These NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.
Atoms and Molecules NCERT Solutions for Class 9 Science Chapter 3
Class 9 Science Chapter 3 Atoms and Molecules InText Questions and Answers
Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid ? sodium ethanoate + carbon dioxide + water
Answer:
In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.
Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)
Now, total mass before the reaction = (5.3 + 6) g = 11.3 g
And, total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g
∴ Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.
Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer:
It is given that the ratio of hydrogen and oxygen by mass to form water is 1 : 8.
Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g.
Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3g = 24 g.
Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
The postulate of Dalton’s atomic theory which is a result of the law of conservation of mass is:
Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.
Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
The postulate of Dalton’s atomic theory which can explain the law of definite proportion is:
The relative number and kind of atoms in a given compound remains constant.
Question 5.
Define atomic mass unit.
Answer:
Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.
Question 6.
Why is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.
Question 7.
Write down the formulae of the followings:
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide → Na2O
(ii) Aluminium chloride → AlCl3
(iii) Sodium sulphide → Na2S
(iv) Magnesium hydroxide → Mg(OH)2
Question 8.
Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3
Answer:
(i) Al2(SO4)3 → Aluminium sulphate
(ii) CaCl2 → Calcimn chloride
(iii) K2SO4 → Potassium sulphate
(iv) KNO3 → Potassium nitrate
(v) CaCO3 → Calcium carbonate
Question 9.
What is meant by the term chemical formula?
Answer:
The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound.
For example, from the chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.
Question 10.
How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?
Answer:
(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO43- ion, five atoms are present; one of phosphorus and four of oxygen.
Question 11.
Calculate the molecular masses of H2, O2, Cl2, CO2, CH4; C2H6, C2H4, NH3, CH3OH.
Answer:
Molecular mass of H2 = 2 × Atomic mass of H
= 2 × 1
= 2u
Molecular mass of O2 = 2 × Atomic mass of O
= 2 × 16
= 32 u
Molecular mass of Cl2 = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u
Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u
Molecular mass of C2H6 = 2 × Atomic mass of C+ 6 × Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u
Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28u
Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1
= 17u
Molecular mass of CH3OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O
= 12 + 4 × 1 + 16
= 32 u
Question 12.
Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer:
Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u
Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u
Formula unit mass of K2CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 138 u
Question 13.
If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer:
One mole of carbon atoms weighs 12 g
(Given)
i. e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 1023 number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon 12
= \(\frac{12}{6.022 \times 10^{23}} \mathrm{~g}\)
= 1.9926 × 10-23 g
Question 14.
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:
Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 1023
number of atoms
Thus, 100 g of Na contains \(\frac{6.022 \times 10^{23}}{23} \times 100\)
number of atoms
= 2.6182 × 1024 number of atoms
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = 6.022 × 1023 number of atoms
Thus, 100 g of Fe contains = \(\frac{6.022 \times 10^{23}}{56} \times 100\)
= 1.0753 × 1024 number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.
Class 9 Science Chapter 3 Atoms and Molecules Textbook Questions and Answers
Question 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:
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage or boron by weight in the compound = \(\frac{0.096}{0.25}\) × 100%
= 40%
And, percentage of oxygen by weight in the compound = \(\frac{0.144}{0.25}\) × 100%
= 60%
Question 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:
Carbon + Oxygen → Carbon dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g 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 unreactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Question 3.
What are polyatomic ions? Give examples?
Answer:
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (NH4+), hydroxide ion (OH–), carbonate ion (CO32-), sulphate ion (SO42-).
Question 4.
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride → MgCl2
(b) Calcium oxide → CaO
(c) Copper nitrate → Cu (NO3)2
(d) Aluminium chloride → AlCl3
(e) Calcium carbonate → CaCO3
Question 5.
Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
| Compound | Chemical formula | Elements present |
| Quick lime | CaO | Calcium, Oxygen |
| Hydrogen bromide | HBr | Hydrogen bromide |
| Baking powder | NaHCO3 | Sodium, Hydrogen, Carbon, Oxygen |
| Potassium sulphate | K2SO4 | Potassium, Sulphur, Oxygen |
Question 6.
