Lakhmir Singh Manjit Kaur Class 9 Chemistry 3rd Chapter “Atoms And Molecules” solution

Lakhmir Singh Manjit Kaur Class 9 Chemistry 3rd Chapter “Atoms And Molecules” solution

Lakhmir Singh and Manjit Kaur Chemistry solution: “Atoms And Molecules” Chapter 3. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Chemistry solution Chapter 3 . Here we have given Chapter 3 all solution of Class 9. Its help you to complete your homework.

  • Board – CBSE
  • Text Book – Chemistry
  • Class – 9
  • Chapter – 03

Lakhmir Singh Manjit Kaur Class 9 Chemistry 3rd Chapter Solution (First Part)

VERY SHORT ANSWER TYPE QUESTIONS

1] Write the full form of IUPAC.

Ans 1- International union of pure and Applied chemistry.

2] Name the scientist who gave :

a) law of conservation of mass.

Ans -2 a) Antoine Lavoisier

b) law of constant proportions.

b) Joseph Proust.

3] Name the law of chemical combination:

a) Which was given by Lavoisier.

Ans 3- a) Law of conservation of mass.

b) which was given by Proust.

Ans. Law of constant proportion.

4] Name the scientist who gave atomic theory of matter.

Ans 4- John Dalton.

5] Which postulate of Dalton’s atomic theory is the result of law of conservation of mass given by Lavoisier?

Ans 5- The postulates of Dalton atomic theory that explain the law of conservation of mass is “atoms can neither be created, nor destroyed”.

6] Which part of Dalton’s atomic theory came from the law of constant proportions given by Proust?

Ans 6- The postulates of Dalton atomic theory that explain the law of constant proportion is “every element is made up of small particles called atoms which have fixed mass and that the number and kind of atom of a given compound is fixed”.

7] Which ancient Indian philosopher suggested that all matter is composed of very small particles? What name was given by him to these particles?

Ans 7- The Indian philosopher was Maharishi Kanad and he called them “parmanu”.

8] Name any two laws of chemical combination.

Ans 8- •Law of conservation of mass.

  • Law of constant proportion.

9] “If 100grams of pure water taken from different sources decompose by passing electricity , 11 grams of hydrogen and 89 grams of oxygen are always obtained”. Which chemical law is illustrated by this statement?

Ans 9- If we repeat this experiment again by taking water from different water bodies, we get the same mass of hydrogen and oxygen that is 11:89. Hence, it obey Law of  constant proportion.

10] “If 100gms of calcium carbonate (whether in the form of marble of chalk) are decomposed completely, then 56 grams of calcium oxide and 44grams of carbon dioxide are obtained”. Which law of chemical combination is illustrated by this statement?

Ans 10- CaCO3 (100g) + heat gives Ca(56g) and CO2(44g).

In this Calcium carbonate has 100g, after decomposition the sum of masses of two product is also 100g. Hence, it obey law of conservation of mass.

11] What are the building blocks of matter?

Ans 11- Atoms are the building blocks of matter.

12] How is the size of an atom indicated?

Ans 12- The size of an atom indicated as very very small.

13] Name the unit in which the radius of an atom is usually expressed.

Ans 13- nanometre/nm.

14] Write the relation between nanometre and metre.

Ans 14- 1nm=1/10^9

= 1 nm = 10^-9m.

15] The radius of an oxygen atom is 0.073nm. What does the symbol ‘nm’ represent?

Ans 15- ‘nm’ represent the unit of measurement that is nanometre.

16] Why is it not possible to see an atom even with the most powerful microscope?

Ans 16- It is because of small size of atoms which make them undetectable under high power microscopes also.

17] State whether the following statement is true or false:

The symbol of element cobalt is CO.

Ans 17- False, because the first letter is in capital letter and second letter is written in small letter. The correct symbol is ‘Co’.

18] Define ‘molecular mass’ of a substance .

Ans 18- The molecular mass is the relative mass of its molecule as compared with the mass of carbon-12 isotope.

19] What is meant by saying that ‘the molecular mass of oxygen is 32’?

Ans 19- It means that Oxygen molecule is 32 times heavier than 1/12 of a carbon-12 isotope.

