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Chapter: 10th Science : Chapter 7 : Atoms and Molecules

Book Back Questions with Answers

I. Choose the best answer. II Fill in the blanks III. Match the following IV. True or False: (If false give the correct statement) V. Assertion and Reason VI. Short answer questions VII. Long answer questions VIII. HOT question IX. Solve the following problems

Atoms and Molecules (Science)


I. Choose the best answer.

 

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

 

2. Which of the following is a triatomic molecule?

a. Glucose

b. Helium

c. Carbon dioxide

d. Hydrogen

 

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

 

4. Mass of 1 mole of Nitrogen atom is

a. 28 amu

b. 14 amu

c. 28 g

d. 14 g

 

5. Which of the following represents 1 amu?

a. Mass of a C – 12 atom

b. Mass of a hydrogen atom

c. 1/12th of the mass of a C – 12 atom

d. Mass of O – 16 atom

 

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.

 

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

 

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

 

9. The gram molecular mass of oxygen molecule is

a. 16 g

b. 18 g

c. 32 g

d. 17 g

 

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

 

II Fill in the blanks


1. Atoms of different elements having same mass number, but different atomic numbers are called isobars.

2. Atoms of different elements having same number of neutrons are called isotones.

3. Atoms of one element can be transmuted into atoms of other element by artificial transmutation

4. The sum of the numbers of protons and neutrons of an atom is called its mass number

5. Relative atomic mass is otherwise known as standard atomic weight

6. The average atomic mass of hydrogen is 1.0079 amu.

7. If a molecule is made of similar kind of atoms, then it is called Homo atomic molecule.

8. The number of atoms present in a molecule is called its atomicity

9. One mole of any gas occupies 22400 ml at S.T.P

10   Atomicity of phosphorous is 4

 

III. Match the following

1. 8 g of O2 - 4 moles

2. 4 g of H2 - 0.25 moles

3. 52 g of He - 2 moles

4. 112 g of N2 - 0.5 moles

5. 35.5 g of Cl2 - 13 moles

Answer:

1. 8 g of O2 - 0.25 moles

2. 4 g of H2 - 2 moles

3. 52 g of He - 13 moles

4. 112 g of N2 - 4 moles

5. 35.5 g of Cl2 - 0.5 moles


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


1.  Two elements sometimes can form more than one compound. - True

2.  Noble gases are Diatomic - False

Noble gases are monoatomic.

3.  The gram atomic mass of an element has no unit - False

The gram atomic mass of an element is exposed in grams.

4.  1 mole of Gold and Silver contain same number of atoms - True

5.  Molar mass of CO2 is 42g. - False

Molar mass of CO2 is 44g.

 

V. Assertion and Reason:

Answer the following questions using the data given below:

i) A and R are correct, R explains the A.

ii) A is correct, R is wrong.

iii) A is wrong, R is correct.

iv) A and R are correct, R doesn’t explains A.

1. Assertion: Atomic mass of aluminium is 27

Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C – 12 atom.

Answer: i) A and R are correct, R explains the A.

2. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.

Reason: The natural abundance of Chlorine isotopes are not equal.

Answer: i) A and R are correct, R explains the A.

 

VI. Short answer questions


1. Define: Relative atomic mass.

The Relative Molecular Mass of a molecule is the ratio between the mass of one molecule of the substance to 1/12th mass of an atom of Carbon-12.


2. Write the different types of isotopes of oxygen and its percentage abundance.

Oxygen exists as a mixture of three stable isotopes in nature.

Isotope: Mass (amu) : % abundance

8o16 : 15.9949 : 99.757

8O17 : 16.9991 : 0.038

8O18 : 17.9992 : 0.205


3. Define: Atomicity

The number of atoms present in the molecule is called as its atomicity.


4. Give any two examples for heterodiatomic molecules.

Examples of heterodiatomic HCl, CO.


5. What is Molar volume of a gas?

(i) The volume occupied by one mole of any gas at S.T.P is called molar volume.

(ii) One mole of any gas occupies = 22.4 litre or 22400 ml at S.T.P.


6. Find the percentage of nitrogen in ammonia.

 % of nitrogen in NH3 

= [ Mass of nitrogen / Molar mass of NH3 ] x 100

= (14/17) X 100 = 82%


 

VII. Long answer questions


1. Calculate the number of water molecule present in one drop of water which weighs 0.18 g.

Answer:

Mass of water = 0.18 g

No of moles = Mass / Molecular Mass

 = 0.18 / 18

 = 0.01 mole

Number of molecules = No of moles × Avogadro number

= 0.01 × 6.023 × 1023

= 0.06023 × 1023

= 6.023 × 1021 molecules


2. N2 + 3 H2 → 2 NH3

(The atomic mass of nitrogen is 14, and that of hydrogen is 1)

1 mole of nitrogen ( 28 g) +

3 moles of hydrogen ( 6 g) →

2 moles of ammonia ( 34 g)

Answer: 1 mole of nitrogen ( 28 g) + 3 moles of hydrogen ( 6 g) → 2 moles of ammonia ( 34 g)

Answer: 28, 6, 34.


