Solved Exercise, Chem-11, Ch-01

BASIC CONCEPTS

Here is the complete solution to textbook exercise of chapter-01, Chemistry-11:

Q.1: Select the most suitable answer from the given ones in each question.

(i) Isotopes differ in:

(a) properties which depend upon mass

(b) arrangement of electrons in orbitals

(c) chemical properties

(d) the extent to which they may be affected in electromagnetic field.

Ans: (a) properties which depend upon mass

EXPLANATION: Isotopes have same atomic number but different atomic masses. So, they differ in those properties which depend upon masses such as density, diffusion etc.

(ii) Select the most suitable answer from the given ones in each question:

Ans: (c) Isotopes with even atomic masses and even atomic numbers are comparatively abundant.

EXPLANATION: The elements of even atomic number usually have larger number of isotopes and isotopes whose mass numbers are multiples of four are particularly abundant. For example, ^l6O, ²⁴Mg, ²⁸Si, ⁴⁰Ca & ⁵⁶Fe form nearly 50% of the earth’s crust. Out of 280 isotopes that occur in nature, 154 have even mass number and even atomic number.

(iii) Many elements have fractional atomic masses. This is because:

(a) the mass of the atom is itself fractional.

(b) atomic masses are average masses of isobars.

(c) atomic masses are average masses of isotopes.

(d) atomic masses are average masses of isotopes proportional to their relative abundance.

Ans: (d) atomic masses are average masses of isotopes proportional to their relative abundance.

EXPLANATION: The atomic mass of an element is not the mass of an individual atom of that element. Rather, it is the average of all its isotopic masses relative to their natural abundances. That’s why, its value is in fractions. Moreover, the presence of various subatomic particles in the atom also makes the relative atomic mass of an element fractional.

(iv) The mass of one mole of electrons is:

(a) 1.008 mg

(b) 0.55 mg

(c) 0.184 mg

(d)1.673 mg

Ans: (b) 0.55 mg

EXPLANATION: One mole of electrons means 6.02 ✕ 10²³ electrons. The mass of one electron is about 9.1095 ✕ 10^-31 kg. So, by multiplying the number of electrons in one mole with the mass of one electron, and then changing the unit of kilogram into milligrams, we get the answer 0.55 mg.

(v) 27 g of Al will react completely with how much mass of O₂ to produce Al₂O₃:

(a) 8 g of oxygen

(b) 16 g of oxygen

(c) 32 g of oxygen

(d) 24 g of oxygen

Ans: (d) 24 g of Oxygen

EXPLANATION: According to the equation 4Al + 3O₂ ⟶ 2Al₂O₃, 4 moles (108 g) of Al react with 3 moles (96 g) of O₂. So, 27 g of Al will react with 24 g of O₂ (96/108 ✕ 27 = 24 g).

(vi) The number of moles of CO₂ which contain 8.0 g of oxygen:

(a) 0.25

(b) 0.50

(c) 1.0

(d)1.50

Ans: (a) 0.25

EXPLANATION: One mole of CO₂ (44 g) contains one mole (32 g) of O₂. So, 8 grams of oxygen will be present in 0.25 moles of CO₂. (Formula: 1/32 ✕ 8 = 0.25 moles of CO₂.

(vii) The largest number of molecules are present in:

(a) 3.6 g of H₂O

(b) 4.8 g of C₂H₅OH

(c) 2.8 g of CO

(d) 5.4 g of N₂O₅

Ans: (a) 3.6 g of H₂O

EXPLANATION: The greater the number of moles is, the greater is the number of molecules of the compound and vice versa. First, calculate the number of moles of each compound and then multiplying with Avogadro’s number (N_A= 6.02 x 10²³), we reach at the conclusion that 3.6 grams of H₂O has greater number of molecules. (Formula: No. of molecules = Mass/Molar mass ✕ Avogadro’s Number OR N=m/M ✕ N_A)

(viii) One mole of SO₂ contains:

(a) 6.02 x 10²³ atoms of Oxygen

(b) 18.1 x 10²³ molecules of SO₂

(c) 6.02 x 10²³ atoms of Sulphur

(d) 4 gram atoms of SO₂

Ans: (c) 6.02 x 10²³ atoms of Sulphur

EXPLANATION: One mole of SO₂ contains one Avogadro’s number (6.02 x 10²³) molecules of SO₂, one Avogadro’s number atoms of S and two Avogadro’s number atoms of O₂. So, according to this information, option (c) is correct.

