Relative Atomic Mass Of Hydrogen

Students can download 10th Science Chapter 7 Atoms and Molecules Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

What Is Relative Mass
Dihydrogen Monoxide Formula
Relative Atomic Mass Of Hydrogen Is 1.008
Relative Atomic Mass Of Hydrogen Gas

Examples of Atomic Mass. The atomic mass of carbon is 12.011. Most carbon atoms consist of six protons and six neutrons. The atomic mass of hydrogen is 1.0079. Hydrogen (atomic number 1) is the element that has the lowest atomic mass. The most common isotope of hydrogen is protium, an atom that consists of a proton or a proton and an electron. I am a student of Science from Pakistan. The Atomic Mass of Hydrogen is 1 and not 2 because of its nucleus. It's nucleus contains 1 proton and it has no neutrons in its nucleus and also has 1 electron which is revolving around its nucleus. Relative Atomic Mass On the Periodic Table we note that the atomic mass for Hydrogen is listed as 1.01 g. This may appear odd as we know that the Hydrogen atom contains 1 electron (which has no significant mass) and 1 proton (which is a significant mass unit). Ta is the sixth period element in the VB family, with an atomic number of 73, relative atomic mass of 180.95, density of 16.6 g/cm3, and melting point of about 3,000°C (2,980 ± 20°C), which is a little lower than W and Re 135. Ta is a rare refractory metal with a hardness of HV120 and good ductility.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules

Samacheer Kalvi 10th Science Atoms and Molecules Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
Which of the following has the smallest mass?
(a) 6.023 × 10²³ atoms of He
(b) 1 atom of He
(c) 2 g of He
(d) 1-mole atoms of He
Answer:
(b) 1 atom of He

Question 2.
Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen.
Answer:
(c) Carbon dioxide
Hint:
(a) Glucose = C₆H₁₂O₆ (Polyatomic molecule)
(b) Helium = He (Monoatomic molecule)
(c) Carbon dioxide = CO₂ (Triatomic molecule)
(d) Hydrogen = H₂ (Diatomic molecule)
So, (c) is the correct answer.

Question 3.
The volume occupied by 4.4 g of CO₂ at S.T.P:
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre
Answer:
(b) 2.24 litre

Question 4.
Mass of 1 mole of Nitrogen atom is _____.
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g.
Answer:
(b) 14 amu
Hint: Atomic mass of Nitrogen is 14.00674 grams. It is equal to 1 mole of Nitrogen atoms.
So, answer (b) is correct.

Question 5.
Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) 1/12 th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer:
(c) 1/12 th of the mass of a C – 12 atom

Question 6.
Which of the following statement is incorrect?
(a) One gram of C – 12 contains Avogadro’s number of atoms.
(b) One mole of oxygen gas contains Avogadro’s number of molecules.
(c) One mole of hydrogen gas contains Avogadro’s number of atoms.
(d) One mole of electrons stands for 6.023 × 10²³ electrons.
Answer:
(a) One gram of C – 12 contains Avogadro’s number of atoms.
Hint: 12 g of Carbon contains 6.023 × 10²³ atoms,
1 g of Carbon contain (frac{6.023 times 10^{23}}{12}) = 5.018 × 10²² atoms and its is not Avogadro’s number of atoms.
So (a) is the incorrect statement.

Question 7.
The volume occupied by 1 mole of a diatomic gas at S.T.P is:
(a) 11.2 litre
(b) 5.6 litre
(c) 22.4 litre
(d) 44.8 litre
Answer:
(c) 22.4 litre

Question 8.
In the nucleus of ₂₀Ca⁴⁰, there are
(a) 20 protons and 40 neutrons
(b) 20 protons and 20 neutrons
(c) 20 protons and 40 electrons
(d) 40 protons and 20 electrons
Answer:
(b) 20 protons and 20 neutrons

Question 9.
The gram molecular mass of oxygen molecule is_____.
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g.
Answer:
(c) 32 g
Hint: By definition, the gram molecular mass of oxygen molecule O₂ is 32 g.
So the answer (c) is correct.