3. Calculate the number of moles in

i) 27g of Al ii) 1.51 × 1023 molecules of NH4Cl

Answer:

(i) 27 g of Al

No of moles = Mass / Atomic Mass

27 / 27 = 1 mole

(ii) 1.51 x 1023 molecules of NH4C1

No of moles = Number of molecules / Avogadro number

= (1.51 × 1023) / (6.023 × 1023)

= 0.25 mole


4. Give the salient features of “Modern atomic theory”.

Answer:

‘The main postulates of modern atomic theory’ are as follows:

(i) An atom is no longer indivisible

(ii) Atoms of the same element may have different atomic mass. Example – isotopes l7Cl35 , l7Cl37.

(iii) Atoms of different elements may have same atomic masses. Example – Isobars 18Ar40, 20Ca40.

(iv) Atoms of one element can be transmuted into atoms of other elements.

(v) Atoms may not always combine in a simple whole number ratio. Eg: Glucose C6H12O6.

(vi) Atom is the smallest particle that take part in a chemical reaction.

(vii) Mass of an atom can be converted into energy. E = mc2.


5. Derive the relationship between Relative molecular mass and Vapour density.

Answer: Relative molecular mass :

(i) (Hydrogen scale): 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.

(ii) Vapour Density : Vapour density is the ratio of the mass of a 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.

Vapour Density

(VD) = Mass of given volume of gas or vapour at S.T.P / Mass of same volume of hydrogen

According to Avogadro's law, equal volumes of all gases contain equal number of molecules.

Thus, let the number of molecules in one volume = n, then

V.D at S.T.P

= Mass of ´n´molecules of gas or vapour at S.T.P / Mass of n molecules of hydrogen

Cancelling 'n' which is common, we get

V.D = Mass of 1 molecule of gas or vapour at S.T.P / Mass of 1 molecule of hydrogen

However, since hydrogen is diatomic

V.D = Mass of 1 molecule of a gas or vapour at S.T.P / Mass of 2 atoms of hydrogen

V.D = ( Mass of 1 molecule of a gas or vapour at S.T.P ) / ( 2 × Mass of 1 atom of hydrogen)

V.D = Molecular Mass / 2

2 × vapour density  = Molecular Mass of a gas

Molecular Mass  = 2 × Vapour density.

 

VIII. HOT question


1. Calcium carbonate is decomposed on heat-ing in the following reaction

CaCO3 → CaO + CO2

i. How many moles of Calcium carbonate are involved in this reaction?

ii. Calculate the gram molecular mass of calcium carbonate involved in this reaction

iii. How many moles of CO2 are there in this equation?

CaCO3 → CaO + CO2

(i) No of moles of CaCO3 involved in this reaction = 1

(ii) Molar mass of calcium = 40

Molar mass of carbon = 12

Molar mass of oxygen = 16

Gram molecular mass of CaCO3 = Molar mass of calcium + Molar mass of carbon + Molar mass of oxygen x 3

= 40 + 12 + (16 x 3)

= 100 g / mol.

(iii) No of moles of CO2 in the reaction = 1

 

IX. Solve the following problems


1. How many grams are there in the following?

i. 2 moles of hydrogen molecule, H2

ii. 3 moles of chlorine molecule, Cl2

iii. 5 moles of sulphur molecule, S8

iv. 4 moles of phosphorous molecule, P4

(i) 2 moles of hydrogen molecule

Molecular mass of H2 = 2

Mass = no of moles x molecular mass

= 2 x 2 = 4

(ii) 3 moles of Chlorine molecule

Molecular mass of Cl2 = 70.9

Mass = no of moles x molecular mass

= 3 x 70.9

= 212.7 g

(iii) 5 moles of sulphur molecule S8

Molecular mass of sulphur molecule

= 32 x 8 = 256

Mass = no of moles x molecular mass

= 5 x 256 = 1280 g

(iv) 4 molecules of P4 molecule

Molecular mass of P4 molecule = 31 x 4 = 124

Mass = no of moles x molecular mass

 = 4 x 124 = 496 g


2. Calculate the % of each element in calci-um carbonate. (Atomic mass: C-12, O-16, Ca -40)

Molar mass of CaCO3 = 40 + 12 + (3 x 16)

= 100

% of C in CaCO3 = [ Mass of carbon / Molar Mass of CaCO] x 100

= [12/100] x 100 =12%

% of O in CaCO3 = [ Mass of oxygen / Molar Mass of CaCO] x 100

= [16x3 / 100] x 100 =48%

% of Ca in CaCO, = [ Mass of calcium / Molar Mass of CaCO] x 100

= [40/100] x 100 =40%


3. Calculate the % of oxygen in Al2(SO4)3. (Atomic mass: Al-12, O-16, S -32)

Molar mass of Al2 (SO4)3

= (27 x 2) + (32 x 3) + (16 x 12)

= 342

% of Oxygen in Al2 (SO4)3

= [ Mass of oxygen in compound / Molar mass of Al2 (SO4)3 ] x 100

= [192/342] x 100

= 56.1 %


4. Calculate the % relative abundance of B -10 and B -11, if its average atomic mass is 10.804 amu.

Let,

% Abundance of B10 = x

% Abundance of B-11 = y

x+y = 100

x = 100-y

Atomic mass unit of B10 = 10.01294

Atomic mass unit of B11 = 11.009305

amu 10.804 =  [10.01294 x +11.009305 y] / 100

Substitute x= 100 – y

10.804 = [10.01294 (100 – y) +11.009305 y] / 100

1080.4 = 1001.294 + 0.996365 y

79.106 = 0.996365 y

y = 79.364%

x = 20.636%


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