Atoms and Molecules Class 9 MCQ Chapter 3 Chemistry

Atoms and Molecules Class 9 MCQ with Solutions

1. The ratio by mass of C and O in $C{O_2}$ is :

(A) 1 : 2

(B) 3 : 14

(D) 3 : 11

Ans. (C)

Sol. Ratio by mass of C and O in CO₂ = 12 : 32 i.e. 3 : 8.

2. Law of multiple proportions will remain valid when nitrogen combines with oxygen to form N₂O₅ and :

(A) N₂O

(B) N₂O₃

(D) All of these

Ans. (D)

Sol. In given examples N₂O, N₂O₃, N₂O₄, N₂O₅

Thus different weight of O which combine with a fixed weight of N(14g) in all these oxides are in simple whole number ratio and hence supports the law.

3. When 63.5 g of copper combine with different amount of oxygen to form CuO and Cu₂O, the ratio of oxygen combining in two cases is :

(A) 1 : 2

(B) 1 : 1

(D) 2 : 1

Ans. (D)

Sol. Ratio of weight of ‘O’ combining with 63.5g of Cu fixed weight in CuO and Cu₂O are different which are found to bear a simple whole number ratio i.e. 16 : 8 = 2 : 1.

4. The discovery of isotopes & isobars is the failure of :

(A) Law of conservation of mass

(B) Dalton’s atomic theory

(D) Law of constant proportions

Ans. (B)

Sol. On the basic of discovery of isotopes (atoms of same element can have different masses) and isobars (even atoms of different elements can have the same mass) the postulates of Dalton’s atomic theory have been discarded.

5. The balancing of chemical equation is based upon :

(A) Law of combining volumes

(B) Law of multiple proportions

(D) Law of definite proportion

Ans. (C)

Sol. Balancing of chemical equation is based upon law of conservation of mass awarding to which matter can neither be created nor destroyed during any physical or chemical change.

6. The law of constant composition is applied to :

(A) Any element

(B) Any chemical compound

(D) None of these

Ans. (C)

Sol. The law of constant composition is applied to a pure chemical compound as it always contains same element combined together in the same definite proportion by weight.

7. The percentage of oxygen in ozone molecule is:

(A) 33.3%

(B) 50.0 %

(D)100.0%

Ans. (D)

Sol. O₂ + [O] → O₃ % of O in O₃ is 100%.

8. The percentage of hydrogen in water is :

(A) 1.11%

(B) 11.11%

(D) 88.9 %

Ans. (B)

Sol. 18 g of H₂O contains 2g H₂

∴ 100 g of H₂O = $\frac{{\rm{2}}}{{{\rm{18}}}} \times 100$ = 11.11%

9. The law of constant composition was given by :

(A) Dalton

(B) Berzelius

(D) Lavoisier

Ans. (C)

Sol. The law of constant composition was given by proust.

10. The law of multiple proportions is illustrated by the two compounds :

(A) Sodium chloride and sodium bromide

(B) Ordinary water and heavy water.

(D) Sulphur dioxide and sulphur trioxide

Ans. (D)

Sol. Law of multiple proportion is illustrated by SO₂ and SO₃ in which different weight of oxygen (i. e 32 : 48) combining with same weight of sulphur (32g) bear a simple whole number ratio (2 : 3) with one another.

11. The atomic symbols for Krypton and Antimony are respectively :

(A) Sn and Pb

(B) Kr and Sb

(D) P and Mn

Ans. (B)

Sol. Krypton = Kr; Antimony = Sb

12. The symbol of hydrogen given by Dalton is :

(A)

(B)

(C)

(D)

Ans. (B)

Sol. Symbol of hydrogen given by Dalton =

13. Dalton’s symbol for nitrogen is :

(A)

(B)

(C)

(D)

Ans. (D)

Sol. Dalton’s symbol for nitrogen =

14. Natrium is the latin name of :

(A) Sodium

(B) Nitrogen

(D) Mercury

Ans. (A)

Sol. Latin name of Sodium = Natrium.

15. Stibium is latin name of :

(A) Arsenic

(B) Sulphur

(D) Antimony

Ans. (D)

Sol. Latin name of antimony = Stibium.