20] Fill in the blanks

Ans 20- a) In water, the proportion of oxygen and hydrogen is 8:1 by mass.

As 0 : H = 16 : 2

=8:1

b) In a chemical reaction, the sum of the masses of the reactants and the products remains unchanged. This is called law of conservation of mass.

SHORT ANSWER TYPE QUESTIONS

21] a] Name the element used as a standard for atomic mass scale.

Ans21- Ans. Carbon element.

b) Which particular atom of the above element is used for this purpose?

Ans. Carbon-12 atom.

c) What value has been given to the mass of this reference atom?

Ans. 12 u.

22] Give one major drawback of Dalton’s atomic theory of matter?

Ans 22- One of the major drawbacks of Dalton atomic theory is that atoms are indivisible particles.

23] Dalton’s atomic theory says that atoms are indivisible . Is that statement still valid? Give reasons for your answer.

Ans 23- No, as atoms are further divided into electron, proton and neutron.

24] Is it possible to see atoms these days? Explain your answer.

Ans 24- Yes, we can see atoms now-a-days by using of scanning tunnelling microscope.

25] What is meant by the symbol of an element? Explain with examples.

Ans 25- Symbol of element is the short way to represent the element.

For example:- Dalton symbol of element – Carbon● , Oxygen○

Modern symbol of elements- Carbon ‘C’,  Oxygen ‘O’.

26] a) Give two symbols which have been derived from the “English names” of the elements.

Ans 26- a) Lithium-Li, Chlorine-Cl.

b) Give two symbols which have been derived from from the “Latin names” of the elements.

b) Sodium-Na, Potassium-K.

27] Give the masses and symbols of five familiar substances which you think are elements.

Ans 27- Iron-Fe , Gold-Au, Silver-Ag, Hydrogen-H and Copper-Cu.

28] State the chemical symbols for the following elements:

Ans 28- Sodium-Na

Potassium-K

Iron-Fe

Copper-Cu

Mercury-Hg

Silver-Ag

29] Name the elements represented by the following symbols

Ans 29–  Hg– Mercury

Pb– lead

Au- Gold

Ag– Silver

Sn– tin

30] What is meant by atomicity ? Explain with two examples.

Ans 30-The number of atoms present in one molecule of an element is called its atomicity.

Eg: Hydrogen (H2)- Diatomic or atomicity is 2.

Neon (Ne)- monoatomic or atomicity is 1.

31] What is the atomicity of the following?

Ans 31 – a) Oxygen

= 2   (exist as O2)

b) Ozone

= 3 (exist as O3)

c) Neon

= 1 (exist as Ne)

d) Sulphur

=8           (exist as S8)

e) Phosphorus

= 4 (exist as P4)

f) Sodium

= 1 (exist as Na)

32] What is meant by chemical formula?

Ans 32- A chemical formula represent the composition of a molecule of the substance in term of the symbol of element present in the molecule.

Eg: Chemical formula of an element- Hydrogen is H

Chemical formula of molecule- water is H2O.

33] Write the formulae of the following compounds. Also name the elements present in them.

Ans33- a) H2O

Element :- Hydrogen and Oxygen

b) NH3

Element:- Nitrogen and Hydrogen

c) CH4

Element:- Carbon and Hydrogen

d) SO2

Element:-Sulphur and Oxygen

 e) C2H5OH

Element:- Carbon, Oxygen and Hydrogen.

34] Explain the difference between 2N and N2.

Ans34- 2N represents two single nitrogen atoms whereas N2 represents two  nitrogen atoms bonded together to form a nitrogen molecule.

35] What do the following abbreviations stand for?

Ans 35- i) O- represent oxygen atom or nascent oxygen

ii) 2O- 2 seperate atoms of oxygen

iii) O2– It denotes a molecule of oxygen

iv) 3O2– represent three molecules of oxygen gas.

36] What do the symbols, H2 , S and O4 mean the in formula H2SO4?

Ans 36- H2- a molecule of hydrogen or hydrogen gas

S- Sulphur atom

O4 denotes four oxygen atom in its molecule.

37] a) In what form does oxygen gas occur in nature?

Ans 37- a) O2– Diatomic in nature.

b) In what form do noble gases occur in nature?

b) Noble gas- monoatomic in nature.

