Chemistry class 9 Chapter 1 Review questions, Federal board

 

1.       Encircle the correct answers.

                       i.            D

                     ii.            D

                    iii.            C

                    iv.            D

                     v.            B

                    vi.            B

                  vii.            A

                 viii.            B

                    ix.            E

                     x.            B

 

2.       Give short answers.

                      I.            Differentiate between ion and free radical.

Ion	Free Radical
An ion is a charged specie formed from an atom Or chemically bonded group of atom by adding Or removing electron. Ions having positive charge are called cation Whereas those having negative charge are called anion.   For Example: Na+ , Ca+2 , O-2 , CN- etc	A free radical is an atom which has an unpaired electron and bears no electrical charge. Halogens when exposed to light form free radical. They are represented by putting dot ( • ) over the symbol.       For example: H• , Cl• , C•N etc

 

                    II.            What do you know about corpuscular nature of matter?

            Ans. In 1924 De-Brogile put forward the theory of dual nature of matter i.e., it has both the properties of wave as well as particle. By evidence he proved that every moving object is attached with wave and every wave has particle nature as well. He explained that these two systems cannot remain detached from each other. This concept is known as corpuscular nature of matter.

                  III.            Differentiate between analytical chemistry and environmental chemistry.

Analytical chemistry	Environmental Chemistry
The branch of chemistry that deals with the methods And instruments for determining the composition of Matter is called Analytical chemistry.	The branch of chemistry that deals with the chemical and toxic Substances that pollute the environment and their adverse effects On human beings is called environmental chemistry.

 

                  IV.            What is mole ?

 Ans. A mole is an amount of substance that contain 6.022 x 10²³ particles of that substance. This experimentally determined value is called Avogadro’s number. It is represented by N_A.

For example:  1 mole of C = 6.022 x 10²³ atoms.

                        1 mole of H₂O = 6.022 x 10²³ molecules.

                    V.            Differentiate between empirical formula and molecular formula.

Empirical Formula	Molecular Formula
The empirical formula of compound is the chemical Formula that gives the simplest whole-number Ratio of atoms of each element.   For Example:  Empirical formula of hydrogen Peroxide(H2O2) is HO.	A molecular formula gives the actual whole number Ratio of atoms of each element present in a compound.     For Example: Molecular formula of hydrogen peroxide Is H2O2.

 

                  VI.            What is the number of molecules in 9.0 grams of steam ?

Ans. Mass in grams = 9.0g

Molar mass of steam = 18 g/mol

No. of molecules = ?

No. of molecules = No.of moles x N_A

No. of molecules = ( Mass in grams / molar mass ) x N_A

=  ( 9 / 18 ) x 6.022 x 10²³

=  0.5 x 6.022 x 10²³

= 3.011 x 10²³ molecules.

      

                VII.            What are the molar masses of uranium-238 and uranium-235 ?

          Ans. Molar mass is the mass of one mole of substance.

          Molar mass of uranium-238 is 238grams. Whereas molar mass of uranium-235 is 235g/mol.

              VIII.            Why one mole of hydrogen molecules and one mole of H-atoms have different masses ?

        Ans.  1 mole of H-atoms have mass= 1g

                1 mole of Hydrogen molecule has mass = 2g

      Since molar mass of hydrogen molecule is twice the molar mass of hydrogen atoms, therefore they have different masses.

 

3.       Define :

 

Ion:- An ion is a charged specie formed from an atom

Or chemically bonded group of atom by adding

Or removing electron.

Ions having positive charge are called cation

Whereas those having negative charge are

called anion.

 

For Example: Na⁺ , Ca⁺² , O^-2 , CN^- etc

Molecular Ion:- When a molecule loses or gains electrons the resulting species is called a molecular ion.

   For example: O₂ when loses one electron it forms O₂⁺ ion, but when it gains an electron it forms O₂^- ion.

Free Radical:- A free radical is an atom which has an unpaired electron

and bears no electrical charge.

Halogens when exposed to light form free radical.

They are represented by putting dot ( • ) over the symbol.

     For example: H^• , Cl^• , C^•N etc

 

Atomic number:- The number of protons in the nucleus of an atom is known as its atomic number. It is represented by symbol Z.

       For example: Carbon has 6 protons in its nucleus so its atomic number is 6.

 

Mass Number:- The total number of protons and neutrons in an atom is known as mass number. It is represented by symbol A. It is also known as nucleon number.