16. Buckminster fullerene has the molecular formula of :

(A) C₇₀

(B) C₆₅

(D) none of the above

Ans. (C)

Sol. Molecular formula of Buckminster fullerene = C₆₀

17. Molecules of helium and phosphorus are :

(A) Monoatomic and triatomic

(B) Monoatomic and diatomic

(D) Monatomic and tetraatomic

Ans. (D)

Sol. Molecule of Helium = He monatomic

Molecule of phosphorus = P₄ tetratomic

18. The symbol belongs to :

(A) Sulphur

(B) Mercury

(D) Nitrogen

Ans. (B)

Sol. Symbol of Mercury =

19. Elements present in lime stone are :

(A) Carbon, hydrogen, oxygen

(B) Calcium, carbon, oxygen

(D) Calcium, oxygen

Ans. (B)

Sol. Elements in limestone CaCO₃= Calcium + Carbon + oxygen.

20. Valency of sulphur in sulphur dioxide and sulphur trioxide is ______ and ________respectively :

(A) 3 and 6

(B) 2 and 3

(D) 4 and 6

Ans. (D)

Sol. Valency of S in SO₂ = 4, Valency of S in SO₃= 6.

21. Which of the following molecules is not formed by transfer of electrons :

(A) KCI

(B) MgCl₂

(D) HCl

Ans. (D)

Sol. HCl is formed by sharing of electrons.

22. Which of the following bond in compounds cause highest melting point :

(A) Ionic bond

(B) Covalent bond

(D) Coordinate bond

Ans. (A)

Sol. Ionic bond (electrovalent bond) in compounds causes highest melting point because of three dimensional electrostatic force between cations and anions arranged in particular manner.

23. Compound which is not soluble in water :

(A) NaCl

(B) MgO

(D) NaF

Ans. (C)

Sol. CS₂is a covalent non-polar compound hence it is not soluble in water.

24. The term covalent bond was introduced by :

(A) Thomson

(B) Lavoisier

(D) None of these

Ans. (C)

Sol. Langmuir introduced the term covalent bound.

25. Number of lone pairs in a molecule of Ammonia :

(A) 1

(B) 2

(D) 4

Ans. (A)

Sol. Number of lone pair of electrons in NH₃ is 1.

= one lone pair present

26. Which of the following compound do not contain lone pair :

(A) CH₄

(B) H₂O

(D) None of these

Ans. (A)

Sol. (A) ≡ No. L.P. of e^–

(B) ≡ 2 L.P. of e^–

27. Molecule containing only single covalent bond is :

(A) N₂

(B) O₂

(D) NH₃

Ans. (D)

Sol. (A) N₂ = ${\bf{N}} \equiv {\bf{N}}$ = triple covalent bond

(B) O₂ = O = O = double covalent bound

(D) NH₃ = = Single covalent bound between N and H-atom

28. Chemical bond is formed between atoms to get :

(A) Stable electron arrangement

(B) To form ions

(D) None

Ans. (A)

Sol. Chemical bond is formed between atoms in order to acquire stability through gaining or loosing or sharing electrons for stable electronic arrangement.

29. The bond which is formed by transfer of electrons is :

(A) Covalent bond

(B) Coordinate bond

(D) Hydrogen bond

Ans. (C)

Sol. Ionic bond is obtained by total transfer of electrons from one atom to another atom.

30. Electrovalent bond is formed between :

(A) Two metals

(B) Two nonmetals

(D) two metalloids

Ans. (C)

Sol. Electrovalent bond is formed between a metal and a non-metal . anion e.g.

31. The number of electrons in outer most shell of non metals are :

(A) 1, 2 or 3

(B) 4, 5 or 6

(D) 3, 4 or 5

Ans. (C)

Sol. Number of electrons in outermost shell of non-metals are 5, 6 or 7.

32. Bond formed by the sharing of electrons is :

(A) Ionic bond

(B) Covalent bond

(D) Metallic bond

Ans. (B)

Sol. Covalent bond is formed by sharing of equal member of electrons between atoms.

33. Number of electrons shared between two oxygen atoms in O₂ molecule :

(A) 2

(B) 4

(D) 6

Ans. (B)

Sol. O₂ molecule = i.e. electrons (2 pairs of e^–) are shared between two oxygen atoms.

34. Number of lone pair of electrons on nitrogen in HCN is :

(A) 1

(B) 2

(D) 4

Ans. (A)

Sol. HC ≡ 1 L.P. of e^– on N-atom in HCN molecule.