38] What is the difference between 2H and H2.

Ans. 2H represents two separate hydrogen atoms whereas H2 represents two H-atoms bonded together to form hydrogen molecule.

39]What do the following denote?

Ans39- i) N- single nitrogen atom

ii) 2N- two separate nitrogen atom

iii) N2– a molecule of nitrogen or nitrogen gas

iv) 2N2– two molecules of nitrogen gas.

40] What is the significance of the formula of a substance?

Ans 40- Significance of formula of the substance:

i) represent the name of all the element present in the molecule.

ii) represent the name of the substance

iii) gives number of atom of each element present in the molecule.

iv) represent one molecule of the substance.

v) represent the definite mass of the substances.

41] What is the significance of the formula H2O?

Ans 41- Significance of  the formula H2O:

i) H2O represent water.

ii) tell us that water contain two elements- hydrogen and oxygen

iii) one molecule of water contains 2 hydrogen atom and 1 oxygen atom

iv) H20 represents 18 g of water.

42] The molecular formula of glucose is C6H12O6. Calculate its molecular mass. (Atomic masses: C = 12u ; H = 1u ; O = 16u)

Ans 42- Molecular mass of C6H12O6 =

=6×C+12×H+6×O

=6×12+12×1+6×16

=72+12+96

=180u.

43] Calculate the molecular masses of the following :

Ans43-i) H2– H+H

=1+1 = 2u

ii) O2– 2×O =2×16

=32u

iii) Cl2 2×Cl =2×35.5

=71u

iv) NH3= 1×N+3×H

=14+3×1  =17u

44] Calculate the molecular masses of the following compounds

Ans 44- a) CH4– 1×C+4×H= 1×12+4×1

=16u

b) C2H6= 2×C+6×H

=2×12+6×1 =30u

c) C2H4= 2×C+4×H

=2×12+4×1 = 28u

d) C2H2= 2×C+2×H

=2×12+2×1 = 26u

45] Calculate the molecular masses of the following compounds

Ans 45– i) Methanol CH3OH=

=1×C+4×H+1×O

=1×12+4×1+1×16

=32u

ii) Ethanol C2H5OH-

=2×C+6×H+1×O

=2×12+6×1+1×16

=46u

46] Calculate the molecular mass of ethanoic acid, CH3COOH.

(Atomic masses ; C = 12 ; H=1 ; O = 16

Ans 46- Ethanoic acid CH3COOH

= 2×C+4×H+2×O

=2×12+4×1+2×16

=24+4+32

=60u.

47] Calculate the molecular mass of nitric acid, HNO3.( Atomic masses: H=1u ; N=14u ; O=16u).

Ans 47- Nitric acid HNO3= 1×H+1×N+3×O

=1×1+1×14+3×16

=1+14+48

=63 u.

48.) Calculate the molecular mass of chloroform (CHCl3). ( Atomic masses: C= 12u ; H=1u ; Cl = 35.5).

Ans 48- Chloroform CHCl3 =1×C+1×H+3×Cl

=1×12+1×1+3×35.5

=12+1+106.5

=119.5u.

49] Calculate the molecular mass of hydrogen bromide (HBr). (Atomic masses: H=1u ; Br=80u).

Ans 49- HBr=1×H+1×Br

=1×1+1×80

=81u.

50] Calculate the molecular masses of the following compounds:

Ans50– i) H2S =2×H+1×S=2+32=34u

ii) CS2-1×C+2×S

=1×12+2×32

=76 u.

LONG ANSWER TYPE QUESTIONS

51] State the law of conservation of mass. Give one example to illustrate this law.

Ans 51- “Mass can neither be created nor destroyed” .

Eg: If 1 gm of A reacts with 8 gm of B, then the product form will have 9g of C.

52] State the law of constant proportions. Give one example to illustrate this law.      

Ans 52- In a chemical substance the element are always present in a fixed proportion by their mass.

Eg: If 5 gm of X reacts with 25 gm of Y, then by the law of definite proportion 10 g of X must react with  50 g of Y.

53] a) State the various postulates of Dalton’s atomic theory of matter.