Question 10.
1 mole of any substance contains molecules.
(a) 6.023 × 10²³
(b) 6.023 × 10^-23
(c) 3.0115 × 10²³
(d) 12.046 × 10²³
Answer:
(a) 6.023 × 10²³

II. Fill in the blanks:

1. Atoms of different elements having ……… mass number, but ………. atomic numbers are called isobars.
2. Atoms of different elements having same number of ………. are called isotones.
3. Atoms of one element can be transmuted into atoms of other element by ………….
4. The sum of the numbers of protons and neutrons of an atom is called its …………
5. Relative atomic mass is otherwise known as …………
6. The average atomic mass of hydrogen is ……….. amu.
7. If a molecule is made of similar kind of atoms, then it is called ……….. atomic molecule.
8. The number of atoms present in a molecule is called its ………….
9. One mole of any gas occupies ………… ml at S.T.P
10. Atomicity of phosphorous is …………
Answer:
1. same, different
2. neutrons
3. artificial transmutation
4. mass number
5. standard atomic weight
6. 1.008
7. homo
8. atomicity
9. 22, 400
10. four

III. Match the following:

Answer:
A. (ii)
B. (iii)
C. (v)
D. (i)
E. (iv)

IV. True or False: (If false give the correct statement)

Two elements sometimes can form more than one compound.
Nobel gases are diatomic.
The gram atomic mass of an element has no unit.
1 mole of Gold and Silver contain same number of atoms.
Molar mass of CO₂ is 42 g.

Answer:

True
False – Noble gases are Monoatomic.
False – The unit of gram atomic mass of an element is gram.
True
False – Molar mass of CO₂ is 44 g.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12 th of the mass of the C-12 atom.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(c) Assertion is wrong, Reason is correct.

VI. Short answer questions:

Question 1.
Define: Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to (frac{1}{12^{th}})part of the mass of a carbon-12 atom. It is denoted as A_r.
[OR]

Question 2.
Write the different types of isotopes of oxygen and its percentage abundance.
Answer:
Oxygen has three stable isotopes. They are

Question 3.
Define Atomicity.
Answer:
The number of atoms present in the molecule is called its ‘Atomicity’.

Question 4.
Give any two examples for heteroatomic molecules.
Answer:
HI, HCl, CO, HBr, HF.

Question 5.
What is Molar volume of a gas?
Answer:
One mole of any gas occupies 22.4 litres.
(or)
22400 ml at S.T.R This volume is called as molar volume.

Question 6.
Find the percentage of nitrogen in ammonia.
Answer:
Molar mass of NH₃ = 1(14) + 3(1) = 17 g

VII. Long answer questions:

Question 1.
Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer:
The molecular mass of water (H₂O) is 18.
18 g of water molecule = 1 mole.
0. 18 g of water = (frac{1}{18} times 0.18) = 0.01 mole.
1 mole of water (Avogadro’s number) contains 6.023 × 10²³ water molecules.
0. 01 mole of water contain (frac{6.023 times 10^{23}}{1} times 0.01) = 6.023 × 10²¹ molecules.

Question 2.
N₂ + 3 H₂ → 2 NH₃
(The atomic mass of nitrogen is 14, and that of hydrogen is 1)
1 mole of nitrogen (……..g) +
3 moles of hydrogen (………g) →
2 moles of ammonia (………g)
Answer:
1 mole of nitrogen (28 g) +
3 moles of hydrogen (6 g) →
2 moles of ammonia (34 g)

Question 3.
Calculate the number of moles in
(i) 27 g of Al;
(ii) 1.51 × 10²³ molecules of NH₄Cl.
Answer:
(i) 27 g of Al
Given mass atomic mass = (frac{Given Mass}{Atomic Mass}) = (frac{27}{27})
= 1 mole

(ii) 1.51 x 10²³ molecules of NH₄Cl
Number of moles

Question 4.
Give the salient features of “Modern atomic theory”.
Answer:
The salient features of “Modem atomic theory” are,

An atom is no longer indivisible.
Atoms of the same element may have different atomic mass.
Atoms of different elements may have the same atomic masses.
Atoms of one element can be transmuted into atoms of other elements. In other words, an atom is no longer indestructible.
Atoms may not always combine in a simple whole-number ratio.
Atom is the smallest particle that takes part in a chemical reaction.
The mass of an atom can be converted into energy [E = mc²].

