Q1
Calculate the molar mass of the following substances.
(a) Ethyne, C _{ 2 } H _{ 2 }
(b) Sulphur molecule, S _{ 8 }
(c) Phosphorus molecule, P _{ 4 } (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO _{ 3 }
Answer
Molar mass of C = 12g
Molar mass of H = 1g
Molar mass of S = 32g
Molar mass of P = 31g
Molar mass of Cl = 35.5g
Molar mass of N = 14g
Molar mass of O = 16g
(a) Ethyne, C _{ 2 } H _{ 2 }
Molar mass of C _{ 2 } H _{ 2 } = 2 x Molar mass of C + 2 x Molar mass of H
= 2x 12g + 2 x 1g
= 24g + 2g
= 26g
(b) Sulphur molecule, S _{ 8 }
Molar mass of S _{ 8 } = 8 x Molar mass of S
= 8x 32g
= 256g
(c) Phosphorus molecule, P _{ 4 }
Molar mass of P _{ 4 } = 4 x Molar mass of P
= 4 x 31g
= 124g
(d) Hydrochloric acid, HCl
Molar mass of HCl = 1 x Molar mass of H + 1 x Molar mass of Cl
= 1x 1g + 1 x 35.5g
= 1g + 35.5g
= 36.5g
(e) Nitric acid, HNO _{ 3 }
Molar mass of HNO _{ 3 } = 1 x Molar mass of H + 1 x Molar mass of N + 3 x Molar mass of O
=1x 1g + 1 x 14g + 3 x 16g
= 1g + 14g + 48g
= 63g
Q2
Find number of moles of H _{ 2 } in 10g of H _{ 2 }
Answer
Given mass of H _{ 2 } = m =10g
Molar Mass of H _{ 2 } = M= 2g
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
= m/M
= 10g / 2g
= 5 moles
Q3
How many moles is 12.044 x 10 ^{ 23 } atoms of He
Answer
Number of atoms of the He = N = 12.044 x 10 ^{ 23 }
Avogadro's Constant = N O = 6.022 x 10 ^{ 23 }
We know that,
Number of moles of a substance = Number of particles of the substance / Avogadro's Constant
= N/N O
= 11.044 x 10 ^{ 23 } / 6.022 x 10 ^{ 23 }
= 2 moles of He atoms
Q4
How many moles are there in 52g of Helium?
Answer
Atomic Mass of Helium = 4 u
So Molar Mass of Helium = 4 gram
It means that Mass of 1 mole of Helium = 4 grams
We can also write as
4 grams = 1 mole
1 gram =1/4 mole
52 grams = (¼) x 52 mole
52 grams = 13 moles
So,
n = Given mass / Molar mass
= 52 / 4
= 13 moles
Q5
Calculate the Molar Mass of Calcium Carbonate (CaCO _{ 3 } )
1 molecule of CaCO _{ 3 } consists of : 1 Calcium (Ca) atom
1 Carbon (C ) atom
3 Oxygen (O) atoms
Answer
M of CaCO _{ 3 } = (1 x mass of Ca) + (1 x mass of C) + (3 x mass of O)
M of CaCO _{ 3 } = (1 x 40) + (1 x 12) + (3 x 16)
M of CaCO _{ 3 } = (1 x 40) + (1 x 12) + (3 x 16)
M of CaCO _{ 3 } = 100g/mol
Q6
What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na _{ 2 } SO _{ 3 } )?