Ans 53- a) i) All the matter is made up of very small particles called matter.

ii) Atoms cannot be divided.

iii) Atoms can neither be created nor destroyed.

iv) All the atoms of a given element are identical in every aspect such as size, mass and chemical properties.

v) During chemical combination, atom of different element combine in small whole numbers to form compound.

vi) Atoms of the same elements can combine in more than one ratio to form more than one compound.

b) Which postulate of Dalton’s atomic theory can explain the law of conservation of mass?

b) The postulates is “mass can neither be created nor destroyed”.

c) Which postulate of Dalton’s atomic theory can explain the law of constant proportions?

c) The postulates is “Every element is made up of small particles called atoms which have fixed mass, number and kind of atom of a given compound.

54] a) What is the significance of the symbol of an element? Explain with the help of an example.

Ans 54- a) i)symbol represent name of the element.

ii) It represent one atom of an element.

iii) It represent one mole of atom of the element i.e 6.022×10^23 atoms •Let us take an example of symbol N:

i) Symbol represent nitrogen element.

ii) Symbol N represent one atom of nitrogen element.

iii) Symbol N represent 14 gms of nitrogen.

b) Explain the significance of the symbol H?

b) i) Symbol H represent hydrogen

ii) Symbol H represent one atom of hydrogen element.

iii) Symbol H represent 1gm of hydrogen.

55] a) What is an atom? How do atoms actually exist?

Ans 55- a) Atoms is a smallest indivisible particles which cannot be seen through naked eyes.

Atoms exist in two forms:

i) in the form of molecules

ii) in the form of ions.

b) What is a molecule? Explain with an example.

b) A molecule is electrically neutral group of two or more similar atoms chemically bonded together.

Eg: H2, N2, NH3,H20 etc

c) What is the difference between the molecule of an element and the molecule of a compound? Give one example each.

 c)• Molecules of an element-

i) Contains two or more same atoms chemically combine together.

ii) Homogenous in nature.

iii) Eg:N2, H2,O3,Cl2.

  • Molecules of compound-

i) Contains two or more different atoms chemically combine together.

ii) Heterogenous in nature.

iii) Eg: HCl, CO2, NH3, SO2 etc.

56] a) Define atomic mass unit. What is its symbol?

Ans56-a) Atomic mass unit is equals to the 1/12th of the mass of an carbon-12 atom.

It is denoted by amu.       

b) Define atomic mass of an element.

b)The atomic mass of an element is the relative mass of its atom as compared with the mass of a carbon-12 atom.

c) What is meant by saying that ‘the atomic mass of oxygen is 16’?

c) It means that one atom of oxygen is 16 times heavier than 1/12 of a carbon-12 atom.

Lakhmir Singh Manjit Kaur Class 9 Chemistry 3rd Chapter Solution (Second Part)

VERY SHORT ANSWER TYPE QUESTIONS

1] Solution:

Ans. Particles which have more or less electron than the normal atom are called ions. Particles with less electron are cation and particles with more electron are called anions.

2] Solution:

a) more electrons than the normal atoms?

Ans. Particles with more electrons are called as anions.

b) less electrons than the normal ones?

Ans. Particles with less electrons are called as cations.

3.] Solution:

Ans. The sum of atomic masses of all the atoms present in the compound is called as formula mass.

Example: for CaSO4=  Ca(1) + S(1)+ O(4)

=   40     +   32 +  64

=136 amu

4] Solution:

a) by the gain of electrons by atoms?

Ans. By the gain of electrons anions are formed.

Ex: X +  e =  X–  (anion)

b) by the loss of electrons by atoms?

Ans.By the loss of electrons cations are formed.

Ex: X – e = X

5] Solution:

a) FALSE. Sodium ion has less electron than a neutral atom and that is why sodium ion forms a cation.

b) TRUE. Chloride ion is formed due to gain of one electron in the neutral atom of Chlorine.

6] Solution:

a) Calcium oxide :

Ca     O

+2      -2

= CaO

b) Magnesium hydroxide:

Mg    OH

+2      -1

=  Mg(OH)2

7] Solution:

Ans. Element with valency 3 will form the following oxide:

Z                  O

+3                -2

= Z2O3

8] Solution:

Ans. Sodium is the particle which has 10 electrons, 11 protons (atomic number) and 12 neutrons and its mass number is 23 (11+12).

