13. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?
Solution:
Given: 1 mole of carbon weighs 12g
1 mole of carbon atoms = 6.022 x 1023
Molecular mass of carbon atoms = 12g = an atom of carbon mass
Hence, mass of 1 carbon atom = 12 / 6.022 x 1023 = 1.99 x 10-23g
14. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23u, Fe = 56 u)?
Solution:
(a) In 100 grams of Na:
m = 100g, Molar mass of Na atom = 23g, N0 = 6.022 x 1023, N = ?
N = (Given mass x N0)/Molar mass
N = (100 x 6.022 x 1023)/ 23
N = 26.18 x 1023 atoms
(b) In 100 grams of Fe:
m = 100 g, Molar mass of Fe atom = 56 g, N0 = 6.022 x 1023, N = ?
N = (Given mass x N0)/ Molar mass
N = (100 x 6.022 x 1023)/ 56
N = 10.75 x 1023 atoms
Therefore, the number of atoms are more in 100 g of Na than in 100 g of Fe.
15. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:
Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g
To calculate percentage composition of the compound:
Percentage of boron = mass of boron / mass of the compound x 100
= 0.096g / 0.24g x 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%
16. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Solution:
When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.
Given that
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
Find out
We need to find out the mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.
Solution
First, let us write the reaction taking place here
C + O2 → CO2
As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.
3g + 8g → 11 g ( from the above reaction)
The total mass of reactants = mass of carbon + mass of oxygen
= 3g + 8g
= 11g
The total mass of reactants = Total mass of products
Therefore, the law of conservation of mass is proved.
Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.
Thus it further proves the law of constant proportions.
3 g of carbon must also combine with 8 g of oxygen only.
This means that (50 − 8) = 42g of oxygen will remain unreacted.
The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11 g of carbon dioxide will be formed
The above answer is governed by the law of constant proportions.
17. What are polyatomic ions? Give examples.
Solution:
Polyatomic ions are ions that contain more than one atom but they behave as a single unit
Example: CO32-, H2PO4–
18. Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Solution:
The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
19. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Solution:
The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)
20. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Solution:
Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C2H2 = 2 x Mass of C + 2 x Mass of H = (2 × 12) + (2 ×1) = 24 + 2 = 26g
(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g
(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1 + 35.5 = 36.5g
(e) Molar mass of Nitric acid, HNO3 = Mass of H + Mass of Nitrogen + 3 x Mass of O = 1 + 14 + 3 × 16 = 63g
21. 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 (Na2SO3)?
Solution:
The mass of the above mentioned list is as follows:
(a) Atomic mass of nitrogen atoms = 14u
Mass of 1 mole of nitrogen atoms= Atomic mass of nitrogen atoms
Therefore, mass of 1 mole of nitrogen atom is 14g
(b) Atomic mass of aluminium =27u
Mass of 1 mole of aluminium atoms = 27g
1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g
(c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O = (2 x 23) + 32 + (3 x 16) = 46 + 32 + 48 = 126g
Therefore, mass of 10 moles of Na2SO3 = 10 x 126 = 1260g