Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5 - ICSEHELP (2023)

Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5. Step by Step Solutions of Practice Problems , Exercise and Objective Type Questions of Goyal Brothers Prakashan Chapter-5 Mole Concept and Stoichiometry for ICSE Class 10 Chemistry.

Practice Problems Numerical based on Volume to Volume relation, Volume to Weight relation and Weight to Weight relation of mole concept and Stoichiometry for ICSE Class-10 Goyal Brother Prakashan Chemistry. Visit official Website CISCE for detail information about ICSE Board Class-10 Chemistry .

Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5

-: Select Topics :-

Practice Problem

Exercise 1

Exercise 2

Objective Type Questions

Practice Problems Mole Concept and Stoichiometry Ch-5 ICSE Class-10

(Numerical Problems Based on Gay Lussac’s Law )

Page-71 Practice Problems-1

Answer

(1)

Thus, gaseous composition after reaction is:

Excessoxygen = 200 – 160 = 40 cm³

CO₂= 80 cm³

Water = negligible

(2)

2 C₂H₂ + 5 O₂ ======> 4 CO₂ + 2 H₂O

100 cm³ of acetylene is mixed with 300 cm³ of pure oxygen and ignited ,

According to the equation :

2 volumes of C₂H₂ + 5 volumes of O₂ ===> 4 volumes of CO₂ + 2 volumes of H₂O

Now it is given that C₂H₅ has 100 cm³ volume .

So 2 volumes = 100 cm³ .

⇒ 1 volume = 100 cm³/2

⇒ 1 volume = 50 cm³

300 cm³ is ignited .

Given 5 volumes of oxygen react .

⇒ 5 × 50 cm³ react .

⇒ 250 cm³ react .

Excess oxygen can be calculated by subtracting the amount of reacted oxygen by total volume of oxygen ignited .

Unused Oxygen = 300 cm³ – 250 cm³

⇒ Unused oxygen = 50 cm³ .

Total volume of the resulting mixture :

4 volumes of CO₂ + 2 volumes of H₂O

⇒ 4 × 50 cm³ + Nil

⇒ 200 cm³ + Nil

200 cm³

Page-72 Practice Problems-2

Answer

(1)

Given react ion is :
N₂+ O₂→ 2NO
According to Gay -Lussac’s law in the above reaction 1 volume of nitrogen combines with 1 volume of oxygen to produce 2 volumes of nitric oxide.
i .e. N₂+ O₂→ 2NO
1 vol. 1 vol. 2 vol.
The volume of nitric oxide produced is = 1400cm³.
Let the volumes of nitrogen and oxygen gases be = x
Then, N₂+ O₂→2NO
x x 1400cm3
So, x + x = 1400
2x = 1400
x= 1400/ 2 = 700cm³
Hence the volumes of reacting gases i .e. nitrogen and oxygen is 700 cm³each.

(2)

2C₂H₂(g) + 5O₂(g) → 4CO₂(g)+ 2H₂O(g)

4 V CO₂is collected with 2 V C₂H₂

So, 200cm³CO₂will be collected with = 100cm³C₂H₂

Similarly, 4V of CO₂is produced by 5 V of O₂

So, 200cm³CO²will be produced by = 250 ml of O₂

Numerical Problems Based on Avogadro’s Law

(Class-10 Mole Concept and Stoichiometry Ch-5)

Page-79 Practice Problems-1

Answer

(1)

At STP,1molegas occupies22.4litrevolume.

So,56cm3gas have22.4×100056​moles

Number of moles of dry Nitrogen at STP=2.5 moles

Number of molecules of dry Nitrogen=2.5×10−3×NA​

NA​=Avogardo Number=6×10 23

So, Nitrogen have15×1020moleculesor1.5×1021molecules

(2)

According to Avogadro’s law,
If 112cc of dry CO contains P molecules,then 112 cc of dry Cleaners will also contain P molecules.

1dm3=1000 cc

so 336 dm3 =1000 x 336 cm3
Applying unitary method,
112cc => P molecules
336000cc => 336000 x P/ 112 = 3000P.

Page-79 Practice Problems-2

Answer

Numerical Problems Based on Molecular Weight

(Class-10 Mole Concept and Stoichiometry Ch-5)

Page-80 Practice Problems-1

Answer

(i)

Al+3 (O+ H)

=27+3(16+1)

=27+51

=78 amu

(ii)

= 2 K+ 2 Cr+ 7 O

= 2x 39 + 2 x 52 + 7 x 16

= 78 + 104 + 112

=294 amu

(iii)

The molar mass of ammonium nitrate= sum of the molar mass of all its constituents in the correct proportion.

Molecular weight of nitrogen = 14.

Molecular weight of hydrogen= 1.

