Avogadro constant
=6.644×1023×6.022×1023
= 40 g
So, the atomic mass of Y = 40
No. of gram-atoms (or moles) of X
=mass of Yatomic mass
=40×100040=1000


Question 10
Find the number of moles and number of atoms of H and S in 10 mole of H2S.

Answer

1 mole of H2S contains 2 mole of H, 1 mole of S
Therefore
10 mole of H2S contains
20 mole of H = 20×6.022×1023=12.044×1024 H atoms
10 mole of S = 10×6.022×1023=6.022×1024 S atoms


Question 11
Calculate the number of atoms of each element in 245 g of KClO3.

Answer

Molecular mass of KClO3= 39+35.5+3×16=122.5
No. of mole of KClO3= 245 g/122.5g = 2 mole
2 mole of KClO3contains
2 mole of K = 2×6.022×1023=12.044×1023 K atoms
2 mole of Cl = 2×6.022×1023=12.044×1023 Cl atoms
6 mole of O = 6×6.022×1023=1.806×1024 O atoms.


Question 12
If 4 g of NaOH dissolves in 36 g of H2O, calculate the Mole fraction of each component in the solution. Also, determine the Molarity of solution (specific gravity of solution is 1 gm/L)

Answer

Mass of NaOH = 4 g
Number of moles of = 4 g NaOH/40 g = 0.1 mol
Mass of H2O = 36 g
Number of moles of = 36 g/18= 2 mol
Mole fraction of water=Number of moles of water No. of moles of water+No. of moles of NaOH =22+.1=0.95
Mole fraction of NaOH=Number of moles of NaOHNo. of moles of NaOH+ No. of moles of water =0.12+0.1=0.047
Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
Volume of solution = 40×1=40 mL (Since specific gravity of solution is = 1 g m/L)
Molarity of solution=Number of moles of solute Volume of solution in litre=0.10.04=2.5 M


Question 13
Calculate the number of atoms in 12.2 litres of the below gas at NTP
(i) mono-atomic
(ii) diatomic gas

Answer

No. of moles in 12.2 L gas at NTP = 12.2/22.4 = 0.50
No. of molecules in 12.2 L gas
=0.50×6.022×1023
=3.011×1023 molecules
In mono-atomic gases, No. of atoms = No. of molecules =3.011×1023.
In diatomic gases, No. of atoms = 2×No. of molecules=2×3.011×1023=6.022×1023.


Question 14
If a mole were to contain 1×1024 particles, what would be the mass of
(i) one mole of oxygen
(ii) a single oxygen molecule?

Answer

Mass of one mole of oxygen molecule (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
=321×1024
=3.2×1023 g


Question 15
Calculate the standard molar volume of oxygen gas. The density of O2 gas at NTP is 1.429 g/L.

Answer

Standard molar volume is Volume of 1 mole of gas i.e 32 gm O2 gas
Now
Density=Massvolume
Volume=MassDensity
= 32/1.429
=22.39 Litres.


Question 16
if 2 mol of Calcium Carbonate (Formula Weight =100) occupies a volume of 67.0 ml, Find the density ?
Mass of Calcium Carbonate = 2 * 100 = 200 g
Volume = 67.0 ml
Density=Massvolume=20067=2.99 g/mL


Question 17
Calculate how many methane molecules and how many carbon and hydrogen atoms are there in 25 g of Methane?

Answer

Molar mass pf methane =16
Number of Moles =25/16
No of methane molecules = 2516×6.022×1023=9.411×1023
No of carbon molecules = 1×9.411×1023=9.411×1023
No of hydrogen molecules = 4×9.411×1023=3.74×1024


Question 18
A metal M of atomic mass 54.94 has a density of 7.42g/cc. Calculate the apparent volume occupied by one atom of the metal.

Answer

Mass of 1 mole metal atoms = 54.94 g
Mass of 1 metal atom = 54.946.022×1023=9.12×1023g
Volume occupied by one metal atom
=Mass of one metal atomdensity
=9.12×10237.42
=1.23×1023 cc.


Question 19
Calculate the number of moles of Ca, C and oxygen atoms and its mass in 200 g of CaCO3.