Question 5.
Derive the relationship between Relative molecular mass and Vapour density.
Answer:
Relative molecular mass : The relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of hydrogen.
Vapour density : Vapour density is the ratio of the mass of certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
According to Avogadro’s law equal volumes of all gases contain equal number of molecules.
Let the number of molecules in one volume = n, then
When cancelling ‘n’ which is common at STP, we get
Since hydrogen is diatomic,
2 × Vapour density = Relative Molecular mass of a gas
[OR]
Relative Molecular Mass = 2 × Vapour density

VIII. HOT Question:

Question 1.
Calcium carbonate is decomposed on heating in the following reaction
CaCO₃ → CaO + CO₂

How many moles of Calcium carbonate is involved in this reaction?
Calculate the gram molecular mass of calcium carbonate involved in this reaction.
How many moles of CO₂ are there in this equation?

Answer:
CaCO₃ → CaO + CO₂

1 mole of CaCO₃ is involved in this reaction.
Gram molecular mass of calcium carbonate
CaCO₃ = (40 + 12 + 3 × 16) = 52 + 48 = 100 g
1 mole of CO₂ is in this equation.

IX. Solve the following problems:

Question 1.
How many grams are there in the following?
(i) 2 moles of a hydrogen molecule, H₂
(ii) 3 moles of chlorine molecule, Cl₂
(iii) 5 moles of sulphur molecule, S₈
(iv) 4 moles of a phosphorous molecule, P₄
Solution:
(i) 2 moles of a hydrogen molecule, H₂
Mass of 1 mole of hydrogen molecule = 2 g
Mass of 2 moles of hydrogen molecule = 2 × 2 = 4 g.

(ii) 3 moles of chlorine molecule, Cl₂
Mass of 1 mole of chlorine molecule = 71 g
Mass of 3 moles of chlorine molecules = 71 × 3 = 213 g.

(iii) 5 moles of sulphur molecule, S₈
Mass of 1 mole of sulphur molecule = 32 g
Mass of 5 moles of sulphur molecules = 32 × 5 = 160 g.

(iv) 4 moles of the phosphorous molecule, P₄
Mass of 1 mole of phosphorous molecule = 30.97 g
Mass of 4 moles of phosphorous molecules = 30.97 × 4 = 123.88 g.

Question 2.
Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca – 40)
Answer:
Formula to find % of each element

Question 3.
Calculate the % of oxygen in Al₂(SO₄)₃.
(Atomic mass: Al – 27, O – 16, S – 32)
Answer:
Formula:
Molar mass of Al₂(SO₄)₃ = [2(Atomic mass of Al) + 3(Atomic mass of S) + 12(Atomic mass of O)]
= 2(27) + 3(32) + 12(16) = 342 g
% of Oxygen = (frac{12(16)}{342}) × 100 = 56.14%.

Question 4.
Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.
Answer:
% of relative abundance can be calculated by the formula.
Average atomic mass of the element
= Atomic mass of 1st isotope × abundance of 1st isotope + Atomic mass of 2nd isotope × abundance of 2nd isotope
∴ Average atomic mass of Boron
= Atomic mass of B – 0 × abundance of B -10 + Atomic mass of B – 11 × abundance of B – 11
Let the abundance of B – 10 be ‘x’ and B – 11 be (1 – x)
So, 10.804 = 10 × x + 11 (1 – x)
10.804 = 10x + 11 – 11x
x = 11 – 10.804
x = 0.196
1 -x = 1 – 0.196 = 0.804
Therefore % abundance of B – 10 is 19.6% and B – 11 is 80.4%
[OR]
Let the % of the isotope B – 10 = x
Then the % of the isotope B – 11 = 100 – x
1100 – x = 1080.4
x = 19.6
% abundance of B – 10 = 19.6%
% abundance of B – 11 = 80.4%

Samacheer Kalvi 10th Science Atoms and Molecules Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The first scientific theory of an atom was proposed by:
(a) Ruther Ford
(b) Newland
(c) John Dalton
(d) Neils Bohr
Answer:
(c) John Dalton