(ix) The volume occupied by 1.4 g of N₂ at S.T.P is:

(a) 2.24 dm³

(b) 22.4 dm³

(c) 1.12 dm³

(d) 112 cm³

Ans: 1.12 dm³

EXPLANATION: One mole of N₂ (28 g) at STP contains volume 22.414 dm³. So, 1.4 g of N₂ will occupy volume 1.12 dm³. (Formula: 22.414 dm³/28 g ✕ 1.4 g = 1.12 dm³)

(x) A limiting reactant is the one which:

(a) is taken in lesser quantity in grams as compared to other reactants.

(b) is taken in lesser quantity in volume as compared to the other reactants.

(c) gives the maximum amount of the product which is required.

(d) gives the minimum amount of the product under consideration.

Ans: (d) gives the minimum amount of the product under consideration.

EXPLANATION: Limiting reactant is present in less amount as required according to balanced chemical equation. So, it is consumed earlier in the reaction, causes to stop the reaction, and ultimately gives minimum amount of the product under consideration.

Q.02: Fill in the blanks:

(i) The unit of relative atomic mass is_________________.

(amu)

(ii) The exact masses of isotopes can be determined by________________ spectrograph.

(Aston’s mass)

(iii) The phenomenon of isotopy was first discovered by ____________ .

(Soddy)

(iv) Empirical formula can be determined by combustion analysis for those compounds which have ________ and _________ in them.

(Carbon & Hydrogen)

(v) A limiting reagent is that which controls the quantities of __________ .

(Products)

(vi) 1 mole of glucose has ____________ atoms of carbon, ______________ of oxygen and ______________ of H.

(3.612 ✕ 10²⁴, 3.612 ✕ 10²⁴, 7.224 ✕ 10²⁴)

(vii) 4 g of CH₄ at 0°C and 1 atm pressure has _____________ molecules of CH₄.

(1.505 ✕ 10²³)

(viii) Stoichiometric calculations can be performed only when __________ is obeyed.

(Law of conservation of mass & Law of definite proportions)

Q.03: Indicate true or false as the case may be:

(i) Neon has three isotopes and the fourth one with atomic mass 20.18 amu.

Ans: (False)

Neon has three isotopes, there is no fourth one with atomic number 20.18 amu. Actually, 20.18 amu is the relative atomic mass of Neon which is the average mass of all its three isotopes, relative to their natural abundance.

(ii) Empirical formula gives the information about the total number of atoms present in the molecule.

Ans: (False)

Empirical formula gives the simple whole number ratio between the atoms of elements in a compound.

(iii) During combustion analysis Mg(ClO₄)₂ is employed to absorb water vapours.

Ans: (True)

(iv) Molecular formula is the integral multiple of empirical formula and integral multiple can never be unity.

Ans: (False)

Molecular formula is the integral multiple of empirical formula and integral multiple can be unity for several molecules which have same empirical and molecular formulas. However, it never be zero.

(v) The number of atoms in 1.79 g of gold and 0.023 g of sodium are equal.

Ans: (True)

(vi) The number of electrons in the molecules of CO and N₂ are 14 each, so 1 g of each gas will have same number of electrons.

Ans: (True)

(vii) Avogadro’s hypothesis is applicable to all types of gases i.e., ideal and non-ideal.

Ans: (False)

Avogadro’s hypothesis is applicable only to ideal gases because non-ideal gases don’t obey gas laws.

(viii) Actual yield of a chemical reaction may be greater than the theoretical yield.

Ans: (False)

Actual yield can never be greater than theoretical. Rather, it is mostly smaller than theoretical yield because some of the product may be lost during the preparation or separation processes.

Q.4: What are ions? Under what conditions are they produced?

Ans:

IONS: “Those species which carry either positive or negative charge are called ions.”

Examples: K⁺, Mg²⁺, Al³⁺, Cl^–, O^2-, CO₃^2- etc.