     For example: Carbon has 6 protons and 6 neutrons so its mass number is 6+6 = 12.

 

Atomic mass unit:- One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one C-12 atom.

                                       1amu = mass of one C-12 atom / 12

 

4.       Differentiate between

a.       Atom and Ion

Atom	Ion
An atom is the smallest particle of element That retain the properties of the element. Atom has equal number of protons and Neutrons. So, it has no charge.       For example: C , H , N , Cl , etc	An ion is a charged specie formed from an atom Or chemically bonded group of atom by adding Or removing electron. Ions having positive charge are called cation Whereas those having negative charge are called anion.   For Example: Na+ , Ca+2 , O-2 , CN- etc

 

b.       Molecular ion and free radical

Molecular ion	Free radical
When a molecule loses or gains electrons the resulting species is called a molecular ion.      For example: O2 when loses one electron it forms O2+ ion, but when it gains an electron it forms O2- ion.	A free radical is an atom which has an unpaired electron and bears no electrical charge. Halogens when exposed to light form free radical. They are represented by putting dot ( • ) over the symbol.        For example: H• , Cl• , C•N etc

 

 

5.       Describe how Avogadro’s number is related to a mole of substance.

 

One mole of any substance contains 6.022 x 10²³ particles of that substance. Just a dozen of eggs represents twelve eggs likewise a mole of substance represents 6.022 x 10²³ particles. This number of representative particles in one mole of substance is called Avogadro’s number.

              For example:

                                                  1 mole of C = 6.022 x 10²³ atoms of C

                                                  1 mole of H₂O = 6.022 x 10²³ molecules of H₂O     , etc.

Relationship between mole and Avogadro’s number can also be explained by this formula:

                                                  No. of moles = No of particles / 6.022 x 10^23.

 

 

 

 

 

 

 

 

6.       Calculate the number of moles of each substance in sample with the following masses:
(a) 2.4 g of He

      Mass in grams = 2.4

     Molar mass of He = 4 g/mol

     No. of moles = Mass in grams / Molar mass

                            = 2.4 / 4

                            = 0.6 moles.

(b)      250mg of carbon

Mass in grams = 250 / 1000

                          = 0.25 g

Molar mass of C = 12 g/mol

No. of moles = Mass in grams / Molar mass

                       = 0.25 / 12

                       = 0.02 moles.

(c)      15g of sodium chloride

Mass in grams = 15 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

No. of moles = Mass in grams / Molar mass

                       = 15 / 58.5

                       = 0.25 moles

(d)     40 g of sulphur

Mass in grams = 40 g

Molar mass of sulphur = 32 g/ mol

No. of moles = Mass in grams / molar mass

                       = 40 / 32

                       = 1.25 moles

(e)     1.5kg of MgO

Mass in grams = 1.5 x 1000 = 1500 g

Molar mass of MgO = 24 + 16 = 40 g/mol

No. of moles = Mass in grams / Molar mass

                        = 1500 / 40

                        = 37.5 moles

7.       Calculate the mass in grams of each of the following samples:

(a)     1.2 Moles of K

No. of moles = 1.2 moles

Molar mass of K = 39 g/mol

Mass in grams = No. of moles x Molar mass

                          = 1.2 x 39 = 46.8 grams

(b)     75 moles of H₂

No. of moles = 75

Molar mass of H₂ = 2 x 1.008 = 2.016 g/mol

Mass in grams = No. of moles x Molar mass

                          = 75 x 2.016 = 151.2 g.

(c)      0.25 moles of steam

No. of moles = 0.25 moles

Molar mass of H₂O (steam) = 2 x 1.008 + 16 = 18.016 g/mol

Mass in grams = No. of moles x Molar mass

                           = 0.25 x 18.016 = 4.5 g

(d)     1.05 moles of CuSO₄.5H₂O

No. of moles = 1.05 moles

Molar mass of CuSO₄.H₂O = 63.5 + 32 + 4 x 16 + 5 ( 2 x 1 + 16)

                                              = 63.5 + 32 + 64 + 5(18)

                                              = 159.5 + 90 = 249.5 g/mol

Mass in grams = No. of moles x Molar mass

                          = 1.05 x 249.5 = 261.96 g

(e)     0.15 moles of H₂SO₄

No. of moles = 0.15

Molar mass of H₂SO₄ = 2 x 1.008 + 32 + 4 x 16

                                      = 2.016 + 32 + 64 = 98.0 g/mol

Mass in grams = No. of moles x Molar mass

                          = 0.15 x 98 = 14.7 g

8.       Calculate the number of molecules present in each of the following samples:

(a)     2.5 moles of carbon dioxide

No. of moles = 2.5 moles

Avogadro’s Number = N_A =6.022 x 10²³

No. of molecules = No. of moles x N_A

No. of molecules = 2.5 x 6.022 x 10²³

                               = 15.055 x 10²³

                               = 1.505 x 10²⁴ molecules.