Question 2.
Identify the pair that indicates isobars among the following _____.
(a) (_{1} mathrm{H}^{2},_{1} mathrm{H}^{3})
(b) (_{17} mathrm{Cl}^{35},_{17} mathrm{Cl}^{37})
(c) (text { (c) }_{18} mathrm{Ar}^{40},_{18} mathrm{Ca}^{40})
(d) (_{6} mathrm{C}^{13},_{7} mathrm{N}^{14}).
Answer:
(c) (text { (c) }_{18} mathrm{Ar}^{40},_{18} mathrm{Ca}^{40})

Question 3.
Which one of the following represents 180 g of water?
(a) 5 moles of water
(b) 90 moles of water
(c) 6.023 × 10²⁴ molecules of water
(d) 6.023 × 10²² molecules of water
Answer:
(c) 6.023 × 10²⁴ molecules of water

Question 4.
The isotope of Carbon-12 contains _____.
(a) 6 protons and 7 electrons
(b) 6 protons and 6 neutrons
(c) 12 protons and no neutrons
(d) 12 neutrons and no protons.
Answer:
(b) 6 protons and 6 neutrons

Question 5.
Which contains the greatest number of moles of oxygen atoms?
(a) 1 mol of water
(b) 1 mole of NaOH
(c) 1 mole of Na₂CO₃
(d) 1 mole of CO
Answer:
(c) 1 mole of Na₂CO₃

Question 6.
The mass of proton or neutron is approximately _____.
(a) 1 amu
(b) 1.609 × 10^-19 g
(c) 1 g
(d) 6.023 × 10^-23 g.
Answer:
(a) 1 amu

Question 7.
The natural abundance of C-12 and C-13 are 98.90% and 1.10% respectively. The average atomic mass of carbon is:
(a) 12 amu
(b) 12.011 amu
(c) 14 amu
(d) 12.90 amu
Answer:
(b) 12.011 amu

Question 8.
The relative atomic mass of magnesium-based on C – 12 scale is _____.
(a) 24 g
(b) 24
(c) 24 amu
(d) 24 kg
Answer:
(b) 24

Question 9.
If 1.5 moles of oxygen combine with Al to form Al₂O₃, the mass of Al in g (atomic mass of Al = 27) used in the reaction is:
(a) 2.7
(b) 54
(c) 40.5
(d) 81
Answer:
(b) 54

Question 10.
The atomicity of methane is:
(a) 5
(b) 4
(c) 3
(d) 6
Answer:
(a) 5

Question 11.
Find the odd one out _____.
(a) (_{8} mathrm{O}^{16})
(b) (_{8} mathrm{O}^{17})
(c) (_{6} mathrm{O}^{12})
(d) (_{8} mathrm{O}^{18}).
Answer:
(c) (_{6} mathrm{O}^{12})

Question 12.
The volume occupied by 3 moles of HCl gas at STP is:
(a) 22.4 L
(b) 44.8 L
(c) 2.24 L
(d) 67.2 L
Answer:
(d) 67.2 L

Question 13.
The mass percentage of hydrogen in ethane (C₂H₆) is:
(a) 25%
(b) 75%
(c) 80%
(d) 20%
Answer:
(d) 20%

Question 14.
Which one of the following is a homo diatomic molecule?
(a) H₂
(6) CO
(c) NO
(d) O₃.
Answer:
(a) H₂

Question 15.
The percentage of nitrogen in urea is about:
(a) 38.4
(b) 46.6
(c) 59.1
(d) 61.3
Answer:
(b) 46.6

Question 16.
Out of the following the largest number of atoms are contained in:
(a) 11 g of CO₂
(b) 4 g of H₂
(C) 5 g of NH₃
(d) 8 g of SO₂
Answer:
(b) 4 g of H₂

Question 17.
Which of the following is an example of a homo triatomic molecule?
(a) Phosphorous
(b) Sulphur
(c) Bromine
(d) Ozone.
Answer:
(d) Ozone.

Question 18.
For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce:
(a) 5 moles of C
(b) 4 moles of C
(c) 8 moles of C
(d) 13 moles of C
Answer:
(b) 4 moles of C

Question 19.
The vapour density of a gas is 32. Its relative molecular mass will be:
(a) 32
(b) 16
(c) 64
(d) 96
Answer:
(c) 64

Question 20.
Find the odd one out _____.
(a) Silver
(b) Potassium
(c) Iron
(d) Phosphorous.
Answer:
(d) Phosphorous.