9] Solution:

Ans. Chlorine is the particle which has 17 protons (atomic number) and 18 neutrons and 18 electrons and its mass number is 35.5( average of isotopes).

SHORT ANSWER TYPE QUESTIONS

11] Solution:

Ans. Water is made up of oxygen and hydrogen.

Valency of hydrogen: +1  , Valency of oxygen: -2

Chemical formula:

H            O

+1            -2

=  H2O

12] Solution:

Ans.  Formula for the compound is :

H                              N

1                               3

= NH3

13] Solution:

Ans. Formula for the compound is :

S                                 O

4                                  2

= S2O4   whose mean compressed formula is = SO2

14] Solution:

Ans. Formula for this compound is :

C                                   S

4                                   2

=  C2S4 whose mean compressed formula is = CS2 ( Carbon di sulphide)

15] Solution:

Ans. Formula for this compound is :

X                                 Y

4                                 1

= XY4

16] Solution:

Ans.a) Formula for the first oxide

B                                 O

4                                   2

= B2O4 whose mean compressed formula is BO2

b) Formula for the second oxide

B                                   O

6                                    2

=  B2O6 whose mean compressed formula is BO3

17] Solution:

Ans. Formula for the compound formed

X                                    Y

3                                    2

= X2Y3

18] Solution:

Ans. Formula for the compound will be

Mg12 = 2,8,2 valency (+2)

HCO3 = -1               valency (-1)

Mg                                  HCO3

+2                                        -1

= Mg (HCO3)2

19]  Solution:

a) bromide of the element.

X                                          Br

2                                            1

= XBr2

b) oxide of the element

X                                             O

2                                              2

= X2O2 whose mean compressed formula is = XO

20] Work out the formula for the following compounds

a) Sodium oxide:

Na11 (2,8,1)       O8 (2,6)

+1                        -2

= Na2O

b) Calcium carbonate:

Ca20 (2,8,8,2)        CO3

+2                            -2

= CaCO3

21.] Solution:

i) Sodium oxide , Na2O: 2 (Na) + 1(O)

2 (23)  + 1 (16)

= 62 amu

ii) Aluminiumoxide , Al2O3: 2 (Al) + 3 (0)

2 (26) + 3 (16)

= 100 amu

22] Solution:

a) CuSO4

Name: Copper sulphate.

Ions   : Cu2+   and    SO42-

b) (NH4)2SO4

Name: Ammonium sulphate

Ions : NH41+ and SO42-

c) Na2O

Name: Sodium oxide

Ions: Na1+ and O2-

d) Na2CO3

Name : Sodium carbonate

Ions: Na+1 and  CO32-

e) CaCl2

Name : Calcium chloride.

Ions : Ca2+  and  Cl1-

23] Write the cations and anions present, if any, in the following

Ans.

       COMPOUND             CATIONS            ANIONS
        CH3COONa                    Na+            CH3COO

NaCl

                   Na+

Cl

          H2                    –                –
         NH4NO3       NH4+                NO3

24] Solution:

a) Calcium and fluorine:

Ca20 (2,8,8,2)        F9 (2,7)

2                           1

= CaF2  (Calcium fluoride)

b) hydrogen and sulphur :

H1 (1)                  S16 (2,8,6)

1                           2

= H2S ( Hydrogen sulphide)

c) nitrogen and hydrogen:

N7 (2,5)           H1(1)

3                         1

= NH3 ( Ammonia)

d) Carbon and chlorine:

C6(2,4)              Cl17(2,8,7)

4                            1

= CCl4( Carbon tetrachloride)

e) Sodium and oxygen:

Na11(2,8,1)       O8(2,6)

1                         2

=  Na2O  (Sodium oxide)

f) Carbon and oxygen:

C6(2,4)       O8(2,6)

4                    2

= C2O4 whose mean compressed formula is CO2 (carbon dioxide)

25] Solution:

i) Ionic compounds : Compounds formed due to loss and gain of electrons are called as ionic compounds.

Ex. i) Sodium loses one electron to chlorine and NaCl is formed.

ii) Mg loses two electrons to two oxygen and MgO is formed.

ii) Molecular compounds: Compounds formed due to sharing of electrons are called as molecular/ covalent compounds .