Answer
We know that,
Mass of 1 mole of a substance = Molar mass of that substance
(a) 1 mole of nitrogen atoms
Molar mass of Nitrogen (N) = M =14g
Therefore, mass of 1 mole Nitrogen atoms = Molar mass of nitrogen = 14g
(b) 4 moles of aluminium atoms
Molar Mass of Aluminium = 27g
Mass of 1 mole Aluminium atoms = Molar mass of aluminium = 27g
Mass of 4 moles Aluminium atoms = 4 x Molar mass of aluminium
= 4 x 27g
= 108g
(c) 10 moles of sodium sulphite (Na 2 SO 3 )
Mass of 1 mole Na _{ 2 } SO _{ 3 } atoms = 2 x Molar mass of Na + 1 x Molar Mass of S + 3 x Molar Mass of O
= 2 x 23g + 1 x 32g + 3 x 16g
= 46g + 32g + 48g
= 126g
Mass of 10 moles Na _{ 2 } SO _{ 3 } atoms = 10 x Molar mass of Na _{ 2 } SO _{ 3 }
= 10 x 126g
= 1260g
Q7
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Answer
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
(a) 12 g of oxygen gas
Given mass of O 2 = m = 12g
Molar Mass of O 2 = M = 32g
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
= m/M
= 12g / 32g
= 6 /16 moles
= ⅜ moles
(b) 20 g of water
Given mass of H _{ 2 } O = m = 20g
Molar Mass of H _{ 2 } O = M = 18g
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
= m/M
= 20g/18g
= 10/9 moles
(c) 22 g of carbon dioxide
Given mass of CO _{ 2 } = m = 22g
Molar Mass of CO _{ 2 } = M = 44g
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
= m/M
= 22g/44g
= 1/2 moles
= 0.5 moles
Q8
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
Therefore,
Mass of the substance = Number of moles x Molar mass of the substance
(a) 0.2 mole of oxygen atoms
Given mass of O = m = ?
Molar Mass of O = M = 16g
Number of moles = n = 0.2 = 2/10
We know that,
Mass of the substance = Number of moles x Molar of the substance
m = n x M
= (2 /10) x 16
= 32 /10
= 3.2g
(b) 0.5 mole of water molecules
Given mass of H _{ 2 } O = m = ?
Molar Mass of H _{ 2 } O = M = 18g
Number of moles = n = 0.5 = 5/10 = 1/2
We know that,
Mass of the substance = Number of moles x Molar of the substance
m = n x M
= (1/2) x 18
= 18/2
= 9g
Q9
Calculate the number of molecules of sulphur (S 8 ) present in 16 g of solid sulphur.
Answer
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
n = m/M ----- (1)
Also,
Number of moles of a substance = Number of particles of the substance / Avogadro's Constant
n = N / N _{ O }
n = N / 6.022 x 10 ^{ 23 } ------- (2)
From the above 2 formulae, we can say that,
n = m / M = N / 6.022 x 10 ^{ 23 } ---- (3)
Given mass of sulphur molecule(S _{ 8 } ) = m = 16g
Molar mass of Sulphur Molecule (S _{ 8 } ) = M = 256g
We need to find Number of molecules, ie, N
Putting values in (3), we get
16 / 256 = N / 6.022 x 10 ^{ 23 }
N = 6.022 x 10 ^{ 23 } x (16 / 256)
= 6.022 x 10 ^{ 23 } / 16
= 3.011 x 10 ^{ 23 } /8
= 0.37 x 10 ^{ 23 } molecules of Sulphur(S _{ 8 } )
= 3.7 x 10 ^{ 22 } molecules of Sulphur (S _{ 8 } )
Q10
3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 x10 ^{ 23 }
(b) 6.09 x10 ^{ 22 }
(c) 6.022 x 10 ^{ 23 }
(d) 6.022.x 10 ^{ 21 }
Answer
(a) 6.68 x10 ^{ 23 }
Explanation:
Number of moles = Given mass of substance / Molecular mass of substance
n _{ Sucrose } = m / M
n _{ Sucrose } = 3.42 g / 342g
n _{ Sucrose } = 0.01 mole
1 mol of sucrose → 11 x N _{ 0 } atoms of O
0.01 mol of sucrose → 0.01 x 11 x N _{ 0 } atoms of O
→ 0.11 x N _{ 0 } atoms of O
n _{ H20 } = m / M
n _{ H20 } = 18 g / 18g
n _{ H20 } = 1 mole
1 mol of water → N _{ 0 } atoms of O
Total number of O atoms = (0.11 + 1) x N _{ 0 }
= 1.11 x N _{ 0 }
= 1.11 x 6.022 x 10 ^{ 23 }
= 6.68 x 10 ^{ 23 }
Q11
If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer
Given,
Mass of 1 mole of Carbon = Molar mass of C = 12g
1 mole of carbon = 6.0 ^{ 22 } x 10 ^{ 23 } atoms of Carbon
Therefore,
Mass of 6.0 ^{ 22 } x 10 ^{ 23 } atoms of Carbon = 12 g
Mass of 1 atom of Carbon = 12 / 6.0 ^{ 22 } x 10 ^{ 23 }
≈ 2 x 10 - 23 g
Q12
If 1.4g of Calcium Oxide is formed by the complete decomposition of Calcium Carbonate, then the amount of Calcium Carbonate taken and the amount of Carbon Dioxide formed will be respectively?