II. Fill in the blanks.

The volume occupied by 16 g of oxygen is ………..
One mole of a triatomic gas contains ………… atoms.
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of …………
The mass of an atom can be converted into energy by using the formula …………
The percentage composition is useful to determine the ………… formula and ………… formula.

Answer:

11.2 L
3 × 6.023 × 10²³
molecules
E = me²
empirical, molecular

III. Match the following:

Question 1.
Match the Column I with Column II.
Answer:
A. (ii)
B. (iii)
C. (i)
D. (v)
E. (iv)

Question 2.
Match the Column I with Column II.
Answer:
A. (iii)
B. (i)
C. (iv)
D. (v)
E. (ii)

Question 3.
Match the Column I with Column II.
Answer:
A. (iv)
B. (v)
C. (ii)
D. (iii)
E. (i)

Question 4.
Match the Column I with Column II.
Answer:
A. (v)
B. (iii)
C. (iv)
D. (i)
E. (ii)

IV. True or False: (If false give the correct statement)

Atoms always combine in a simple whole number ratio.
2 × RMM = VD
The average atomic mass of Beryllium is 9.012 because of the presence of isotopes.
The noble gases are diatomic.
The number of atoms present in one mole of phosphorus(P₄) is 4 × 6.023 × 10²³

Answer:

False -Atoms may not combine always in a simple whole number ratio.
False – 2 × VD = RMM
True
False – The noble gases are mono atomic.
True

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: The standard unit for expressing mass of atom is amu.
Reason: Atomic mass unit is one-twelth of the mass of a C-12 atom
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: The volume occupied by 44 g of CO₂ is 22.4 L
Reason: The volume occupied by one mole of any gas is 22.4 L
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

VI. Short answer questions:

Question 1.
Define vapour density.
Answer:
Vapour density is the ratio of the mass of certain volume of a gas or vapour, to the mass of an equal volume of hydrogen measured under the same condition of temperature and pressure.
[OR]

Question 2.
What are isobars? Give an example.
Answer:
Atoms of different elements that have same atomic masses but different atomic numbers are called isobars.
e.g., (_{18} mathrm{Ar}^{40}), (_{20} mathrm{Ca}^{40}).

Question 3.
Write the differences between an atom and a molecule.
Answer:

Question 4.
What is artificial transmutation?
Answer:
Atoms of one element can be transmuted into atoms of other elements. In other words, the atom is no longer indestructible. It is known as artificial transmutation.

Question 5.
Classify the following based on atomicity.
Answer:
(i) Bromine
2 – Diatomic

(ii) Argon
1 – Monoatomic

(iii) Ozone
3 – Triatomic

(iv) Sulphur
8 – Polyatomic

Question 6.
Define atomic mass unit.
Answer:
Atomic mass unit is one-twelfth of the mass of carbon – 12 atom, as an isotope of carbon which contains 6 protons and 6 neutrons. It is amu.

VII. Long answer questions:

Question 1.
Explain how Avogadro hypothesis is used to derive the value of atomicity.
Answer:
(i) The Avogadro’s law states that “equal volumes of all gases under similar conditions of temperature and pressure contain the equal number of molecules”.

(ii) Let us consider the reaction between hydrogen and chlorine to form hydrogen chloride gas.
H_{2 (g)} + Cl_{2 (g)} → 2HCl _(g)
⇒ 1 volume + 1 volume → 2 volumes.

(iii) According to Avogadro’s law, 1 volume of any gas is occupied by “n” number of molecules,
“n” molecules + “n” molecules → “2n” molecules
If “n” = 1, then
1 molecule + 1 molecule → 2 molecules.
(frac { 1 }{ 2 }) molecule + (frac { 1 }{ 2 }) molecule → 1 molecule.

(iv) 1 molecule of hydrogen chloride gas is made up of (frac { 1 }{ 2 }) molecule of hydrogen and (frac { 1 }{ 2 }) molecule of chlorine.

