Answer the following questions for class 9 chapter 3 atoms and molecules from exercises
Answers
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Exercises
1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer
Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100% = 40%
And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%
2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?
Answer
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. 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
Page No: 44
3. What are polyatomic ions? Give examples?
Answer
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, Nitrate (NO3-), hydroxide ion (OH-).
4. Write the chemical formulae of the following:
(a) Magnesium chloride
Answer
MgCl2
(b) Calcium oxide
Answer
CaO
(c) Copper nitrate
Answer
Cu (NO3)2
(d) Aluminium chloride
Answer
AlCl3
(e) Calcium carbonate
Answer
CaCO3
5. Give the names of the elements present in the following compounds:
(a) Quick lime
Answer
Calcium and oxygen
(b) Hydrogen bromide
Answer
Hydrogen and bromine
(c) Baking powder
Answer
Sodium, hydrogen, carbon, and oxygen
(d) Potassium sulphate
Potassium, sulphur, and oxygen
6. Calculate the molar mass of the following substances:
(a) Ethyne, C2H2
Answer
Molar mass of ethyne, C2H2 = 2×12 + 2 × 1 = 26 g
(b) Sulphur molecule, S8
Answer
Molar mass of sulphur molecule, S8 = 8×32 = 256 g
(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)
Answer
Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g
(d) Hydrochloric acid, HCl
Answer
Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Nitric acid, HNO3
Answer
Molar mass of nitric acid, HNO3 = 1 + 14 + 3×16 = 63 g
7. 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)?
Answer
(a) The mass of 1 mole of nitrogen atoms is 14 g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is
10 × [2×23 + 32 + 3×16] g = 10×126 g = 1260 g
8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer
(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12 / 32 mole = 0.375 mole
(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole = 1.111 mole
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22 / 44 mole = 0.5 mole
9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer
(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g
(b) Mass of one mole of water molecule = 18 g
Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g
10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer
1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16 g of solid sulphur contains = 6.022 × 1023 / 256 = 16 molecules
= 3.76375 × 1022 molecules
11. 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
Mole of aluminium oxide (Al2O3) = 2×27 + 3×16 = 102 g
i.e., 102 g of Al2O3= 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains = 6.022 × 1023 / 102 × 0.051 molecules
= 3.011×1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020