Ex.i) H2O is formed due to sharing of electrons between hydrogen and oxygen.

ii) CO2 is formed due to sharing of electrons between carbon and oxygen as carbon can neither gain nor lose electrons.

LONG ANSWER TYPE QUESTIONS

26] a) Solution:

Ans. Ions are charged species of atoms  and  are formed due to gain or loss of electrons. There are two types of ions which are cation and anion.

Cations are formed due to loss of electrons :i) Na+ = Na + 1e– 

ii) Mg2+ = Mg + 2e

 

Anions are formed due to gain of electrons: i) Cl + 1e= Cl

ii) O + 2e= O2-

b) The valencies (or charges) of some of the ions are given below

                            ION                   VALENCY(CHARGE)

   Sodium ion

                        1+
Ammonium ion

                          1+

Calcium ion

                          2+
Lead ion

                          2+

Bromide ion

                           1-
Hydroxide ion

                           1-

Sulphate ion

                           2-
Phosphate ion

                           3-

Using this information, write down the formulae of the following compounds:

i) Sodium phosphate :

Na     PO4

+1        3-

= Na3PO4

ii) Ammonium sulphate:

NH4                          SO4

+1                              -2

= (NH4)2SO4

iii) Calcium hydroxide :

Ca                               OH

+2                                -1

= Ca(OH)2

iv) Lead bromide :

Pb                                 Br

2+                                 -1

= Pb(Br)2

27] a) Solution:

Ans. Cation and anion both are types of ions but cation is formed due to loss of electron whereas anion is formed due to gain of electrons. Losing and gaining of electrons occurs for the bond formation and to acquire stable configuration.

Cation formation: Calcium loses two electrons to form calcium cation

Ca = Ca2+ + 2e

Anion formation: Nitrogen gains 3 electrons to form nitrite ion

N + 3e = N3-

b) The valencies (or charges) of some of the ions are given below:

                         ION      VALENCY ( CHARGE)
                  Sodium ion               +1
                 Copper ion               +2
                 Nitrate ion                -1

Sulphide ion

               -2

Using this information, write down the formula of :

i) Sodium sulphide :

Na         SO2

+1           -2

= Na2SO2

ii) Copper nitrate :

Cu                    NO3

+2                       -1

= Cu(NO3)2

28] Solution:

Ans. i) Formation of sodium ion ( Na11 : 2,8,1)

Sodium loses the last shell electron to acquire a stable configuration of 8e in L-shell. Thus losing one electron from the neutral atom decreases the number of electrons to 10, but number of protons remain same i.e 11 and that is why sodium ion carries a positive charge due to more number of protons than electrons.

Na = Na+ + 1e

ii) Formation of chloride ion : (Cl17: 2,8,7)

Chlorine gains one electron in the outermost shell to acquire a stable configuration of 8e in the M-shell. Thus gaining one electron in the neutral atom increases the number of electrons to 18, but number of protons remains same i.e 17 and that is why chloride ion carries a negative charge due to more number of electrons than the protons.

Cl + 1e= Cl

29] a) Solution:

Ans.

                   SIMPLE IONS               COMPOUND IONS
                       Na+                    NH4+
                       O2- Cl

b) An element Y has a valency of 4. Write the formula for its :

i) Chloride

Y    Cl

4     1

= YCl4

ii) Oxide

Y      O

4       2

= Y2O4  whose mean compressed formula will be YO2

iii) Sulphate

Y                       SO4

4                          2

=Y2(SO4)4 whose mean compressed formula will be Y(SO4)2

iv) Carbonate

Y      CO3

4         2

=  Y2(CO3)4 whose mean compressed formula will be Y(CO3)2

V) Nitrate

Y         NO3

4           1

= Y (NO3)4

30] a) Solution:

Ans Formula unit is the simplest form in which a compound can be expressed which in general language is called as empirical formula.

i) NaCl is the formula unit of sodium chloride.

ii) MgCl2 Is the formula unit of magnesium chloride.

b) Calculate the formula masses of the following compounds:

i) Calcium chloride (CaCl2).