(v) (frac { 1 }{ 2 }) molecule of hydrogen contains 1 atom.
So, 1 molecule of hydrogen contains 2 atoms.
So, hydrogen atomicity is 2. Similarly, chlorine atomicity is also 2.
So, H₂ and Cl₂are diatomic molecules.

Question 2.
Write a note on the following,
(i) Isotopeos
(ii) Isobars
(iii) Relative atomic mass.
Answer:
(i) Isotopes : Atoms of same element with different mass number. Eg: ₁₇Cl³⁵, ₁₇Cl³⁷.
(ii) Isobars : Atoms of different elements with same mass number. Eg: ₁₈Ar⁴⁰, ₂₀Ca⁴⁰
(iii) Relative Atomic Mass (RAM) :

Question 3.
Sodim bicarbonate breaks down on heating as follows:
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
(Atomic mass of Na = 23, H = 1, C = 12, O = 16)
(i) How many moles of NaHCO₃ are there in the equation?
Answer:
2 moles.

(ii) What is the mass of CO₂ produced in the equation?
Answer:
44 g

(iii) How many moles of water molecules are produced in the equation?
Answer:
1 mole.

(iv) What is the mass of NaHCO₃ used in this equation?
Answer:
= 2[23 + 1 + 12 + 3(16)]
= 2[84]
= 168 g

(v) What is the volume occupied by CO₂ in this equation?
Answer:
22.4 lit.

VIII. Hot Questions

Question 1.
Why do we take an atomic mass of Carbon – 12 as standard?
Answer:
Carbon – 12 is the standard while measuring the atomic masses. Because no other nuclides other than C – 12 have exactly whole-number masses in this scale. This is due to the different mass of neutrons and protons acting to change the total mass in nuclides with proton/neutron ratio other than 1 : 1 ratio of carbon – 12.

Question 2.
The cost of common salt (NaCl) is Rs 18 per kg. Calculate the cost of the salt per mole.
Answer:
Gram molar mass of NaCl = 23 + 35.5
= 58.5 g
1000 g of NaCl costs = Rs 18
∴ 58.5 g of NaCl costs = (frac{18}{1000}) × 58.5
= Rs 1.053
The cost of one mole of NaCl = Rs 1.053

Question 3.
What will be the mass of one (^{12} mathbf{C}) atom in g?
Answer:
1 mol of (^{12} mathbf{C}) atoms = 6.022 × 10²³ atoms = 12 g
Thus, 6.022 × 10²³ atoms of (^{12} mathbf{C}) have mass = 12 g
∴ 1 atom of (^{12} mathbf{C}) will have mass = (frac{12}{6.022 times 10^{23}}) g = 1.9927 × 10^-23 g.

IX. Solve the following problems:

Question 1.
Calculate the average atomic mass of carbon, if the natural abundance of C – 12 and C – 13 are 98.90% and 1.10% respectively.
Solution:
Average atomic mass of carbon
(=left(12 times frac{98.9}{100}right)+left(13 times frac{1.1}{100}right))
= (12 × 0.989) +(13 × 0.011)
= 11.868 + 0.143
= 12.011 amu.

Question 2.
Find how many moles are there in
(a) 98 g of H₂SO₄
Answer:
(a) 98 g of H₂SO₄
GMM of H₂SO₄ = 2(1) + 32 + 4(16)
= 98 g
Number of moles = (frac{Given Mass}{Mol. Mass})
= (frac{98}{98})
= 1 mole

(b) 18.069 × 10²³ atoms of calcium
Answer:
Number of moles
= 3 moles

Question 3.
Calculate the number of moles in
(i) 12.046 × 10²³ atoms of copper
(ii) 27.95 g of iron
(iii) 1.51 × 10²³ molecules of CO₂
Answer:
(i) 12.046 × 10²³ atoms of copper
6.023 × 10²³ atoms of copper = 1 mole
12.046 × 10²³ atoms of copper = (frac{1 times 12.046 times 10^{23}}{6.023 times 10^{23}}) = 2 moles of copper

(ii) 27.95 g of iron
55.9 g of iron = 1 mole
27.95 g of iron = (frac{1}{55.9}) × 27.95 = 0.5 mole of iron.