35. Number of lone pair of electrons on each oxygen atom in CO₂ molecule :

(A) 1

(B) 2

(D) 4

Ans. (B)

Sol. CO₂ = = 2 L.P. of e^– on each O-atom in CO₂molecule.

36. Atoms or group of atoms, which are electrically charged, are known as :

(A) Anions

(B) Cations

(D) Atoms

Ans. (C)

Sol. Ions are atoms or group of atoms which are either positively charged (cation e.g. Na⁺, NH₄⁺) or negatively charged (Anion e.g. Cl ^– NO₃^–).

37. The charge on cation M is +2 and anion A is –3. The compound has the formula :

(A) MA₂

(B) M₃A₂

(D) M₂A

Ans. (B)

Sol. M²⁺A^3– = M₃A₂

38. Which of the following is the formula of Nickel bisulphate :

(A) NiHSO₄

(B) Ni₂HSO₄

(D) Ni(HSO₄)₂

Ans. (D)

Sol. Nickel bisulphate Ni²⁺ HSO₄^– = Ni(HSO₄)

39. Which of the following does not contain electrovalent bond :

(A) K₂O

(B) H₂O

(D) NaCl

Ans. (B)

Sol. H₂O molecule contains covalent bonding

40. During the formation of electrovalent bond :

(A) Metal gets reduced

(B) Non metal gets oxidized

(D) None of these

Ans. (D)

Sol. During electrovalent bond formation metal gets oxidised and non-metal gets reduced.

41. The formula of a chloride of a metal M is MCl₃, the formula of the phosphate of metal M will be :

(A) M₂(PO₄)₃

(B) M₂PO₄

(D) MPO₄

Ans. (D)

Sol. In MCl₃ metal M is trivalent

∴ Metal M phosphate will be M³⁺ PO₄^3– = MPO₄

42. The chemical formula of a compound of radium is RaBr₂ .The formula of its sulphate is :

(A) Ra(SO₄)₂

(B) Ra2(SO₄)₄

(D) Ra₂SO₄

Ans. (C)

Sol. RaBr₂ = (Ra²⁺ Br^–)

∴ Formula of radium sulphate = Ra²⁺ SO₄^2– = RaSO₄

43. The formula of chromium sulphite is :

(A) Cr₂(SO₄)₃

(B) Cr₃(SO₄)₂

(D) None of these

Ans. (D)

Sol. Formula of chromium sulphate = Cr³⁺ SO₃^2– = Cr₂(SO₃)₃

44. Which one of the following is penta-atomic molecule :

(A) CO₂

(B) PCl₅

(D) C₂H₆

Ans. (C)

Sol. POCl₃ is penta–atomic molecule as it contains 5 atoms.

45. The valency of cation in cupric sulphate is :

(A) 1

(B) 2

(D) can’t determined

Ans. (B)

Sol. Valency of cation (Cu²⁺) in cupric sulphate CuSO₄ is 2 i.e. $\mathop {{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}}\limits_{{\rm{(cation)}}\,{\rm{(anion)}}}$

46. The molar mass of CH₃OH is same as that of :

(A) C₂H₅OH

(B) H₂O

(D) O₂

Ans. (D)

Sol. Molar mass of CH₃OH = 12 + 3 + 16 + 1 = 32

Which is same at that of O₂ = 32

In others molar mass of N₂O = 28 + 16 = 44

Molar mass of CO = 12 + 16 = 28

Molar mass of C₂H₅OH = 24 + 5 + 16 + 1 = 46

47. Which of the following represents a polyatomic ion :

(A) Sulphide

(B) Chloride

(D) Nitride

Ans. (C)

Sol. Palyatomic ion is sulphate SO₄^2–

Others are – sulphide S^–2, Chloride Cl^–, Nitride N^3–

48. An ion is a …………. species :

(A) Positively charged

(B) Negatively charged

(D) either A or B

Ans. (D)

Sol. An ion can be either positively charged (e.g. Na⁺ cation) or negatively charged (e.g. Cl^– anion) species.

49. A cation is :

(A) Negatively charged atom

(B) Neutral atoms

(D) Negatively charged atom or group of atoms

Ans. (C)

Sol. A cation is either positively charged atom (e.g. Na⁺) or group of atoms (e.g. NH₄⁺).

50. An anion is :

(A) Positively charged atom

(B) Neutral atoms

(D) Positively charged atom or group of atoms

Ans. (C)

Sol. An anion is either negatively charged atom (e.g.Cl^–) or group of atoms (e.g. SO₄^2–).