- 2.2g and 1.1g
- 1.1g and 2.5g
- 2.5g and 1.1g
- 5.0 and 1.1g
Answer
- c) 2.5g and 1.1g
Calcium Carbonate → Calcium Oxide + Carbon Dioxide
CaCO _{ 3 } → CaO + CO _{ 2 }
X → 1.4g + Y
Molecular mass of CaO = 56g
Number of moles of CaO = 1.4/56
= 0.025
Molecular mass of CaCO _{ 3 } = 100g
Molecular mass of CO _{ 2 } = 44g
So, 0.025 moles of CaCO _{ 3 } = 0.025 x 100g
= 2.5g of CaCO _{ 3 }
And, 0.025 moles of CO _{ 2 } = 0.025 x 44g
= 1.1g of CO _{ 2 }
Q13
Which of the following correctly represents 360 g of water?
(i) 2 moles of H _{ 2 } O
(ii) 20 moles of water
(iii) 6.022 x 10 ^{ 23 } molecules of water
(iv) 1.2044 x10 ^{ 25 } molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer
(d) (ii) and (iv)
Explanation:
(i) 2 moles of H _{ 2 } O → 1 mole of H _{ 2 } O = 18g → 2 moles = 36g ≠ 360g
(ii) 20 moles of water → 1 mole of H _{ 2 } O = 18g → 20 moles = 360g
(iii) 6.022 x 10 ^{ 23 } molecules of water → 1 mole of water molecules = 18g ≠ 360g
(iv) 1.2044 x10 ^{ 25 } molecules of water → 20 moles of water = 360g
Q14
Which of the following pairs have the same number of atoms?
A - 16 g of O _{ 2 } (g) and 4 g of H _{ 2 } (g)
B - 16 g of O _{ 2 } and 44 g of H _{ 2 }
C - 28 g of N _{ 2 } and 32 g of O _{ 2 }
D - 12 g of C(s) and 23 g of Na(s)
Answer
C - 28 g of N _{ 2 } and 32 g of O _{ 2 }
D - 12 g of C(s) and 23 g of Na(s)
Explanation:
One mole of any species (atoms,molecules, ions or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.
1 mole of an item = 6.022 x 10 ^{ 23 } items
In 28 g of N _{ 2 } and 32 g of O _{ 2 }
28 g of N _{ 2 } = 1 mole of N 2 = 2 x 6.022 x 10 ^{ 23 } atoms
32 g of O _{ 2 } = 1 mole of O 2 = 2 x 6.022 x 10 ^{ 23 } atoms
In 12 g of C(s) and 23 g of Na(s)
12 g of C(s) = 1 mole of C = 6.022 x 10 ^{ 23 } atoms
23 g of Na(s) = 1 mole of C = 6.022 x 10 ^{ 23 } atoms
Q15
Which of the following contains the greatest number of atoms?
A - 1g of butane
B - 1g of Nitrogen
C - 1g of silver
D - 1g of water
Answer
A - 1g of butane
Explanation :
Molecular mass of Butane (C _{ 4 } H _{ 10 } ) is 58g/mol
There are 14 atoms in the molecule
So, 1g of butane means 1/58 moles = 0.0172 moles
Total number of atoms = 14 x 0.0172 x 6.022 x 10 ^{ 23 } = 1.45 x 10 ^{ 23 } atoms
Molecular mass of Nitrogen (N _{ 2 } ) is 28g/mol
There are 2 atoms in the molecule
So, 1g of nitrogen means 1/28 moles = 0.0357 moles
Total number of atoms = 2 x 0.0357 x 6.022 x 10 ^{ 23 } = 0.43 x 10 ^{ 23 } atoms
Molecular mass of Silver (Ag) is 108g/mol
There are is 1 atom in the molecule
So, 1g of silver means 1/108 moles = 0.0092 moles
Total number of atoms = 0.0092 x 6.022 x 10 ^{ 23 } = 0.054 x 10 ^{ 23 } atoms
Molecular mass of Water (H _{ 2 } O) is 18g/mol
There are 3 atoms in the molecule
So, 1g of water means 1/18 moles = 0.055 moles
Total number of atoms = 3 x 0.055 x 6.022 x 10 ^{ 23 } = 1.003 x 10 ^{ 23 } atoms
Thus 1g of butane has greatest number of atoms
Q16
Which sample contains the largest number of atoms?