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
Answer:
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 26g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g
(c) Molar mass of phosphorusjnolecule, P4 = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63 g
Question 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
Question 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 = \(\frac {12}{32}\) = 0.375 mole
(b) 18 g of water = 1 mole
Then, 20 g of water= \(\frac {20}{18}\) = 1.11 moles (approx)
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = \(\frac {22}{44}\) = 0.5 mole
Question 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 = 3.2g
(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
Question 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 \(\frac{6.022 \times 10^{23}}{256} \times 16\) molecules
= 3.76 × 1022 molecules (approx)
Question 10.
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:
1 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
= \(\frac{6.022 \times 10^{23}}{102} \times 0.051\)
= 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 pxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
Class 9 Science Chapter 3 Atoms and Molecules Additional Important Questions and Answers
Multiple Choice Questions
Choose the correct option:
Question 1.
Which of the following correctly represents 360 g of water?
(i) 2 moles of H2O
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv) 1.2044 × 1025 molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(d) (ii) and (iv)
Question 2.
Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch
Answer:
(a) Atoms are not able to exist independently
Question 3.
The chemical symbol for nitrogen gas is
(a) Ni
(b) N2
(c) N+
(d) N
Answer:
(b) N2
Question 4.
The chemical symbol for sodium is
(a) So
(b) Sd
(c) NA
(d) Na
Answer:
(d) Na
Question 5.
Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C12H22O11)
(b) 2 moles, of CO2
(c) 2 moles of CaCO3
(d) 10 moles of H2O
Answer:
(c) 2 moles of CaCO3
Question 6.
Which of the following has maximum number of atoms?
(a) 18g of H2O
(b) 18g of O2
(c) 18g of CO2
(d) 18g of CH4
Answer:
(d) 18g of CH4
Question 7.
Which of the following contains maximum number of molecules?
(a) 1g CO2
(b) 1g N2
(c) 1g H2
(d) 1g CH4
Answer:
(c) 1g H2
Question 8.
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 × 1023
(b) 6.09 × 1022
(c) 6.022 × 1023
(d) 6.022 × 1021
Answer:
(a) 6.68 × 1023
Question 9.
A change in the physical state can be brought about
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change
Answer:
(c) when energy is either given to, or taken out from the system
Very Short Answer Type Questions
Question 1.
Define law of conservation of mass.
Answer:
In a chemical reaction mass can neither be created nor destroyed.
E.g., 2Na + Cl2 → 2NaCl
2 × 23 + 2 × 35.5 → 2(23 + 35.5)
Question 2.
Explain law of constant proportion.
Answer:
In a chemical substance the elements are always present in definite proportions by mass.
E.g., In water, the ratio of the mass of hydrogen to the mass of oxygen H : O is always 1 : 8
Question 3.
Who coined the term atom?
Answer:
John Dalton coined the term atom.
Question 4.
Define atom.
Answer:
The smallest particle of matter, which can take part in a chemical reaction is called atom.
Question 5.
Define molecule.
Answer:
The smallest particle of an element or compound which can exist independently is called molecule.
Question 6.
Define atomicity.
Answer:
The number of atoms constituting a molecule is known as its atomicity.
Question 7.
What is atomic mass unit?
Answer:
The sum of the atomic masses of all die atoms in a molecule of the substance is expressed in atomic mass unit.
E.g., H2O = (1 × 2) + 16 = 18 amu
Question 8.
How do atoms exist?
Answer:
Atoms exist in the form of atom, molecule or ions.
Question 9.
Give the atomicity of phosphorous and nitrogen.
Answer:
The atomicity of phosphorus is P4 i.e., 4.
The atomicity of nitrogen is N2 i.e., 2.
Question 10.
What is an ion?