Conditions for the Formation of Ions:

Positive ions or cations are formed when an atom loses one or more electrons. This requires energy, which is used to release the electron from the attraction of the nucleus and to bring it out of the atom. e.g.,       

M ➜ M⁺ + e^–     ΔH = Positive

Negative ions or anions are formed when an atom gains one or more electrons. During this process, energy is usually released due to the attraction of the nucleus to the incoming electron. e.g.,                                 

B + e^–  ➜  B^–      ΔH = Negative

Q.05: (a) What are isotopes? How do you deduce the fractional atomic masses of elements from the relative isotopic abundance? Give two examples in support of your answer.

Ans:

ISOTOPES: “The atoms of the same element having same atomic number but different atomic masses are called isotopes.”

Isotopes have same position in the periodic table. They have same chemical but different physical properties.

Fractional Atomic Masses from Relative Isotopic Abundance:

Fractional or average atomic mass of an element can be calculated by multiplying the relative abundance with isotopic mass, and then dividing their sum by 100. For example:

(1) Neon has three isotopes, Ne-20, ne-21 & Ne-22 present in the percentage of 90.92%, 0.26% & 8.825 respectively. Its average atomic mass can be calculated as:

(2) Similarly, the average atomic mass of silver can be calculated from its isotopic abundances as:

(b) How does a mass spectrograph show the relative abundance of isotopes of an element? What is the justification of two strong peaks in the mass spectrum for Bromine while for Iodine only one peak at 127 amu is indicated?

(c) What is the justification of two strong peaks in the mass spectrum for Bromine while for Iodine only one peak at 127 amu is indicated?

Ans: In mass spectrum, the number of peaks represents the number of isotopes, while the heights of the peaks represent their natural abundance. Two equally strong peaks in the mass spectrum of bromine indicate that bromine has two isotopes which have almost equal abundance. While, only one peak for iodine at 127 amu indicates that iodine has only one isotope with atomic mass 127 amu.

Q.6: Silver has atomic number 47 and has 16 known isotopes but two occur naturally i.e., Ag-107 & Ag-109. Given the following mass spectrometric data, calculate the average atomic mass of silver.

Isotopes	Mass(amu)	% Abundance
107Ag	106.90509	51.84
109Ag	108.90476	48.16

Solution:

Q.07: Boron with atomic number 5 has two naturally occurring isotopes. Calculate the percentage abundance of ¹⁰B & ¹¹B from the following information: Average at. mass of boron = 10.81 amu / Isotopic mass of ¹⁰B= 10.0129 amu / Isotopic mass of ¹¹B= 11.0093 amu.

Solution:

Q.08: Define the following terms and give three examples of each: (i) Gram atom (ii) Gram molecular mass (iii) Gram formula (iv) Gram ion (v) Molar volume (vi) Avogadro’s number (vii) Stoichiometry (viii) Percentage yield.

Ans:

GRAM ATOM: “The atomic mass of an element expressed in grams is called one gram atom.” It is also called one gram mole or simply one mole of that element.

Examples:

1 gram atom of Hydrogen=1.008 g                  

1 gram atom of Carbon=12.000 g

GRAM MOLECULAR MASS: “The molecular mass of a substance expressed in grams is called one gram molecule or one gram mole or simply one mole of that substance.”

Examples:

1 gram molecule of water = 18 g   

1 gram molecule of H₂SO₄ = 98 g

GRAM FORMULA: “The formula unit mass of an ionic compound expressed in grams is called one gram formula of the substance.”

Examples:

1 gram formula of NaCl = 58.5 g  

1 gram formula of Na₂CO₃ = 106 g

GRAM ION: “The ionic mass of an ionic species expressed in grams is called one gram ion or one mole of ions.”

Examples:

1 gram ion of OH^– =17 g       

1 gram ion of SO₄^2- = 96 g

MOLAR VOLUME: “One mole of any gas at standard temperature and pressure contains volume 22.414 dm³. This volume of 22.414 dm³ is called molar volume.” It is true when the gas is ideal.