=1(Ca) + 2(Cl)

= 1(40) + 2(35.5)

= 111 amu

ii) Sodium carbonate (Na2CO3)

= 2(Na) + 1(C) + 3(O)

= 2(23) + 1(12) + 3(16)

= 106 amu

3rd Part Solution

VERY SHORT ANSWER TYPE QUESTIONS

1] Solution:

Ans 1:- one  mole.

2] Solution:

Ans 2- One mole.

3] Solution:

Ans 3- 6.022 × 10^23.

4] Solution:

Ans 4- One gram atomic mass = 6.022×10^23 atom.

5] Solution:

Ans 5- One gram molecular mass= 6.022×10^23 molecules.

6] Solution:

Ans 6- Avogadro number given by the  scientistAmedeo Avogadro.

7] Solution:

Ans 7-  Moles=given mass/ molecular mass

=12÷32         [O2 = 32 u]

= 0.375 moles.

8] Solution:

Ans 8- Moles=given mass/molecular mass

= 3.6 ÷ 18

= 0.2 moles.

9] Solution:

Ans 9- Given mass= moles × molecular mass

=0.2 × 16

= 3.2g

10] Solution:

Ans10- Given mass= moles × molecular mass

= 2 × 14

=  28 g

Short Answer Type Questions Answers:

12] (a) Solution:

a) Given mass/molar mass= No.of atoms/Avogadro number

12 ÷ 12 = x ÷ 6.022 × 10^23

= x = 6.022× 10^23 atom

(b) Solution:

b) Avogadro number

 (c) Solution:

C) 1 mole.

13] Solution:

Ans 13- Given mass/molar mass= No. of molecules/Avogadro number

= x÷32=12.044×10^25÷6.022×10^23

X = 64×10^2g   OR

X = 6.4 Kg

14] Solution:

Ans 14- Moles = no.of particles /Avogadro number

1.5 = x ÷ 6.022 × 10^23

x = 9.033 × 10^23molecules

15] Solution:

Ans15-Molecular mass of CaCO3 = Ca+C+3×O

= 40 +12 + 3 × 16

= 100 u

Moles = given mass/molecular mass

10 ÷ 100 = 0.1 moles.

16] Solution:

Ans16- Moles = No.of molecules/ Avogadro number

= 1.20 × 10^22 ÷ 6.022 × 10^23

= 0.199 × 10^-1 moles

= 0.0199 moles.

17] Solution:

Ans17- given mass/molecular mass = no.of molecules/Avogadro number

= x/28 = 1/6.022 ×10^23

x = 28 ÷ 6.022 × 10^23

x = 4.64×10^-23g

18] Solution:

Ans18- Moles = given mass ÷ molar mass

=34.5 ÷ 23

= 1.5 mol

19] Solution:

Ans 19- Given mass/molar mass= no.of atoms/Avogadro number

= 10 ÷ 65 = x ÷ 6.022 × 10^23

X = 10 × 6.022 × 10^23÷65

X = 0.926 × 10^23 atom

X = 9.26×10^22 atom

20] Solution:

Ans20- given mass/molar mass= no. of atom/Avogadro number

x/12 = 3.011 × 10^24 ÷ 6.022 × 10^23

= x = 0.5 × 12

x = 60g

21] Solution:

Ans 21-given mass/molar mass=no. of atom/Avogadro number

X ÷ 16 = 1 ÷ 6.022 × 10^23

X = 16 ÷ 6.022 × 10^23

X = 2.656 × 10^-23 g

22] Solution:

Ans 22- No. of molecules/Avogadro number

So, no. of molecules = moles × Avogadro number

= 0.25 × 6.022 × 10^23

= 1.505×10^23atoms

23] Solution:

Ans 23-Moles = No.of atoms/Avogadro number

= 12.044 × 10^25 ÷ 6.022 × 10^23

=2 × 10^2 mol   OR

= 200 mol

24] Solution:

Ans 24- Molecular mass of CHCl3 = C + H + 3 × Cl

= 12 + 1 + 3 × 35.5 = 119.5u

Given mass/molar mass=no of molecules/Avogadro number

0.0239/119.5 = x/6.022×10^23

x = 0.0239×6.022×10^23÷119.5

x = 0.0012044×10^23  OR

x = 12.044×10^19 molecules.

