Q.1. Calculate the number of hydrogen atoms in 1 mole of H2.
1 molecule of H2 = 2 hydrogen atoms
∴ 1 mole of H2 = 2 mole hydrogen atoms
= 2 × 6.022 × 1023
= 12.044 × 1023
= 1.2044 × 1024 hydrogen atoms.
Q.2. Calculate the number of molecules in 22.4 litres of CH4 gas at STP.
22.4 L of CH4 gas = 1 mole CH4 gas
= 6.022 × 1023 CH4 gas molecules.
Q.3. Calculate the number of Cu atoms in 0.3175 g of Cu.
No. of moles of Cu
= Mass of Cu / Atomic mass
= 0.3175/63.5
= 0.005 mole
No. of Cu atoms
= No. of moles × NA
= 0.005 × 6.022 × 1023
= 3.011 × 1021 Cu atoms.
Q.4. How many molecules of water are there in 54 g of H2O ?
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
Q.5. Calculate the mass of 6.022×1023 molecule of NH4Cl ?
Molar mass (Molecular mass in gram) of NH4Cl
= 14 + 4 + 35.5 = 53.5g
No. of moles of NH4Cl
= 6.022 × 1023 / 6.022 × 1023
= 1 mole
Mass of NH4Cl
=No. of moles × molar mass
=1 × 53.5g
= 53.5g.
Q.6. Calculate the mass of 12.044 × 1023 Oxygen atoms.
No. of moles of Oxygen atoms
=12.044 × 1023 / 6.022 × 1023
= 2 mole
Mass of Oxygen atoms
= No. of moles × atomic mass
= 2 × 16
= 32 g
Q.7. How many atoms of hydrogen are there in 36 g of NH4?
Molar mass (Molecular mass in gram) of NH4
= 14 + 4 = 18 g
No. of moles of NH4
= 36 / 18 = 2 moles
Total Moles of Hydrogen Atoms
= 4 × No. of moles of NH4
= 4 × 2
= 8 moles
No. of atoms of Hydrogen
= 8 × 6.022 × 1023
= 48.176 × 1023
= 4.8176 × 1024
Q.8. a. Find the volume of 1 g of H2 gas in litres at STP.
b. Find the volume of 20g H2 at STP.
c. What is the volume occupied by 6.022 × 1023 molecules of any gas at STP?
a. Molar Mass of H2 gas = 2 gm
Number of moles = Mass / Molar Mass
= 1 / 2 = 0.5
1 mole of H2 = 22.4 L (at STP)
∴ 0.5 mole of H2 = 0.5 × 22.4 = 11.2 litre
b. No. of moles of H2 = 20 / 2 =10
1 mole of H2 = 22.4 L (at STP)
∴ 10 moles = 10 × 22.4
= 224 L
c. 6.022 × 1023 molecules = 1 mole molecules
1 mole molecules of any ideal gas = 22.4 L at STP
Q.9. An atom of some element Y weighs 6.644 × 10−23 g. Calculate the number of gram-atoms in 40 kg of it.
Mass of 1 mole Y atoms
= mass of 1 atom × NA
= 6.644 × 10−23 × 6.022 × 1023
= 40 g
∴ the atomic mass of Y = 40
No. of gram-atoms (or moles) of X
= mass of Y / atomic mass
= 40 × 1000 / 40
= 1000
Q.10. Find the number of moles and number of atoms of H and S in 10 moles of H2S.
1 mole of H2S contains 2 mole of H, 1 mole of S
Therefore,
10 mole of H2S contains
20 moles of H = 20 × 6.022 × 1023 H-atoms
= 1.2044 × 1025 H atoms
10 mole of H2S contains
10 mole of S = 10 × 6.022 × 1023 S-atoms
= 6.022 × 1024 S atoms
Q.11. Calculate the number of atoms of each element in 245 g of KClO3.
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 KClO3 contain
2 moles of K
= 2 × 6.022 × 1023
= 1.2044 × 1024 K atoms
2 mole of KClO3 contain
2 moles of Cl
= 2 × 6.022 × 1023
= 1.2044 × 1024 Cl atoms
2 mole of KClO3 contain
6 moles of O
= 6 × 6.022 ×1023
= 3.6132 ×1024 O atoms.
Q.12. Calculate the number of atoms in 12.2 litres of the below gas at STP
(i) mono-atomic
(ii) diatomic gas
No. of moles in 12.2 L gas at STP = 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
Q.13. 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?