Examples:

• 2.016 g of H₂ = 1 mole of H₂ = 6.02×10²³ molecules of H₂ = 22.414 dm³ of H₂ at S.T.P.
• 16 g of CH₄ =1 mole of CH₄ = 6.02×10²³ molecules of CH₄ =22.414 dm³ of CH₄ at S.T.P

AVOGADRO’S NUMBER “The number of atoms, molecules, formula units and ions in one gram atom of an element, one gram molecule of a compound, one gram formula of an ionic substance and one gram ion of an ionic specie respectively is equal to 6.02×10²³. This number is called Avogadro’s number and is represented by N_A.”

Examples:

• 1.008 g of H=1 mole of hydrogen=6.02×10²³ atoms of H
• 18 g of H₂O =1 mole of water = 6.02×10²³ molecule of water
• 58.5 g of NaCl=1 mole of NaCl=6.02×10²³ formula units of NaCl
• 96 g of = 1 mole of = 6.02×10²³  ions

STOICHIOMETRY: “Stoichiometry is the branch of chemistry which tells us the quantitative relationship between reactants and products in a balanced chemical equation.”

Examples: Following types of relationship can be studied with the help of balanced chemical equation:

Mass–Mass Relationship: If we are given the mass of one substance, we can calculate the mass of the other substances involved in the chemical reaction.

Mass–Mole or Mole–Mass Relationship: If we are given the mass of one substance, we can calculate the moles of the other substances and vice versa.

Mass–Volume Relationship: If we are given the mass of one substance, we can calculate the volume of the other substances and vice versa.

Percentage Yield: The efficiency of a chemical reaction is calculated by comparing the actual and theoretical yields, in the form of percentage yield.                            

The greater is the percentage yield, the greater is the efficiency, and vice versa.

The equation shows that whatever the mass of K₂CrO₄ is, its one formula unit ionizes in water to give two K⁺ ions and one  ion. Thus, total number of ions will always be thrice of the number of formula units.

Q.10(a): Calculate the mass in grams of 2.74 moles of KMnO₄.

Solution:

(b) Calculate the moles of O atoms in 9.00 g of Mg(NO₃)₂.

Solution:

(c) Calculate the number of O atoms in 10.037 g of CuSO₄.5H₂O.

Solution:

(d) Calculate the mass in kilograms of 2.6 × 10²⁰ molecules of SO₂.

Solution:

(e) Calculate the moles of Cl atoms in 0.822 g of C₂H₄Cl₂.

Solution:

(f) Calculate the mass in grams of 5.136 moles of Ag₂CO₃.

Solution:

(g) Calculate the mass in grams of 2.78×10²¹ molecules of CrO₂Cl₂.

Solution:

(h) Calculate the number of moles & formula units in 100 g of KClO₃.

Solution:

(i) Calculate the number K⁺ ions,  ions, Cl atoms & O atoms in (h).

Solution:

Q.11: Aspartame, the artificial sweetener, has a molecular formula of C₁₄H₁₈N₂O₅.

a) What is the mass of one mole of aspartame?

b) How many moles are present in 52 g of aspartame?

c) What is the mass in grams of 10.122 moles of aspartame?

d) How many hydrogen atoms are present in 2.43 g of aspartame?

Solution:

Solution:

Q.13: In each pair, choose the larger of the indicated quantity, or state if the samples are equal.

a) Individual particles: 0.4 moles of oxygen molecules or 0.4 mole of oxygen atoms.

Solution:

b) Mass: 0.4 mole of ozone molecules or 0.4 mole of oxygen atoms.

Solution:

c) Mass: 0.6 mole of C₂H₄ or 0.6 mole of I₂.

Solution:

d) Individual particles: 4.0 g N₂O₄ or 3.3 g SO₂.

Solution:

e) Total ions: 2.3 moles of NaClO₃ or 2.0 moles of MgCl₂.

Solution:

f) Molecules: 11.0 g H₂O or 11.0 g H₂O₂.

Solution:

g) Na⁺ ion: 0.500 moles of NaBr or 0.0145 kg of NaCl.

Solution:

h) Mass: 6.02×10²³ atoms of ²³⁵U or 6.02×10²³ atoms of ²³⁵U.