Answer:
Charged atom is called as an ion. The ion can be positively charged called cation or negatively charged called anion.
Question 11.
Give one example of cation and anion.
Answer:
Cation ⇒ Na+
Anion ⇒ Cl–
Question 12.
Give one difference between cation and anion.
Answer:
Cations are positively charged ion. Anions are negatively charged ion.
Question 13.
Give the chemical formula for ammonium sulphate.
Answer:
Ammonium sulphate NH4+ SO42-
Chemical formula → (NH4)2SO4.
Question 14.
What is Avogadro’s constant?
Answer:
The Avogadro’s constant (6.022 × 1023) is defined as the number of atoms that are present inexactly 12 g of carbon-12.
Question 15.
Find the molecular mass of H2O.
Answer:
Molecular mass of H2O=(2 × 1) + (16) = 2 + 16 = 18 u
Short Answer Type Questions
Question 1.
Give the unit to measure size of atom and give size of hydrogen atom.
Answer:
The unit to measure size of atom is nanometer, size of hydrogen atom is 10-10 m.
Question 2.
What is IUPAC, give its one function?
Answer:
IUPAC stands for International Union of Pure and Applied Chemistry. It approves the names of elements.
Question 3.
Give the Latin name for sodium, potassium, gold and mercury.
Answer:
Sodium → Natrium, Gold → Aurum Potassium → Kalium, Mercury → Hydrargyrum
Question 4.
What is the ratio by mass of combining elements in H2O,CO2 and NH3?
Answer:
H2O ratio by mass of combining elements 2 : 16 → 1 : 8 (H : O)
CO2 ratio by mass of combining elements 12 : 32 → 3 : 8 (C : O)
NH3 ratio by mass of combining elements 14 : 3 → 14 : 3 (N : H)
Question 5.
Define valency and give the valency for the following elements:
Magnesium, Aluminium, Chlorine and Copper.
Answer:
Valency: The combining capacity of an element is called its valency.
Valency of the following elements:
Magnesium — 2
Aluminium — 3
Chlorine — 1
Copper — 2
Question 6.
What is polyatomic ion? Give one example.
Answer:
A group of atoms carrying a charge is known as a polyatomic ion.
E.g., Ammonium — NH4–
Nitrate — NO3–
Question 7.
Write down the formula for:
Copper nitrate, calcium sulphate and aluminium hydroxide.
Answer:
Chemical formula:
Copper nitrate → Cu(NO3)
Calcium sulphate → CaSO4
Aluminium hydroxide → Al(OH)3
Question 8.
What is formula unit mass? How is it different from molecular mass?
Answer:
The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound. The constituent particles of formula unit mass are ions and the constituent particles of molecular mass are atoms.
Question 9.
Find the number of moles in the following:
(i) 50g of H2O
(ii) 7g of Na
Answer:
Number of moles in
(i) Molar mass of H2O = 18g
Given mass of H2O = 50 g
∴ No. of moles in 50g of H2O
= \(\frac {50}{18}\) = 2.78 moles.
(ii) Molar mass of Na = 23g
Given mass of Na = 7g
∴ No. of moles in 7g of Na = \(\frac {7}{23}\) = 0.304 moles.
Question 10.
Find the number of atoms in the following:
Answer:
(i) 0.5 mole of C atom
(ii) 2 mole of N atom
Answer:
(i) 0.5 mole of C atom:
Number of atoms in 1 mole of C atom
= 6.022 × 1023 atoms
Number of atoms in 0.5 mole of C atom
= 6.022 × 1023 × 0.5
= 3.011 × 1023 atoms
(ii) 2 mole of N atom:
Number of atoms in 1 mole of N atom = 6.022 × 1023 atoms
∴ Number of atoms in 2 mole of N atom
= 6.022 × 2 × 1023
= 1.2044 × 1024 atoms
Question 11.