Mass of one mole of oxygen molecule (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
= 32 / 1 × 1024
= 3.2 × 10−23 g
Q.14. Calculate the standard molar volume of oxygen gas. The density of O2 gas at STP is 1.429 g/L.
Standard molar volume is Volume of 1 mole of gas i.e., 32 gm O2 gas
Density = Mass / volume
Volume = Mass / Density
= 32 / 1.429
= 22.39 Litres.
Q.15. 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 = Mass / volume
= 200 / 67
= 2.99 g/mL
Q.16. Calculate how many methane molecules and how many carbon and hydrogen atoms are there in 25 g of Methane?
Molar mass pf methane = 16
Number of Moles = 25 / 16
No of methane molecules
= (25 / 16) × 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
Q.17. A metal M of atomic mass 54.94 has a density of 7.42 g/cc. Calculate the apparent volume occupied by one atom of the metal.
Mass of 1 mole metal atoms = 54.94 g
Mass of 1 metal atom
= 54.94 / (6.022 × 1023)
= 9.12 × 10−23 g
Volume occupied by one metal atom
= Mass of one metal atom / density
= 9.12 × 10−23 / 7.42
= 1.23 × 1023 cc.
Q.18. Calculate the number of moles of Ca, C and oxygen atoms and its mass in 200 g of CaCO3.
Molecular mass of CaCO3
= 40 + 12 + 3 × 16
= 100
No. mole of CaCO3
= 200g / 100g
= 2
Number of Moles of Ca = 2 moles
Number of Moles of C = 2 moles
Number of Moles of O = 6 moles
Q.19. Calculate the total number of electrons present in 3.2 g of CH4.
Molecular mass of CH4
= 12 + 4 × 1
= 16
Moles of CH4
= 3.2 / 16
= 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
= 1.2044 × 1024 electrons.
Q.20. How much Calcium is in the amount of Ca(NO3)2 that contains 20.0g of Nitrogen?
Number of Moles of Nitrogen
= 20 / 14
= 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
= 1.428 / 2
= 0.714 moles
So mass of Ca
= 0.714 × 40
= 28.56 g
Q.21. State the number of atoms in 1 g atom of Aluminium?
1 gm atom = atomic weight
So number of atoms will be Avogadro number
= 6.022 × 1023
Q.22. 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?
The molar ratios are also volume ratios for gases (Avogadro’s principle)
Molecular mass of air
= (78/100) × 28 + (21/100) × 32 + (0.9/100) × 40 + (0.1/100) × 44
= 28.964 u.
Q.23. How many molecules are there in 1.624 gm Ferric chloride(FeCl3)?
Molar Mass of Ferric chloride
= 56 + 3 × 35.5
= 162.5 g
Number Of moles
= 1.624 /162.5
Number Of molecules
= (1.624/162.5) × 6.022 × 1023
= 6.023 × 1021 Molecules
Q.24. What is the mass of 1 Ammonia Molecule ?
Molar Mass of Ammonia
= 14 + 3 × 1
= 17 g/mol
Mass of One molecule
= 17 / (6.022 × 1023)
=2.8 × 10−23 g
Q.25. 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?
No. of mole of P = x / 20
No. of atoms of P = (x / 20) × NA
Therefore, y = (x / 20) × NA
No. of mole of Q = 2x / 40
No. of atoms of Q
= (2x / 40) × NA
= (x / 20) × NA
= y
Q.26. Oxygen is present in a 1-litre flask at a pressure of 7.6 × 10−10 mm of Hg at 0°C. Calculate the number of oxygen molecules in the flask.
Pressure = 7.6 × 10−10 mm Hg
as 1 atm = 760 mm Hg
p = 7.6 × 10−10 / 760
= 10−12 atm
Volume = 1 litre
Temperature = 0°C = 273 K
We know
pV = nRT
or
n = pV / RT
n = (10−12 × 1) / (0.0821 × 273)
= 0.44 × 10−13
= 4.4×10−14
No. of molecules
= no. of moles × NA
= 0.44 × 10−13 × 6.022 × 1023
= 2.65 × 1010
Q.27. What is the ratio of the volumes occupied by 1 mole of O2and 1 mole of O3 in identical conditions?
Volume ratio
= Molar ratio (Avogadro’s principle)
= 1 : 1
Q.28. Calculate the mass of 0.5 moles of CaCO3 in g.
Molar mass (i.e., molecular mass in g)
= 40 + 12 + 3 × 16
= 100 g
Mass of 0.5 moles of CaCO3
= 0.5 × 100
= 50 g
Q.29. 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?