Solution:

Q.14: a) Calculate the percentage of nitrogen in the four important fertilizers i.e. (i) NH₃ (ii) NH₂CONH₂ (iii) (NH₄)₂SO₄ (iv) NH₄NO₃

Solution:

b) Calculate the percentage of nitrogen and phosphorus in each of the following:

(i) NH₄H₂PO₄ (ii) (NH₄)₂HPO₄ (iii) (NH₄)₃PO₄

Solution:

Q.15: Glucose C₆H₁₂O₆ is the most important nutrient in the cell for generating chemical potential energy. Calculate the mass % of each element in glucose and determine the number of C, H and O atoms in 10.5 g of the sample.

Solution:

Q.16: Ethylene glycol is used as automobile antifreeze. It has 38.7% carbon, 9.7% hydrogen and 51.6% oxygen. Its molar mass is 62.1 g mol^-1. Determine its empirical formula?

Solution:

Q.17: Serotonin (Molar mass=176 g mol^-1) is a compound that conducts nerve impulses in brain and muscles. It contains 68.2% C, 6.86% H, 15.09% N, and 9.08% O. What is its molecular formula?

Solution:

Q.18: An unknown metal M reacts with S to form a compound with a formula M₂S₃. If 3.12 g of M reacts with exactly 2.88 g of Sulphur, what are the names of metal M and the compounds M₂S₃?

Solution:

Q.19: The octane present in gasoline burns according to the following equation:

2C₈H₁₈(ℓ) + 25O₂(g) ⟶ 16CO₂(g) + 18H₂O(ℓ)

a) How many moles of O₂ are needed to react fully with 4 moles of octane?

b) How many moles of CO₂ can be produced from one mole of octane?

c) How many moles of water are produced by the combustion of 6 moles of octane?

d) If this reaction is to be used to synthesize 8 moles of CO₂, how many grams of oxygen are needed? How many grams of octane will be used?

Solution:

Q.20: Calculate the number of grams of Al₂S₃ which can be prepared by the reaction of 20 g of Al and 30 g of Sulphur. How much the non-limiting reactant is in excess?

Solution:

Q.21: A mixture of two liquids, hydrazine N₂H₄ and N₂O are used in rockets. They produce N₂ and water vapours. How many grams of N₂ gas will be formed by reacting 100 g of N₂H₄ and 200 g of N₂O₄?         

 2N₂H₄ + N₂O₄ ⟶ 3N₂ + 4H₂O

Solution:

Q.22: Silicon carbide (SiC) is an important ceramic material. It is produced by allowing sand (SiO₂) to react with carbon at high temperature.

                 SiO₂ + 3C ⟶ SiC + 2CO

When 100 kg sand is reacted with excess of carbon, 51.4 kg of SiC is produced. What is the percentage yield of SiC?

Solution:

Q.23: a) What is stoichiometry? Give its assumptions? Mention two important laws, winch help to perform the stoichiometric calculations? b) What is a limiting reactant? How does it control the quantity of the product formed? Explain with three examples?

ANS:

a) STOICHIOMETRY: “Stoichiometry is the branch of chemistry which tells us the quantitative relationship between reactants and products in a balanced chemical equation.”

Basic Assumptions of Stoichiometry:

1. All the reactants are converted into products.
2. No side reaction occurs.

Laws of Stoichiometry: While doing stoichiometric calculations, the following two laws must be obeyed:

1. Law of conservation of mass.
2. Law of definite proportions.

b) LIMITING REACTANT: “The reactant which is consumed earlier in the chemical reaction and controls the amount of the product due to its smaller amount is called limiting reactant.”

Limiting reactant controls the amount of the product because the reaction proceeds as long as there is limiting reactant. When the limiting reactant is consumed, the reaction stops and no more product is formed. So, the amount of the product formed is determined by the limiting reactant.

Examples:

1. If we have 30 “kababs” and 5 breads “having 58 slices”, then we can only prepare 29 “sandwiches”. One “kabab” will be extra (excess reactant) and “slices” will be the limiting reactant.
2. Consider the reaction: 

2H₂(g) + O₂(g) ⟶ 2H₂O(ℓ)

If 2 moles of H₂ are allowed to react with 2 moles of O₂, only 2 moles of water will be formed. This is because 2 moles of H₂ require only 1 mole of O₂ to form 2 moles of water. 1 mole of O₂ will be left unreacted. So, H₂ is the limiting reactant because it is consumed earlier and controls the amount of the product. In the burning of wood in free supply of oxygen, wood is the limiting reactant because burning continues until there is wood. When the wood is consumed, the reaction stops.