 

(b)     3.4 moles of ammonia, NH₃

No. of moles = 3.4 moles

Avogadro’s Number = N_A =6.022 x 10²³

No. of molecules = No. of moles x N_A

No. of molecules = 3.4 x 6.022 x 10²³

                               = 20.5 x 10²³

                               = 2.05 x 10²⁴ molecules

(c)      1.09 moles of benzene, C₆H₆

No. of moles = 1.09 moles

Avogadro’s Number = N_A =6.022 x 10²³

No. of molecules = No. of moles x N_A

No. of molecules = 1.09 x 6.022 x 10²³

                               = 6.56 x 10²³ molecules

(d)     0.01 moles of acetic acid, CH₃COOH

No. of moles = 0.01 moles

Avogadro’s Number = N_A =6.022 x 10²³

No. of molecules = No. of moles x N_A

No. of molecules = 0.01  x 6.022 x 10²³

                               = 6.022 x 10²¹ molecules

9.       Decide whether or not each of the following is an example of empirical formula:

(a)     Al₂Cl₆

The molecular ratio is 2:6 in this compound which is not the simplest ratio. So, Al₂Cl₆ is not the empirical formula.

(b)     Hg₂Cl₂

 

The molecular ratio is 2:2 in this compound which is not the simplest ratio. So, Hg₂Cl₂ is not the empirical formula.

 

(c)      NaCl

 

The molecular ratio is 1:1 in this compound which is the simplest ratio. So, NaCl is the empirical formula.

 

(d)     C₂H₆O

 

The molecular ratio is 2:6:1 in this compound which is the simplest ratio. So, C₂H₆O is the empirical formula.

 

10.   TNT or trinitrotoulene is an explosive compound used in bombs. It contains 7 C-atoms, 5 H-atoms, 3 N-atoms and 6 O-atoms. Write its empirical formula.

Ans. Molecular formula of TNT = C₇H₅N₃O₆

Empirical formula of TNT = C₇H₅N₃O₆

 

11.   A molecule contains four phosphorus atoms and ten oxygen atoms. Write the empirical formula of this compound. Also determine the molar mass of this compound.

Ans. Molecular formula of this compound = P₄O₁₀

         Empirical formula = P₂O₅

         Molar mass of P₄O₁₀ = 4 x 31 + 10 x 16 = 284 g

 

12.   Indigo (C₁₆H₁₀N₂O₂), the dye used to color blue jeans is derived from a compound known as indoxyl (C₈H₇ON). Calculate the molar masses of these compounds. Also write their empirical formulas.

Ans. Molar mass of indigo (C₁₆H₁₀N₂O₂) = 16 x 12 + 10 x 1 + 2 x 14 + 2 x 16 = 192 + 10 + 28 + 32 = 262 g

        Empirical formula of indigo (C₁₆H₁₀N₂O₂) = C₈H₅NO

        Molar mass of indoxyl (C₈H₇ON) = 8 x 12 + 7 x 1 + 16 + 14 = 133 g

        Empirical formula of indoxyl (C₈H₇ON) = C₈H₇ON

13.   Identify substance that has formula mass of 133.5 amu

(a)     MgCl₂

Formula mass of MgCl₂ = 24 + 2 x 35.5 = 95 amu

(b)     S₂Cl₂

Fomula mass of S₂Cl₂ = 2 x 32 + 2 x 35.5 = 64 + 71 = 135 amu

(c)      BCl₃

Formula mass of BCl₃ = 11 + 3 x 35.5 = 11 + 106.5 = 117.5 amu

(d)     AlCl₃

Formula mass of AlCl₃ = 27 + 3 x 35.5 = 27 + 106.5 = 133.5 amu

         Hence , AlCl₃ has formula mass of 133.5 amu.

14.   Calculate the number of atoms in each of the following samples:

(a)     3.4 moles of nitrogen atoms.