A - 1mg of C _{ 4 } H _{ 10 }
B - 1mg of N _{ 2 }
C - 1mg of Na
D - 1ml of H _{ 2 } O
Answer
D - 1ml of H _{ 2 } O
Explanation:
1mg of C _{ 4 } H _{ 10 } (14 atoms in 1 molecule) = (14 x N x 10 ^{ -3 } ) / 58 = 2.41 x 10 ^{ -4 } x N atoms
1mg of N _{ 2 } (2 atoms in 1 molecule) = (2N x 10 ^{ -3 } ) / 28 = 7.14 x 10 ^{ -5 } x N atoms
1mg of Na (1 atom in 1 molecule) = (N x 10 ^{ -3 } ) / 23 = 4.34 x 10 ^{ -5 } x N atoms
1ml of H _{ 2 } O (3 atoms in 1 molecule) = 1g of H _{ 2 } O = 3N / 18 = 2.41 x 10 ^{ -4 } x N atoms
Thus 1ml of H _{ 2 } O has largest number of atoms
Q17
How many moles of Magnesium Phosphate (Mg _{ 3 } (PO _{ 4 } ) _{ 2 } ) will contain 0.25 mole of oxygen atoms?
A - 3.125 x 10 ^{ -2 }
B - 1.25 x 10 ^{ -2 }
C - 2.5 x 10 ^{ -2 }
D - 2 x 10 ^{ -2 }
Answer
A - 3.125 x 10 ^{ -2 }
Explanation:
In one molecule of Mg _{ 3 } (PO _{ 4 } ) _{ 2 } there are 8 Oxygen atoms
So in 1 mole of Mg _{ 3 } (PO 4 ) _{ 2 } molecules , there will be 8 moles of Oxygen atoms.
1 mole of Mg _{ 3 } (PO 4 ) _{ 2 } → 8 moles of O atoms
X → 0.25 moles of O atoms
Cross multiplying,
0.25 x 1 = 8X
X = 0.25/8
X = 3.125 x 10 ^{ -2 }
3.125 x 10 ^{ -2 } moles of Magnesium Phosphate will contain 0.25 mole of oxygen atoms.
Q18
Which has the maximum number of molecules?
A - 7g N _{ 2 }
B - 2g H _{ 2 }
C - 16g NO _{ 2 }
D - 16g O _{ 2 }
Answer
B - 2g H _{ 2 }
Explanation:
Number of moles = Given mass of substance / Molecular mass of substance
n = m / M
7g N _{ 2 } → 7 / 28 = 0.25 moles of molecules
2g H _{ 2 } → 2 / 2 = 1 mole of molecules
16g NO _{ 2 } → 16 / 46 = 0.347 moles of molecules
16g O _{ 2 } → 16 / 32 = 0.5 moles of molecules
Thus 2g H _{ 2 } has the most number of molecules.
Q19
How many gram atoms of H and S are contained in 0.40 mole of H _{ 2 } S?
Answer
1 mole of a substance = 1 gram atom
In this case
H _{ 2 } S has 2 H atoms and 1 S atom
1 mole of H _{ 2 } S → 2 gram atoms of H and 1 gram atom of S
So, 0.4 moles of H _{ 2 } S →
2 x 0.4 = 0.8 gram atoms of H
1 x 0.4 = 0.4 gram atoms of S
Q20
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Answer
To find which one has more atoms, we’ll find the number of moles of both.
Since 1 mole of any substance has the same number of atoms, the element having more number of moles will have more atoms .
Let,
Given mass of Na = m _{ n } =100g
Given mass of Fe = m _{ f } =100g
We know that,
Molar mass of a substance = Atomic mass of the substance (in g)
Therefore,
Molar mass of Na = M _{ n } = 23g
Molar mass of Fe = M _{ f } = 56g
Now,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
For Na, = m _{ n } /M _{ n }
= 100/23
≈ 4.34 moles
For Fe, = m _{ f } /M _{ f }
= 100/56
≈ 1.78 moles
Since, Number of moles of Na > Number of moles of Fe .