Q.24: a) Define yield. How do we calculate the percentage yield of a chemical reaction? b) What are the factors which are mostly responsible for the low yield of the products in chemical reactions?

ANS: a) Yield: “The amount of the products obtained in a chemical reaction is called yield of that reaction.”

The efficiency of a chemical reaction is calculated by comparing the actual and theoretical yields, in the form of percentage yield.                           

The greater is the percentage yield, the greater is the efficiency, and vice versa.

b) In most reactions, actual yield remains less than the theoretical yield because of the following factors:

1. A practically inexperienced worker has many shortcomings and cannot get the expected yield.
2. The processes like filtration, separation by distillation, separation by a separating funnel, washing, drying and crystallization if not properly carried out, decrease the amount of the product.
3. Some competing side reaction may waste some of the reactants

Q.25: Explain the following with reasons.

i) Law of conservation of mass has to be obeyed during stoichiometric calculations.

ANS: Law of conservation of mass has to be obeyed during stoichiometric calculations. This is because according to this law, reactants undergo a chemical change to give an equal amount of products. It means there is no creation or destruction of mass during the reaction. Thus, if we know the masses of the reactants, we can easily calculate the masses of the products and vice versa. Otherwise, if this law is not obeyed, the whole stoichiometric calculations will be wrong.

ii) Many chemical reactions taking place in our surrounding involve the limiting reactants.

ANS: Many chemical reactions taking place in our surrounding involve the limiting reactants. For example, burning of wood, rusting of iron, etc. In these reactions, one of the reactants is in excess. The other reactant which is present in less amount is the liming reactant. In burning of wood, air is excess reactant, while wood is the limiting reactant. In rusting of iron, air and moisture are excess reactants, while iron is the limiting reactant.

iii) No individual neon atom in the sample of the element has a mass of 20.18 amu.

ANS: Neon has three isotopes Ne–20, Ne–21 and Ne–22, present in the percentage of 90.92%, 0.26% and 8.82% respectively. When we calculate the average of all these isotopic masses, it comes out to be 20.18 amu.

So, 20.18 amu is the average atomic mass of Neon, not the mass of any individual Neon atom.

iv) One mole of H₂SO₄ should completely react with two moles of NaOH. How does Avogadro’s number help to explain it.

ANS: One mole of H₂SO₄ should completely react with two moles of NaOH. This is because one mole of H₂SO₄ produces two moles of H⁺ ions.   

H₂SO₄ ⇌  2H⁺ + SO₄^2-       

While one mole of NaOH produces only mole of OH^– ion.

NaOH ⇌  Na⁺ + OH^–

Since, two moles H⁺ ions require two moles OH^– ions for complete neutralization, so one mole of H₂SO₄ requires two moles of NaOH to react completely.

H₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O

v) One mole of H₂O has two moles of bonds, three moles of atoms, ten moles of electrons and twenty-eight moles of the total fundamental particles present in it.

ANS: One mole of H₂O has two moles of bonds, three moles of atoms, ten moles of electrons and twenty-eight moles of the total fundamental particles present in it. This can be explained as:

Since, one molecule of H₂O contains two O—H bonds, so one mole of H₂O has two moles of covalent bonds i.e., 2 ✕ 6.02 ✕ 10²³ bonds.

Since, one molecule of H₂O contains three (2H+1O) atoms, so one mole of H₂O has three moles of atoms i.e., 3 ✕ 6.02 ✕ 10²³ atoms.

Since, one molecule of H₂O contains 2+8=10 electrons, so one mole of H₂O has ten moles of electrons i.e., 3 ✕ 6.02 ✕ 10²³ electrons.

Since, one molecule of H₂O contains 10 electrons (2+8=10), 10 protons (2+8=10) and 8 neutrons (H atoms have no neutrons), so one mole of H₂O has twenty-eight moles of all three fundamental particles. i.e., 28 ✕ 6.02 ✕ 10²³ fundamental particles.

vi) N₂ and CO have the same number of electrons, protons and neutrons.

ANS: N₂ and CO have the same number of electrons, protons and neutrons. This can be verified by adding the number of electrons, protons and neutron in N, C & O atoms in these molecules.