51. The no. of sulphur atoms present in one mole of S₈ :

(A) 4.018 × 10²⁴

(B) 4.818 × 10²⁴

(D) None of the above

Ans. (B)

Sol. 1 molecule of S₈ have = 8 S-atom

1 mole of S₈ have = 8× 6.02×10 ²³S-atom

= 4.818 ×10²⁴

52. The element whose gram – atomic mass and gram – molecular mass are same is :

(A) Hydrogen

(B) Oxygen

(D) Helium

Ans. (D)

Sol. Hydrogen : g atomic mass = 1g: g molecular mass = 2g
Oxygen : g molecular mass = 16g: g molecular mass = 32g
Nitrogen : g atomic mass = 14g: g mol. mass = 28g

Helium :

53. 2 g-atom of Aluminium is : (Al = 27) :

(A) 27 u

(B) 54 u

(D) None of these

Ans. (D)

Sol. 1 g atom of Al = 27g

∴ 2g atom of Al = 27 × 2 = 54 g

54. One atomic mass unit of carbon-12 is :

(A) 1/12 of the mass of one ¹²C atom.

(B) 6.022 × 10²³ g

(D) 1 g

Ans. (A)

Sol. 1 amu mass = $\frac{{\rm{1}}}{{{\rm{12}}}}$ of the mass of one c¹² atom = $\frac{1}{{6.023 \times \,{{10}^{23}}}}$

= 1.66 × 10^–24g

55. The mass of 1 a.m.u is :

(A) $\frac{1}{{12}} \times \frac{1}{{6.023 \times {{10}^{23}}}}g$

(B) $\frac{1}{{6.023 \times {{10}^{23}}}}g$

(D) $6.023 \times {10^{23}}$

Ans. (B)

Sol. Mass of 1 a. m.u. = $\frac{1}{{6.023 \times {{10}^{23}}}}g$

56. The atomic masses of the elements are usually fractional because :

(A) Elements consists of impurities.

(B) These are mixtures of all allotropes

(D) These are mixtures of isotopes

Ans. (D)

Sol. Atomic masses of the elements are usually fractional because it is average atomic mass as most of the element occur in nature as mixture of several isotopes.

57. 1 mol of CH₄ contains :

(A) 6.02 × 10²³ atoms of H

(B) 4 g-atom of hydrogen

(D) 3.0 g of carbon

Ans. (B)

Sol. 1 mole of CH₄contains 4 g atom of hydrogen

= N_A molecules by CH₄

= 4 N_A atoms of H

= N_A atoms of C.

58. Which of the following has smallest mass :

(A) 4 g of He

(B) $6.022 \times \,{10^{23}}\,$ atoms of He

(D) 1 mole of He

Ans. (C)

Sol. (i) 4g He

(ii) 6.022 × 10²³atoms of He = 4g of He

(iii) 6.023 × 10²³ atoms of He = 4g

∴ 1 atom of the = $\frac{{\rm{1}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}{\rm{g = }}\mathop {\frac{{\rm{4}}}{{{{\rm{N}}_{\rm{A}}}}}g}\limits_{{\rm{Smallest mass}}}$

1 mole of He = 4g of He

59. How many grams are present in 1 gram atom of sodium :

(A) 23 g

(B) 13 g

(D) $\frac{1}{{23}}$ g

Ans. (A)

Sol. 1 g atom of N_A ≡ 1 mole of Na ≡ 23 g of Na

60. Weight of 1 atom of hydrogen is :

(A) $1.66 \times \,{10^{ - 24}}g$

(B) ${10^{22}}\,g$

(D) $1 \times {10^{ - 23}}\,g$

Ans. (A)

Sol. wt. of 1 atom of H = $\frac{1}{{6.023 \times {{10}^{23}}}}$ = 1.66 × 10^–24g

ALSO REAd

Matter in our surroundings MCQ Questions | CBSE Science Class 9

Nazism and Rise of Hitler MCQ History Chapter 3

Structure of Atom Class 9 MCQ with Solutions Science

Atoms and Molecules Class 9 MCQ Chapter 3 Chemistry

ALSO REAd

Leave a Comment Cancel Reply

QUICK LINKS

CBSE CLASS 9 QUESTION BANK

CBSE CLASS 10 QUESTION BANK

CBSE ONLINE TEST

ALSO REAd

Related Posts

Leave a Comment Cancel Reply