Find the mass of the following:
(i) 6.022 × 1023 number of O2 molecules
(ii) 1.5 mole of CO2 molecule
Answer:
(i) 6.022 × 1023 number of O2 molecules:
Mass of 1 mole of O2 molecule = 6.022 × 1023 molecules = 32 g
(ii) 1.5 mole of CO2 molecule:
Mass of 1 mole of CO2 molecule = 6.022 × 1023 molecules = 44 g
Mass of 1.5 mole CO2 molecule = 44 × 1.5 = 66 g
Question 12.
What are the rules for writing the symbol of an element?
Answer:
IUPAC → International Union of Pure and Applied Chemistry approves name of elements.
Symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (upper case) and the second letter as a small letter (lower case).
e.g., Hydrogen → H
Helium → He
Some symbols are taken from the names of elements in Latin, German or Greek.
e.g., Symbol of iron is Fe, its Latin name is Ferrum.
Symbol of sodium is Na, its Latin name is Natrium.
Question 13.
Explain relative atomic mass and relative molecular mass.
Answer:
Relative atomic mass: It can be defined as the number of times one atom of given element is heavier than \(\frac {1}{2}\) th of the mass of an atom of carbon-12.
Relative Molecular Mass: It is defined as the number of times one molecule of a substance or given element is heavier than vyth of the mass of one atom of carbon-12.
Question 14.
The formula of carbon-dioxide is CO2. What information do you get from this formula?
Answer:
- CO2 represents carbon-dioxide.
- CO2 is one molecule of carbon-dioxide.
- CO2 is one mole of carbon-dioxide i.e., it contains 6.022 × 1023 molecules of carbon dioxide.
- CO2 contains 1 atom of carbon and two atoms of oxygen.
- CO2 represents 44 g of molar mass.
Question 15.
State 3 points of difference between an atom and an ion.
Answer:
Atom:
- An atom has no change.
- Number of electrons = number
- Atom is reactive.
Ion:
- Anion has either positive or negative charge.
- Number of electrons ≠ number of protons.
- Ion is stable.
Question 16.
Calculate the formula unit mass of NaCl and CaCl2.
(Na = 23, Cl = 35.5, Ca = 40)
Answer:
Formula unit mass of NaCl = 23 + 35.5
= 58.5 u
Formula unit mass of CaCl2 = 40 + (2 × 35.5)
= 40 + 71
= 111 u
Question 17.
Write down the chemical formula for the following compounds:
(a) Aluminium carbonate
(b) Calcium sulphide
(c) Zinc carbonate
(d) Copper phosphate
(e) Magnesium bicarbonate
(f) Aluminium hydroxide.
Answer:
The chemical formula are:
Question 18.
Give the atomicity of the following compounds:
(a) Ca(OH)2
(b) Mg(HCO3)2
(c) Cu2O
(d) H2SO4
(e) Al2(SO4)
(f) MgCl2
Answer:
The atomicity of the molecules are:
(a) Ca(OH)2 → 05
(b) Mg(HCO3)2 → 11
(c) Cu2O → 03
(d) H2SO4 → 07
(e) Al2(SO4)3 → 17
(f) MgCl2 → 03
Question 19.
Explain the difference between 2O, O2 and O3.
Answer:
2O → It represents 2 atoms of oxygen (cannot exist independently).
O2 → It represents one molecule of oxygen (made up of 2 atom) can exist freely.
O3 → It represents one molecule of ozone (made up of 3 atoms) it can exist independently.
Question 20.
1.50 g sample of barium hydroxide was dissolved in water. The total volume of the solution was 100 cm3.
A 25.0 cm3 portion of the barium hydroxide solution was titrated against hydrochloric acid. The volume of hydrochloric acid required was 18.75 cm3.
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
(i) Calculate how many moles of barium hydroxide were in the 25.0 cm3 portion used in the titration.
(ii) Calculate the concentration of the hydrochloric acid used.
Answer:
(i) Moles of barium hydroxide = 0.00219 mol.