No. of moles = 3.4

Avogadro’s number = N_A = 6.022 x 10²³

No. of atoms = No. of moles x N_A

                        = 3.4 x 6.022 x 10²³

                        = 20.5 x 10²³ = 2.05 x 10²³ atoms

(b)     23 g of Na

Mass in grams = 23 g

Molar mass = 23 g/mol

No. of moles = Mass in grams / Molar mass

                        = 23 / 23 = 1 moles

No. of moles = 1 mole

Avogadro’s number = N_A = 6.022 x 10²³

No. of atoms = No. of moles x N_A

                        = 1 x 6.022 x 10²³

                                     = 6.022 x 10²³

(c)      5 g of H atoms

Mass in grams = 5g

Molar mass = 1

No. of moles = Mass in grams / Molar mass

                        = 5 / 1  = 5 moles

No. of moles = 5

Avogadro’s number = N_A = 6.022 x 10²³

No. of atoms = No. of moles x N_A

                        = 5 x 6.022 x 10²³

                        =30.1 x 10²³

                        = 3.01 x 10²⁴ atoms

15.   Calculate the mass of the following

(a)     3.24 x 10¹⁸ atoms of iron

No. of atoms of iron = 3.24 x 10¹⁸ atoms

Molar mass of iron = 56 g

Mass of iron = ?
Avogadro’s number = N_A = 6.022 x 10²³

Mass in grams = (No. of atoms x Molar mass) / N_A

                          = ( 3.24 x 10¹⁸ x 56 ) / 6.022 x 10²³

                          = 181.44 x 10¹⁸ / 6.022 x 10²³

                          = 30.1 x 10^18-23

                          = 30.1 x 10^-5 = 3.01 x 10^-4 g

(b)     2 x 10¹⁰ molecules of nitrogen gas

No. of molecules = 2 x 10¹⁰

Molar mass of nitrogen gas (N₂) = 2 x 14 = 28 g

Mass of N₂ = ?

Avogadro’s number = N_A = 6.022 x 10²³

Mass in grams = (No. of atoms x Molar mass) / N_A

                           = ( 2 x 10¹⁰ x 28 ) / 6.022 x 10²³

                           = 56 x 10¹⁰ / 6.022 x 10²³

                           =9.299 x 10^10-23

                                         = 9.299 x 10^-13 g

 

(c)      1 x 10²⁵ molecules of water

No. of molecules = 1 x 10²⁵
Molar mass of H₂O = 2 x 1 + 16 = 18 g

Mass in grams = ?

Avogadro’s number = N_A = 6.022 x 10²³

Mass in grams = (No. of atoms x Molar mass) / N_A

Mass in grams = (1 x 10²⁵ x 18 ) / 6.022 x 10²³

= 0.055 x 10²⁵ / 6.022 x 10²³

= 0.009 x 10^25-23

= 0.009 x 10²

= 0.9 g

(d)     3 x 10⁶ atoms of Al

 

No. of atoms = 3 x 10⁶

Molar mass of Al = 27 g

Mass in grams = ?

Avogardo’s No. = N_A = 6.022 x 10²³

Mass in grams = (No. of atoms x Molar mass) / N_A

Mass in grams = ( 3 x 10⁶ x 27 ) / 6.022 x 10²³

                          = 0.11 x 10⁶ / 6.022 x 10²³

                         = 0.018 x 10^6-23

                        = 0.018 x 10^-17 g

16.   Identify the branch of chemistry that deals with the following examples:

1)       A cornstalk grows from a seed.

Ans. Biochemistry.

2)       Dynamite (C₃H₅N₃O) explodes to form a mixture of gases.

Ans. Inorganic chemistry

3)       Purple iodine vapour appears when solid iodine is warmed.

Ans. Analytical chemistry

4)       Gasoline (a mixture of hydrocarbons) fumes are ignited in an auto mobile engine.

Ans. Organic chemistry.

5)       A silver articles tarnishes in air.

Ans. Environmental chemistry.

6)       Ice floats on water.

Ans. Physical chemistry

7)       Sulphur dioxide is the major source of acid rain.

Ans. Environmental chemistry.

8)       Many other light chlorinated hydrocarbons in drinking water are carcinogens.

Ans. Environmental chemistry.

9)       In Pakistan most of the factories use wet process for the production of cement.

Ans. Industrial chemistry.

10)   Carbon-14 is continuously produced in the atmosphere when high energy neutrons from space collide with nitrogen.

Ans. Nuclear chemistry.