Answer

Molecular mass of CaCO3= 40+12+3 \times 16 = 100

No. mole of CaCO3= 200g100g=2
Number of Moles of Ca=2 moles
Number of Moles of C=2 moles
Number of Moles of O=6 moles


Question 20
Calculate the total number of electrons present in 3.2 g of CH4.

Answer

Molecular mass of CH4= 12+4×1=16
Moles of CH4=3.216=0.2
No. of electron in 1 molecule of CH4= 6+4 = 10 electrons
Total no. of electrons = 0.2×6.022×1023×10=12.044×1023 electrons.



Question 21
How much Calcium is in the amount of Ca(NO3)2 that contains 20.0g of Nitrogen?

Answer

Number of Moles of Nitrogen = 2014=1.428 moles
From the Molecular formula, it is apparent that, the number of Moles Ca will be half number of Number of Nitrogen
So, number of Moles of Ca=1.4282=.714 moles
So mass of Ca = .714×40=28.56 g


Question 22
State the number of atoms in 1 g atom of Aluminium?

Answer

1 gm atom = atomic weight
So number of atoms will be Avogadro number = 6.022×1023



Question 23
If the components of the air are N2, 78%; O2, 21%; Ar, 0.9% and CO2, 0.1% by volume, what would be the molecular mass of air?

Answer

The molar ratios are also volume ratios for gases (Avogadro’s principle)
Molecular mass of air
=78×28+21×32+0.9×40+0.1×4478+21+0.9+0.1
=28.964.


Question 24
How many molecules are there in 1.624 gm Ferric chloride(FeCl3)?
Molar Mass of Ferric chloride= 56 + 3 * 35.5 = 162.5 gm
Number Of moles = 1.624 /162.5
Number Of molecules = (1.624162.5)×6.022×1023=6.023×1021 Molecules


Question 25
What is the mass of 1 Ammonia Molecule ?

Answer

Molar Mass of Ammonia= 14+3×1=17 gm
Mass of One molecule =17(6.022×1023)=2.8×1023 g


Question 26
The atomic masses of two elements (P and Q) are 20 and 40 respectively. x g of P contains y atoms, how many atoms are present in 2x g of Q?

Answer

No. of mole of P = x20
No. of atoms of P = (x20)×N [N is Avogadro constant]
Therefore, y=x×N20 or x=20yN
Now,
No. of mole of Q = 2x40
No. of atoms of Q
=(2x40)×N
=2N40×20yN
=y


Question 27
Oxygen is present in a 1-litre flask at a pressure of 7.6×1010 mm of Hg at 0°C. Calculate the number of oxygen molecules in the flask.

Answer

Pressure = 7.6×1010 mm Hg
as 1 atm = 760 mm Hg
p=7.6×1010760
=1012 atm
Volume = 1 litre
Temperature = 0°C = 273 K
We know pV=nRT
or
n=pVRT
n=(1012×1)(0.0821×273)=0.44×1013
No. of molecules=no. of moles×Avogadro constant
=0.44×1013×6.022×1023
=2.65×1010 .


Question 28
What is the ratio of the volumes occupied by 1 mole of O2and 1 mole of O3in identical conditions?

Answer

Volume ratio
= Molar ratio (Avogadro’s principle – the molar ratios are also volume ratios for gases)
=1:1


Question 29
Calculate the mass of .5 moles of CaCO3in g.

Answer

Molar mass (i.e., molecular mass in g) = 40+12+3×16=100g
Mass of .5 moles of CaCO3= .5×100=50 g


Question 30
The cost of the Table Salt(NaCl) and Table sugar(C12H22O11) is Rs 10 and Rs 40 per kg. Find the cost of the salt and sugar per mole?

Answer

Molar Mass of NaCl= 23+35.5=58.5 g/mol
Cost of 1 Mole of NaCl =(101000)×58.5=.585 Rs/mole
Molar Mass of C12H22O11= 12×12+22×1+11×16=342 g/mol
Cost of 1 Mole of C12H22O11 = (401000)×342=13.68 Rs/mole


Question 31
Calculate the number of oxygen atoms in 0.2 mole of Na2CO3.10H2O.

Answer

Moles of oxygen atoms in 1 mole of Na2CO3.10H2O = 3+10 = 13
Moles of oxygen atoms in 0.2 mole of Na2CO3.10H2O = 0.2×13=2.6
Therefore, Number of oxygen atoms = 2.6×6.022×1023=1.565×1024.