(ii) Moles of HCl = 2 × 0.00219 = 0.00438.
Question 21.
Analysis of a certain compound showed that 39.348 grams of it contained 0.883 grams of hydrogen, 10.497 grams of Carbon, and 27.968 grams of Oxygen. Calculate the empirical formula of the compound.
Answer:
First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula
Analysis of the ratio shows that the first two are identicaland that the third is twice the other two. Therefore the ratio of H to C to O is 1 to 1 to 2. The empirical formula is HCO2.
Long Answer Type Questions
Question 1.
(a) How do atoms exist?
(b) What is atomicity?
(c) What are polytom1c Ions?
Answer:
(a) Atoms of some elements are not able to exist independently. For such elements atoms
form molecules and ions. In case of metals and inert gases atoms can exist independently.
Atoms of metals and inert gases: E.g.,
\(\frac{\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}}{\text { Metals }} \frac{\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}}{\text { Inert gases }}\)
Non-metals: E.g., H2, Cl2, P4, S8
Exceptional non-metal C
(b) The number of atoms constituting a molecule is known as its atomicity.
E.g., O3 → atomicity is 3 .
O2 → atomicity is 2
(c) Polyatomic ions: When more than two atoms combine together and act like an atom with a charge on it is called polyatomic ion.
E.g., OH–, NO3–,NH4+
Question 2.
Calculate
(a) the mass of one atom 6f oxygen
(b) the mass of one molecule of oxygen
(c) the mass of one mole of oxygen gas
(d) the mass of one ion of oxygen
(e) the number of atoms in 1 mole of oxygen molecule
Answer:
(a) Mass of one atom of oxygen
1 mole of oxygen atom = 16gm = 6.022 × 1023 atoms.
∴ Mass of one atom of oxygen
= \(\frac{16}{6.022 \times 10^{23}}\) = 2.65 × 10-23
(b) Mass of one molecule of oxygen 1 molecule of oxygen = O2
= 2 × 16
= 32 u
(c) Mass of one mole of oxygen gas
1 mole of oxygen gas is O2 = 32 u or 32 g
(d) Mass of one ion of oxygen
One mole of oxygen = 6.022 × 1023 atoms = 16 g
Mass of one ion of oxygen = \(\frac{16}{6.022 \times 10^{23}}\) = 2.65 × 10-23
(e) Number of atoms in one mole of oxygen molecule
1 mole of oxygen molecule i.e.,
O2 = 6.022 × 1023 molecules.
1 molecule of O2 = 2 atoms.
∴ Number of atoms in 1 mole of oxygen molecule = 6.022 × 1023 × 2 atoms
= 1.2044 × 1024 atoms
Question 3.
What is meant by atomic mass, gram atomic mass of an element? Why is the mass have different expressions i.e., ‘u’ and ‘g’?
Answer:
The atoms are very tiny and their individual mass cannot be calculated as it is negligible. Hence the mass of atoms is expressed in units with respect to a fixed standard. Initially hydrogen atom with mass 1 was taken as standard unit by Dalton. Later, it was replaced by oxygen atom (O = 16). But due to the isotopes the masses were found in fractions instead of whole number. Hence, carbon (C = 12) isotope was taken as standard unit and was universally accepted.
The atomic mass unit is equal to one twelfth (\(\frac {1}{12}\)) the mass of an atom of carbon-12, its unit is u.
Gram atomic mass: When the atomic mass of an element is expressed in grams, it is called the gram atomic mass of the element. The mass of atoms, molecules is expressed in ‘u’ and the mass of moles i.e., molar mass is expressed in g.
Question 4.
Define a mole. Give the significance of the mole.
