Numericals on Mole, Molar Volume and Avogadro’s Number
(New Text Book Questions)
Q1. Calculate
the number of moles in 25.5 g of sodium metal.
Solution
Given
Mass of sodium metal = 25.5 g
Molar mass of sodium = 23 g/mol
Required
No. of moles = ?
Parent Formula
No. of moles (n) = mass of substance/molar mass
Calculation
No. of moles (n) = 25.5/23 = 1.11 moles
Q2. Calculate
the mass of 3.25 moles of water (H2O) .
Solution
Given
No. of moles of water (H2O) = 3.25
Molar mass of water; H2O = 2(1) + 16 =
18 g/mol
Required
Mass of water = ?
Parent Formula
No. of moles (n) = mass of substance/molar mass
Derived Formula
Mass of substance = no. of moles x molar mass
Calculation
Mass of substance = 3.25 x 18 = 58.5 g
Q3.Calculate
the number of molecules in 610 g of benzoic acid (C7H6O2).
Solution
Given
Mass of benzoic acid (C7H6O2)
= 610 g
Molar mass of benzoic acid; (C7H6O2)
= 7(12) + 6(1) + 2(16) = 122 g/mol
Avogadro’s number = 6.02 x 1023
Required
number of molecules = ?
Parent Formula
No. of moles (n) = mass of substance/molar mass
No. of moles (n) = no. of particles/Avogadro’s no
Derived Formula
Calculation
No. of
particles = 3.01 x 1024 molecules
Q4. Calculate the
mass of 4.39 x 1024 atoms of gold (Au), molar mass of gold is 197
g/mol.
Solution
Given
number of atoms of gold (Au) =
4.39 x 1024
Molar mass of gold (Au; an element) =
197g/mol
Avogadro’s number = 6.02 x 1023
Required
Mass of given atoms of gold = ?
Parent Formula
No. of moles (n) = mass of substance/molar mass
No. of moles (n) = no. of particles/Avogadro’s no
Q5. Calculate the number of
moles in 2.35 x 1025 atoms of aluminium (Al).
Solution
Given
number of atoms of aluminium (Al) =
2.35 x 1025
Avogadro’s number = 6.02 x 1023
atoms/mole
Required
No. of moles of given atoms of Al = ?
Parent Formula
No. of moles (n) = no. of particles/Avogadro’s no
Calculation
No. of moles (n) = 2.35 x 1025/6.02
x 1023 = 39 moles of Al
Q6. What volume of
oxygen gas (O2) occupied by 1.5 moles at STP.
Solution
Given
number of moles of oxgyen gas =
1.5
Molar volume at STP = 22.4 dm3
Required
Volume of oxygen gas at STP = ?
Parent Formula
No. of moles (n) = volume of gas at STP/molar
volume
Derived Formula
volume
of gas at STP = No. of moles (n) x molar volume
Calculation
volume
of gas at STP = 1.5 x 22.4 = 33.6 dm3
Q7. Graphite is one
of the two crystalline forms of carbon which is a constituent component of lead
pencils. How many atoms of
carbon are there in 360 g of graphite? Also find the number of moles of carbon.
Solution
Given
Mass of graphite (An allotrope of C) = 360 g
Molar mass of C = 12 g mol−1
Avogadro’s number = 6.02 x 1023
atoms/mole
Required
number of atoms of carbon in given
mass = ?
number of moles of carbon in given
mass = ?
Conversion of Mass into No. of atoms
No. of moles (n) = mass of substance/molar mass
No. of moles (n) = no. of particles/Avogadro’s no
Conversion of Mass into No. of moles
No. of moles (n) = mass of substance/molar mass
= 360/12 = 30
moles of C
Q8. 1.6 g of a
sample of a gas occupies a volume of 1.12 dm3 at STP. Determine the
molar mass of the gas.
Solution
Given
Mass of gas = 1.6 g
Volume of gas at STP = 1.12 dm3
Molar volume
= 22.4 dm3/mol
Required
Molar mass of the gas = ?
Tricky Problems on Mole Concept
Solution
Solution
Volume of 1 mole of gas at STP = 22.4 L
Q5. An unknown gas having of 8g is present in a container of 44.8 L. Find out unknown gas and number of atoms in a gas.
Solution
Q6.The number of molecules of O2 present in a container is same as that of number of molecules present in 9.6 g of SO2. Find:
(i) Number of moles of O2 (0.15 mol)
(ii) Number of atoms of oxygen(2 x 0.15NA)
(iii) Volume of oxygen at STP (0.15 x 22.4)
Solution
Q7. In a mixture of gas, a gas ‘X’ is present 0.56% then find number of moles of ‘X’ gas present in 1 L mixture at STP.
Solution
100 L mixture contains = 0.56 L gas X
1 L mixture contains = 0.56/100 = 5.6 x 10−3 L
No of moles = volume of gas/molar volume
= 5.6 x 10−3/22.4
= 0.25 mole
Q8. How many gram of CaCO3 would contain 2.4 g of oxygen?
Solution
100 g (1 mole) of CaCO3 contains = 48 g oxygen
48 g oxygen is present in → 100 g of CaCO3
1 g oxygen is present in → 100/48 g of CaCO3
2.4 g oxygen is present in = 100 x 2.4/48 g of CaCO3
= 5 g CaCO3
Q9. The molecular (in fact formula) weight of NaOH is 40 amu. Find
(i) Its gram formula weight.
(ii) Weight of 3 formula unit in amu.
(iii) Weight of 60200 formula unit in g.
(iv) Weight of 6 gram mole.
Q10. Which one of the following will have largest number
of atoms?
(i) 1 g Au(s)
(ii) 1 g Na(s)
(iii) 1 g Li(s)
(iv) 1 g of Cl2(g)
Solution
(i). The number of mole in 6.54 g.
(ii). The number of atoms in 10.9 g of Zn
(iii). The number of atoms in 0.5 mole
(iv). The mass of 1 atom of zin
(v). The mass of 1.204 x 1024 atoms of Zn in g
Solution
Q2. A sample of oxygen gas at STP has a volume of 1700 cm3. Calculate:
(i) The number of moles in the sample.
(ii) The mass of the sample.
(iii) The number of molecules in the sample.
(iv) The volume of 16 g of oxygen gas in cm3.
(v) The volume of 3.01 x 1024 molecules of the sample in dm3
Solution
Q4. 10 g of H2SO4 has been dissolved in excess of water to dissociate it completely into its ions. Calculate [K.B – 2016]
(i) Number of molecules in 10 g of H2SO4
(ii) Number of positive ions
Solution
Q7. Calculate number of particles in of following
(i) 112 dm3 of H2 gas at stp
(ii) 9 g of Na
(Answer; 3.01 x 1023, 2.35 x 1023)
Q8. 0.98 g of H3PO4 has been dissolved in excess of water to dissociate completely into ions. Calculate the number of molecules in 0.98 g of H3PO4 and also number of positive ions.
(Answer; , 6.02 x 1021, 1.806 x 1022)
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