Question 32
A polystyrene, having the formula Br3C6H3(C3H8)n, was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n.

Answer

Let the mass of polystyrene be 100g,
Therefore, No. of moles of Br in 100g of polystyrene = 10.46/79.9 = 0.1309
From the formula Br3C6H3(C3H8)n, we have,
No. of moles of Br=3×moles ofBr3C6H3(C3H8)n
0.1309=3×massmolecular mass=3×100314.5+44n
n = 44.9 n=45(approx)


Question 33
It has been estimated that 93% of all atoms in the entire universe are hydrogen and that the vast majority of those remaining are helium. Based on only these two elements, estimate the mass percentage composition of the universe.

Answer

It is given that out of 100 atoms, 93 atoms are Hydrogen and 7 atoms are Helium.
Mass of Hydrogen atoms = 93×1=93
Mass of Helium atoms = 7×4=28
Therefore, Mass percentage of Hydrogen = 9393+28×100=76.86%
Mass percentage of Helium = 2893+28×100=23.14%.


Question 34
The molecular weight of haemoglobin is about 65,000 g/mol. Haemoglobin contains 0.35% Fe by mass. How many iron atoms are there in a haemoglobin molecule?

Answer

Mass of Fe in haemoglobin
= 0.35% of 65000
=0.35×65000100
= 227.5
Therefore, No. of Fe atoms in a haemoglobin molecule = 227.556=4


Question 35
Calculate the Number of atoms of oxygen present in 88g of CO2. What would be the weight of the Carbon monoxide having the same number of oxygen atoms?

Answer

Moles of CO2 in 88 g = 88/44 = 2
Number of Oxygen atom=2×6.022×1023×2=20.092×1023
Now
Number of molecules of CO having 20.092×1023 of oxygen atom will be 20.092×1023
Hence of weight of CO = 28×20.092×10236.022×1023|=112g


Question 36
At room temperature, the density of water is 1.0 g/mL and the density of ethanol is 0.789 g/mL. What volume of ethanol contains the same number of molecules as are present in 175 mL of water?

Answer

Let the volume of ethanol containing the same number of molecules as are present in 175mL of water be V mL.
Moles of C2H5OH in V mL = Moles of H2O in 175 mL
Mass of C2H5OH /mol. mass of C2H5OH = Mass of H2O / mol. mass of H2O
Now as Density=Massvolume or Mass=Density×Volume
Therefore
0.789×V46=1×17518
V =566.82 mL.


Question 37
Chlorophyll the green colouring matter of plants responsible for photosynthesis contains 2.68% of Magnesium by weight.Calculate the number of magnesium atoms in 4 g of Chlorophyll

Answer

Mass % of Mg = 2.68%
Now 100 g of Chlorophyll contains 2.68 g of Mg
Then 4 g of Chlorophyll contains = (2.68100)×4=.1072 g
Now
Number of atoms of Mg =MassMolar mass×6.022×1023=1.34×1021 atoms


Question 38
Calculate approximately the diameter of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom. The density of mercury is 13.6 g/cc.

Answer

Suppose the side of cube = x cm = diameter of mercury atom
Therefore, Volume of 1 Hg atom = x3 and
Mass of 1 Hg atom
=density×volume
=13.6×x3
Mass of 1 Hg atom
= Atomic mass/ Avogadro constant
=2006.022×1023
13.6×x3=2006.022×1023
x=(2.44×1023)1/3=2.9×108 cm.


Question 39
How many years it would take to spend Avogadro number of rupees at the rate of Rs 10 laks per sec?

Answer

Rupees= 6.022×1023
Rupees spend rate = 106 /sec
Rupees spent per year =106×3600×24×365
Number of Years required to fully spend the money
=6.022×1023106×3600×24×365
=1.90988×1010 years


Question 40
Find the total number of neutrons present in 7 mg of 14C atoms.

Answer

1 mole of 14C=14 g
No. moles of carbon atoms = 7 mg/14 g = 5×104 Moles
Number of atoms of carbon atoms = 6.022×1023×5×104=3.01×1020
A C-14 atom contains 8 neutrons
Therefore, The total number of neutrons = 8×3.01×1020=2.408×1021.



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