Answer:
Mole-One mole of any species (atoms, molecules, ions or particles) is that quantity or number having a mass equal to its atomic or molecular mass in grams. 1 mole = 6.022 × 1023 in number (atoms, molecules, ions or particles)
Significance of the mole:
1. A mole gives the number of entities present i.e, 6.022 × 1023 particles of the substance.
2. Mass of 1 mole is expressed as M grams.
3. Mass of 1 mole=mass of 6.022 × 1023 atoms of the element.
E.g., 1 mole of O2 = 6.022 × 1023 atoms
1 mole of O2 = 2 × 16 = 32g
6.022 × 2 × 1023= 1.2044 × 104 atoms
1 mole of (compound) HCl = 6.022 × 1023 atoms of H and Cl atoms
(1 + 35.5 = 36.5 g) (6.022 × 1023 molecules of HCl)
Question 5.
Barium carbonate decomposes when heated
BaCO3 (s) → BaO(s) + CO2(g)
(a) A student heated a 10.0 g sample of barium carbonate until it was fully decomposed.
(i) Calculate the number of moles of barium carbonate the student used.
(ii) Calculate the volume of carbon dioxide gas produced at room temperature and pressure. Give your answer in dm3.
(b) The student added 2.00 g of the barium oxide produced to water.
BaO + H2O → Ba(OH)2
Calculate the mass of barium hydroxide that can be made from 2.00 g of barium oxide. The molecular mass of Ba(OH)2 is 171.
Answer:
(a) (i) Relative formula mass BaCO3 = 197,
moles of barium carbonate = 10.0/197 = 0.0508 mol.
(ii) Volume of carbon dioxide = 1.22 dm3
(b) Mass of barium hydroxide = 2.24 g
Question 6.
Magnesium sulphate crystals are hydrated. A student heated some hydrated magnesium sulphate crystals in a crucible and obtained the following results. Mass of hydrated magnesium sulfate crystals = 4.92 g Mass of water removed = 2.52 g
(i) Calculate the number of moles of water removed,
(ii) Calculate the number of moles of anhydrous magnesium sulfate remaining in the crucible. The molecular mass of anhydrous nlagnesium sulfate is 120.
(iii) Calculate the ratio of moles of anhydrous magnesium sulfate: moles of water.
Answer:
(i) Moles of water= 2.52/18 = 0.14. moles.
(ii) Moles of anhydrous magnesium sulfate = 0.02 moles,
(iii) Ratio = 0.02/0.02 : 0.14/0.02 = ratio in whole numbers is 1 : 7.
Question 7.
Calculate the mass of each element in potassium carbonate, K2CO3.
Answer:
First calculate the formula mass for K2CO3. Find the atomic mass of each element from the periodic table. Multiply it by the number, of times it appears in the formula and add up the total
To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass.
Percent of Potassium
K= \(\frac{78.20}{138.21} \times 100\) = 56.58%
Percent of Carbon
C = \(\frac{12.01}{138.21} \times 100\) = 8.69%
Percent of Oxygen
O = \(\frac{48.00}{138.21} \times 100\) = 34.73%
Question 8.
For each of the following calculate the empirical formula:
(a) When 2.20 g of a hydrocarbon, D, is burnt in excess oxygen, 6.90 g of CO2 and 2.83 g of water are produced.
(b) When 1.52 g of compound E, which contains carbon, hydrogen and oxygen only, is burnt in excess oxygen, 3.04 g CO2 and 1.24 g H2O are produced.
Answer:
(a) Moles of CO2 = 6.90 ÷ 44.01 = 0.157 mol
Moles of H2O = 2.83 ÷ 18.02 = 0.157 mol
Moles of C = 0.157 mol moles of H = 2 ÷ 0.
157 = 0.304 mol
Empirical formula = CH2
(b) Mass of C in CO2 = 12.01 ÷ 44.01
3.04 = 0.830 g
Mass of H in H2O = 2.02 ÷ 18.02
1.24 = 0.139g
Mass of O = 1.52 – (0.830 + 0.139) = 0.551 g
Ratio of moles C : H : O = 0.0691 : 0.139 : 0.0344
Whole number ratio = 2.01 : 4.04 : 1
Empirical formula: C2H4O