Molar Mass of NaCl
= 23 + 35.5
=58.5 g/mol
∴ Mass of 1 mole of NaCl = 58.5 g
Cost of 1 Mole (58.5 g) of NaCl
= (58.5 / 1000) × 10
= ₹ 0.585 per mole
Molar Mass of C12H22O11
= 12 × 12 + 22 × 1 + 11 × 16
=342 g/mol
∴ Mass of 1 mole of C12H22O11 = 342 g
Cost of 1 Mole (342 g) of C12H22O11
= (342 / 1000) × 40
= ₹ 13.68 per mole
Q.30. Calculate the number of oxygen atoms in 0.2 mole of Na2CO3.10H2O.
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
Q.31. 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.
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
= 93 / (93 + 28) × 100
= 76.86%
Mass percentage of Helium
= 28 / (93 + 28) × 100
= 23.14%
Q.32. The molecular mass 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?
Mass of Fe in haemoglobin
= 0.35% of 65000
= (0.35 / 100) × 65000
= 227.5 g
Therefore, No. of Fe atoms in a haemoglobin molecule
= 227.5 / 56
= 4
Q.33. 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?
Moles of CO2 in 88 g = 88/44 = 2
Number of Oxygen atom
= 2 × 6.022 × 1023 × 2
= 2.4088 × 1024
Number of molecules of CO having 2.4088 × 1024 of oxygen atom will be
= 2.4088 × 1024
Hence of weight of CO
= [(2.4088 × 1024) / (6.022 × 1023)] × 28
= 4 × 28
= 112 g
Q.34. 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?
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)
As Density = Mass / volume
Mass = Density × Volume
Therefore
(0.789 × V) / 46 = (1 × 175) / 18
V = 566.82 mL
Q.35. 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.
Mass % of Mg = 2.68%
i.e., 100 g of Chlorophyll contains 2.68 g of Mg
∴ 4 g of Chlorophyll contains
= (2.68 / 100) × 4
= 0.1072 g
Number of atoms of Mg
= (Mass / Molar mass) × 6.022 × 1023
= 1.34 × 1021 atoms
Q.36. How many years it would take to spend Avogadro number of rupees at the rate of Rs 10 lakhs per sec?
Rupees = 6.022 × 1023
Rupees spend rate = 106 /sec
Rupees spent per year
= 106 × 365 × 24 × 60 × 60
Number of Years required to fully spend the money
= 6.022 × 1023 / (106 × 365 × 24 × 60 × 60)
= 1.90988 × 1010 years
Q.37. Find the total number of neutrons present in 7 mg of 14C atoms.
1 mole of 14C=14 g
No. moles of carbon atoms
= (7 × 10−3 g) / 14 g
= 5 × 10−4 Moles
Number of atoms of carbon atoms
= 6.022 × 1023 × 5 × 10−4
= 3.011 × 1020
A C-14 atom contains 8 neutrons
Therefore, The total number of neutrons
= 8 × 3.011 × 1020
= 2.408 × 1021
Q.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.
Suppose the side of cube = x cm = diameter of mercury atom
Therefore, Volume of 1 Hg atom = x3
Mass of 1 Hg atom
= density × volume
= 13.6 × x3
Mass of 1 Hg atom
= Atomic mass / Avogadro constant
= 200 / 6.022 × 1023
13.6 × x3 = 200 / 6.022 × 1023
x = (2.44 × 10−23)1/3
= 2.9 × 10−8 cm
Q.39. 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)
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)
= 2 / (2 + 0.1)
= 0.95
Mole fraction of NaOH
= (Number of moles of NaOH) / (No. of moles of NaOH + No. of moles of water)
= 0.1 / (2 + 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
= 0.1 / 0.04
= 2.5 M
Q.40. 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.
Let the mass of polystyrene be 100 g,
∴ No. of moles of Br in 100g of polystyrene
= Mass / Molar mass of Br
= 10.46 / 79.9
= 0.1309
Molecular Mass of polystyrene Br3C6H3(C3H8)n
= 3 × 80 + 6 × 12 + 3 × 1 + n × (3 × 12 + 8 × 1)
= 240 + 72 + 3 + n × 44
= 315 + 44n
From the formula Br3C6H3(C3H8)n,
No. of moles of Br = 3 × moles of Br3C6H3(C3H8)n
0.1309 = 3 × (mass / molecular mass of polystyrene)
= 3 × (100 / 315 + 44n)
n = 44.9
= 45 (approx.)