25] Solution:

Ans 25-Molecular mass of Na2CO3= 2×23+12+3×16=106u

Given mass= moles×molecular mass

= 5 × 106

= 530 g

26] Solution:

Ans 26-Given mass/molar mass=no.of molecules/Avogadro number

=4 ÷ 32 = x ÷ 6.022 × 10^23

= x = 4 × 6.022 × 10^23 ÷ 32

X = 24.088 ÷ 32

x = 0.752×10^23 = 7.527×10^22molecules.

27] Solution:

Ans27- Molecular mass of C6H12O6 = 6×12+12×1+6×16 = 180 u

Moles= given mass/molar mass

=100/180

=0.555 moles.

28] Solution:

Ans28-Molecular mass of H2S = 2×1+32=32 u

Given mass=moles×molecular mass

0.17 × 34 = 5.78g

29] Solution:

Ans 29-•5 moles of CO2

Molar mass of CO2 = 12+32 = 44u

So, given mass=5 × 44 = 220g

  • 5 moles of H20

Molar mass of H2O = 2 × 1 + 16 = 18u

So, 5 ×18 = 90 g

Therefore, they do not have same masses.

The differences in their masses is =220-90=130g.

30] Solution:

Ans30 -Moles of calcium

= given mass/ molecular mass= 240/40 = 6 moles

Moles of Magnesium

= given mass/ molecular mass

= 240/24 = 10 moles

Moles of calcium/Moles of Magnesium =6/10=3/5=3:5.

LONG ANSWER TYPE QUESTIONS

31] (a) Solution:

Ans31- a) It is the unit of measurement used to measure the amount of substance. It is defined as the group as the group of 6.022×10^23 particles (atoms, molecular mass or ions) of a substance or atoms in 12 grams of pure carbon-12 isotope.

It represents:-

  • The amount of substance which is equal to the gram atomic mass of that element.
  • Definite number of particles (atoms,ions or molecules)

(b) Solution:

b) •Moles of Na= given mass/molecular mass

1.5=x/46       {2Na=46}x=1.5×46

Given mass of Na=69 g

  • Mass of Sulphur = 1.5×32 {S=32u}

= 48 g

  • Mass of oxygen

=1.5×48      {3×O = 48u}

= 72 g

32] (a) Solution:

Ans32- a) 1 mole of carbon atom contains exactly 6.022×10^23 carbon-12 atom.

 (b) Solution:

b)•Number of atom of Al

given mass/molecular mass= no.of atoms/Avogadro number

=50/27= no.of atoms/6.022×10^23

=50×6.022×10^23/27

x=11.15 atoms

  • Number of atoms of Iron=50×6.022×10^23/56

=301.1/56

=5.37 atoms

Hence, 50g of Al has more atoms.

33] (a) Solution:

Ans 33- a) The amount of a substance whose mass is numerically equal to its atomic mass is called gram atomic mass.

Atomic mass of oxygen = 16u

Gram atomic mass = 16 grams

(b) Solution:

1.) Al2O3

Ans. According to the formula 3 moles of oxygen atoms are present to form Al2O3.

 2.) CO2  

Ans. 2 moles of oxygen atoms are present to form CO2 molecule.

3.) Cl2O7 

Ans. 7 moles of oxygen atoms are present to form Cl2O7 molecule.

4.) H2SO4 

Ans. 4 moles of oxygen atom are present to form H2SO4 molecule.

5.) Al2(SO4)3

Ans. 12 moles of oxygen atoms are present to form Al2 (SO4)3.

34] (a) Solution:

Ans34- a) The amount of substance whose mass is numerically equal to it molecular mass is called gram molecular mass.

Molecular mass of O2= 32u

Gram molecular weight=32g

(b) Solution:

b) Moles of S8= given mass/molecular mass

Molecular mass=8×32 =256gm

So, 100/256=0.39mol

35] (a) Solution:

Ans35-a) Molar mass of a substance is the mass of 1 mole of its molecules.

SI unit= g/mol

(b) Solution:

1.) Ozone molecule

b)•Molecular mass of ozone= O+O+O

=16+16+16= 48g/mol

2.) Ethanoic acid

  • Molecular mass of ethanoic acid

=2×C+4×H+2×O

2×12+4×1+2×16 =60g/mo

 

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Updated: May 28, 2022 — 3:33 pm

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