(iii) 1.51 × 10²³ molecules of CO₂
No of moles = (frac{text { No. of molecules }}{text { Avogadro number }})
= (frac{1.51 times 10^{23}}{6.023 times 10^{23}})
= 0.25 mole of CO₂

Question 4.
Calculate the number of atoms of mercury present in 1 kg of Mercury. [Atomic mass of Hg = 200.6]
Answer:
200.6 g of Hg contains 6.023 × 10²³ Hg atoms
∴ 1 Kg of Hg will contain = (frac{6.023×10^{23}}{200.6}) × 1000
= 30.02 × 10²³ Hg atoms

Question 5.
How many molecules are present in 7 × 10^-3 m³ of NH₃ at STP?
Answer:
Molar volume = 22.4 dm³ = 2.24 × 10^-2 m³
2.24 × 10^-2 m³ of NH₃ at STP contains 6.023 × 10²³ NH₃ molecules.
:. 7 × 10^-3 m³ of NH₃ will contain
= (frac{6.023×10^{23}}{2.24×10^{-2}}) × 7 × 10^-3
18.82 × 10²² NH₃ molecule

Question 6.
What is the mass in grams of the following?
(a) 3 moles of NaOH
(b) 6.023 × 10²² atoms of Ca
(c) 224 L of CO₂
Answer:
Formula:
(a) 3 moles of NaOH
Mass of 3 moles of NaOH = 3 × mol. mass of NaOH
GMM of NaOH = 23 + 16 + 1 = 40 g
Mass of 3 moles of NaOH = 3 × 40 = 120 g

(b) 6.023 × 10²² atoms of Ca = n × atomic mass of ca
= 4 g

Question 7.
How many grams are therein:
(i) 5 moles of water
(ii) 2 moles of Ammonia
(iii) 2 moles of Glucose
Solution:
(i) 5 moles of water.
Mass of 1 mole of water (H₂O) = 18 g (2 + 16)
Mass of 5 moles of H₂O = 18 × 5 = 90 g.

(ii) 2 moles of ammonia.
Mass of 1 mole of ammonia (NH₃) = 17 g (14 + 3)
Mass of 2 moles of ammonia = 17 × 2 = 34 g.

(iii) 2 moles of glucose.
Mass of 1 mole of glucose (C₆H₁₂O₆) = 180 g (72 + 12 + 96)
Mass of 2 moles of glucose = 180 × 2 = 360 g.

Question 8.
Calculate tbe molar mass of the following compounds.
(a) Urea (NH₂CONH₂)
(b) Ethanol(C₂H₅OH);
(c) Boric acid (H₃BO₃)
[Atomic mass of N – 14, H – 1, C – 12, B – 11, O – 16]
Answer:
(a) Urea (NH₂CONH₂) = 2(14) + 4(1) + 1(16) + 1(12)
= 28 + 4 + 16 + 12
= 60 g

(b) Ethanol(C₂H₅OH) = 2(12) + 6(1) + 1(16)
= 24 + 6 + 16 = 46 g

Question 9.
Mass of one atom of an element is 6.645 × 10^-23 g. How many moles of element are there in 0.320 kg.
Answer:
Mass of one atom of an element = 6.645 × 10^-23 g
∴ Mass of 1 mol of atom = 6.645 × 10^-23 × 6.023 × 10²³ = 40 g
Number of moles =

Related Topics: More Chemistry Lessons
In this lesson, we will learn

how to calculate the relative formula mass or relative molecular mass
how to use the relative molecular mass to calculate the percent mass of an element in a compound
how to use the relative molecular mass to calculate the percent mass of water in a compound

The following diagram shows how to calculate the relative molecular mass or relative formula mass. Scroll down the page for more examples and solutions.
What is relative formula mass and relative molecular mass?

The relative formula mass of a substance is the sum of the relative atomic masses of the elements present in a formula unit. The symbol for relative formula mass is M_r.
If the substance is made of simple molecules, this mass may also be called the relative molecular mass.

How to calculate the relative formula mass?