100g of Na will have more atoms than 100g of Fe .
Q21
Calculate the number of moles and molecules in 22g of Acetic Acid.
Answer
Acetic Acid has the formula CH _{ 3 } COOH
The molecular mass is thus → (12x2 + 1x4 + 16x2)
= 60g
Number of moles n = Given mass / Molar mass
n = 22/60
n = 0.3666 moles
1 mole → 6.022 x 10 ^{ 23 } molecules (according to mole concept)
0.3666 moles → X molecules
X = 0.3666 x 6.022 x 10 ^{ 23 }
X = 2.207 x 10 ^{ 23 } molecules
So, 22g of Acetic acid has 0.3666 moles and 2.207 x 10 ^{ 23 } molecules
Q22
24 g of Carbon reacts with some Oxygen to make 88g of Carbon Dioxide. Find out the amount of Oxygen used.
Answer
The reaction for formation of Carbon Dioxide is;
C + O _{ 2 } → CO _{ 2 }
1 mole of C reacts with 1 mole of O _{ 2 } to give 1 mole of CO _{ 2 }
12g of C reacts with 1g of O _{ 2 } to give 44g of CO _{ 2 }
It is given that 24g of C gives 88g of CO _{ 2 }
Which is double the amount of C and CO _{ 2 }
So even O _{ 2 } will be double the amount → 32g
Q23
In 2 moles of acetaldehyde (CH _{ 3 } CHO) calculate the following
- Number of moles of C
- Number of moles of H
- Number of moles of O
- Number of molecules of CH _{ 3 } CHO
Answer
- 1 mole of acetaldehyde has 2 C atoms ( C H _{ 3 } C HO )
So, 1 mole of acetaldehyde has 2 moles of C atoms
That means, 2 mole of acetaldehyde has 4 moles of C atoms
- 1 mole of acetaldehyde has 4 H atoms ( C H _{ 3 } C H O )
So, 1 mole of acetaldehyde has 4 moles of C atoms
That means, 2 mole of acetaldehyde has 8 moles of C atoms
- 1 mole of acetaldehyde has 1 O atom ( CH _{ 3 } CH O )
So, 1 mole of acetaldehyde has 1 moles of O atom
That means, 2 mole of acetaldehyde has 2 moles of C atoms
- 1 mole → 6.022 x 10 ^{ 23 } molecules (according to mole concept)
2 moles → X molecules
X = 2 x 6.022 x 10 ^{ 23 }
X = 12.044 x 10 ^{ 23 } molecules
Q24
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Answer
We know that,
Number of moles of a substance = Given mass of the substance / Molar mass of the substance
n = m/M ----- (1)
Also,
Number of moles of a substance = Number of particles of the substance / Avogadro's Constant
n = N / N O
n = N / 6.022 x 10 ^{ 23 } ------- (2)
From the above 2 formulae, we can say that,
n = m / M = N / 6.022 x 10 ^{ 23 } ---- (3)
Given mass of Aluminium ion (Al _{ 2 } O _{ 3 } )= m = 0.051g
Molar mass of Aluminium Oxide (Al _{ 2 } O _{ 3 } ) = M = 102g
We need to find Number of ions, ie, N
Putting values in (3), we get
0.051 / 102 = N / 6.022 x 10 ^{ 23 }
N = 6.022 x 10 ^{ 23 } x (0.051 / 102)
= 6.022 x 10 ^{ 23 } x (51 x 10 ^{ -3 } ) / 102
= 3.011 x 10 ^{ (23-3) }
= 3.011 x 10 ^{ 20 } molecules
1 molecule of Al _{ 2 } O _{ 3 } has 2 Al ^{ 3+ } ions
So, number of Al ^{ 3+ } ions in 0.051 molecules of Al _{ 2 } O _{ 3 } is;
3.011 x 10 ^{ 20 } x 2 = 6.022 x 10 ^{ 20 }
Thus, 6.022 x 10 ^{ 20 } Al ^{ 3+ } ions are present in 0.051 molecules of Al _{ 2 } O _{ 3 }