Example:

What is the relative mass formula of hydrogen gas? (Relative atomic mass: H = 1)

Solution:

The formula for hydrogen gas is H₂. Each molecule contains 2 hydrogen atoms.
The relative mass formula of hydrogen gas is
M_r(H₂) = 2 × A_r(H) = 2 × 1 = 2

Example:

What is the relative mass formula of water? (Relative atomic masses: H = 1, O = 16)

Solution:

The formula for water is H₂O. Each molecule contains 2 hydrogen atoms and 1 oxygen atom.
The relative mass formula of water is
M_r(H₂O) = 2 × A_r(H) + A_r(O) = 2 × 1 + 16 = 18

Example:

What is the relative mass formula of sodium chloride? (Relative atomic masses: Na = 23, Cl = 35.5)

Solution:

Sodium Chloride is an ionic solid with the formula Na⁺Cl^-.
The relative mass formula of sodium chloride is
M_r(NaCl) = A_r(Na) + A_r(Cl) = 23 + 35.5 = 58.5
How to find the Percent Mass of Elements in a compound?
How to use M_r to calculate the percent mass of an element in a compound?

Example:

What percentage of the mass of ammonium nitrate is nitrogen? (The formula for ammonium nitrate is NH₄NO₃, Relative atomic masses: H = 1, O = 16, N = 14)

Solution:

M_r(NH₄NO₃) = (2 × 14) + (4 × 1) + (3 × 16) = 28 + 4 + 48 = 80
Mass of nitrogen in the formula = 28
Mass of nitrogen as a fraction of the total =

What Is Relative Mass

Mass of nitrogen as percentage of total mass

Dihydrogen Monoxide Formula

Example:

What percentage of the mass of O in NaNO₃? (Relative atomic masses: Na = 23, O = 16, N = 14)

Solution:

M_r(NaNO₃) = 23 + 14 + (3 × 16) = 23 + 14 + 48 = 85

Mass of O in the formula = 48
Mass of O as a fraction of the total =
Mass of O as percentage of total mass
How to find percent by mass and percent composition?
Example:
Find the percent composition by mass of potassium dichromate (K₂Cr₂O₇)
Step 1: Find the molar mass of the compound.
Relative atomic mass of hydrogen peroxide

Relative atomic mass of hydrogen peroxide

Step 2: Divide the total mass of each element by the molar mass and multiply by 100 to find the % mass.

Show Step-by-step Solutions

How find the percent of an element in a compound?
Finding the percent by mass means finding the mass of the elements in the compound and adding the masses for the total mass.
Example:
You have a 7364 milligrams sample of SO₂. How many grams of sulfur are in the sample?
How to calculate the Mass Percent of an Element in a Compound?
Mass Percent Composition of an Element in a Compound
To calculate the mass percent composition (or simply, the mass percent) of an element in a compound, we divide the mass of the element in 1 mol of the compound with the mass of 1 mol of the compound and multiply by 100%.
Examples:
1. What is the mass percent of carbon in carbon dioxide?
2. What is the mass percent of oxygen in carbon dioxide?

Show Step-by-step Solutions

How to find the Percent Mass of Water in a compound?
How to use M_r to calculate the percent mass of water in a compound?

Example:

What percentage of the mass of magnesium sulfate is water? (Given that the formula for magnesium sulfate is MgSO₄•7H₂O, Relative atomic masses: H = 1, O = 16, S = 32, Mg = 24)

Solution:

Relative Atomic Mass Of Hydrogen Is 1.008

_r(MgSO₄•7H₂O) = 24 + 32 + (4 × 16) + (7 × 18) = 246
M_r(H₂O) = 2 × A_r(H) + A_r(O) = 2 × 1 + 16 = 18
Mass of water in the formula = 18 × 7 = 126
Mass of water as a fraction of the total =
Mass of hydrogen as percentage of total mass
How to calculate the Water of Crystallization?
This video outlines how to determine the percent, by mass, of water trapped in a hydrate's crystal lattice.
Example:
What is the percent of water, by mass, in the hydrate CuSO₄•2H₂O How to calculate the percentage of water in the formula of a hydrated compound?
Example:
Determine the % water in the following hydrates:
a) CuSO₄•5H

Relative Atomic Mass Of Hydrogen Gas

₂O
b) CaCl₂•2H₂O
c) KAl(SO₄)₂•12H₂O

Show Step-by-step Solutions

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.