The molecular weight of oxygen= 16

The molecular weight of ammonium nitrate= 2 * weight of nitrogen atom + 4 * weight of hydrogen atom + 3 * weight of an oxygen atom

= 28 + 4. + 48

= 80 amu

Page-80 Practice Problems-2

Answer

(i)

MM of Cu + S +4 O + 5 (2H + O)

=64+32+ 4* 16 + 5(2*1 +16)

=64+32+64+90

=250amu

(ii)

MM of 2 Na + S +4 O + 10 (2H + O)

= 2 *23 +32 + 4* 16 + 10 (2*1 +16)

= 46+32 +64+ 180

= 322 amu

Numerical Problems Based on Avogadro’s Number

(Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Page-80 Practice Problems-1

Answer

(1)

Number of molecules in 12.8g of sulphur dioxide gas.
Molecular mass of SO₂= 64a.m.u .
So, 64g = 1 mole
12.8g = 12.8/ 64 = 0.2 mole
Now 1 mole of SO₂contains = 6 x 10²³molecules
0 .2 mole of SO₂contains = 0 .2 x 6 x 10²³
= 1.2 x 10²³molecules

(2)

71g of chlorine has 6×10^23 molecules

7.1g has 0.1×6×10^23 molecules

=6×10^22

Page-81 Practice Problems-2

Answer

(1)

We know
One mole contains 6×10^23 molecules
Weight of one mole of nitrogen gas= 28gm
Weight of 2.8×10^24 molecules
= (2.8×10^24/6×10^23) * 28
= 28×28/6
= 130.67 gm

(2)

gram molecular mass of Coso4 is =159

6*x 10²³molecules weight = 159 g

1 molecule weight is = 159/ 6 *x 10²³

2*x 10²³molecules weight is = (159 * 2*x 10²³⁾/ (6 *x 10^{23 )}

^{=5.33 gm}

^{Page-81 Practice Problems-3}

^Answer

⁽¹⁾

O₂is a diatomic molecule.

The mass of an oxygen atom = 16 amu.

∴ Mass of O₂molecule = 2 × 16 = 32 amu.

Mass of one molecule of oxygen

⇒ 32/ Avogadro constant ( 6.0 × 10²³)

⇒ 5.33 × 10^-23g.

⁽²⁾

O₃ is a triatomic molecule.

The mass of an oxygen atom = 16 amu.

∴ Mass of O₃molecule = 3 × 16 = 48 amu.

Mass of one molecule of oxygen

⇒ 48/ Avogadro constant ( 6.0 × 10²³)

⇒ 8 × 10^-23g.

Numerical Based on Gram Molecular Volume Mole Concept and Vapour Density

(Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Page -82 Practice Problem -1

Answer

(1)

Mass of one mole of nitrogen dioxide = 14 + 16 + 16 = 46g
No. of moles in 0.23g on NO2 = 0.005
volume occupied by 1 mole = 22.4 L
volume occupied by 0.005 mole = 22.4*0.005

=0.112 L =0.112 dm3

0.112L OR112mL:

(2)

Mass of one mole of Butane = 4*12+10*1 = 58

No. of moles in 2.9 g butane = 2.9/58
volume occupied by 1 mole = 22.4 L
volume occupied by 2.9/58 mole = (22.4* 2.9)/58

=1.12 dm3

Practice Problem -2 Page -82

Answer

(1)

Molecular mass of co2 = 44 g

22.4 L = 22.4 dm3 = 44 g

hence 1 L = 44/22.4 g

therefore 5.6 dm3 = 5.6 * 44/22.4 g

= 11 g

(2)

Molecular mass of butane = 58 g

22.4 L = 22.4 dm3 = 58 g

hence 1 L = 58/22.4 g

therefore 7.84 dm3 = 7.84 * 58/22.4 g

= 20.3

Page -82 Practice Problem -3

Answer

(1)

(i)

MM of lead nitrate = 207+14+16*3

=207+14+48

=269 g

Mole = Mass /MM

Mole= 6.62/269=0.024 mole ans

(ii)

1 mole =6× 10²³molecule

0.024 mole = 0.024 x 6 × 10²³.

1.2 × 10²²molecule.

(2)

(i)

mole = mass/MM

mole=2/100

=0.02 moles

(ii)

1 mole =6× 10²³molecule

=0.02 mole = 0.02 x 6 × 10²³.

=1.2 × 10²²molecule.

Page-83 Practice Problem -4

Answer

(1)

(i)

(ii)

Weight of N₂= 1.4 gm

No.of molecules = 3×10²²

Moles ofN₂= 0.05 mol

(iii)

……………….

(2)

(i)

…………….

(ii)

Volume of carbon dioxide gas = 44.8 L

Since at STP ; 22.4 L = 1 mole of gas

=2 moles of gas.

(iii)

………………..

Practice Problem -5 Page -83

Answer

(1)

Under similar conditions same volume of two gases have equal no of moles so,number of moles of X = number of moles of hydrogen

VD= Mass of certain gas/ Mass of hydrogen same volume

VD= 24.64/1.12

VD=22

Relative Molar Mass= 2*VD= 2x 22 = 44 amu

(2)

Page-84 Practice Problem -6

Answer :

(1)

MM =2 x VD =2 x 32 = 64

(i) number of moles= 0.56/22.4 =0.025
(ii) Weight in grams

1 mole weight = 64

so 0.025 mole weight = 0.025 x 64 = 1.6 gm
(iii) number of molecules

1 mole = 6.023 x 10²³

0.025 mole = 0.025 x 6.023 x 10²³

1.505 x 10²²

(2)

MM =2 x VD =2 x 8.5 = 17 gm

(i) number of moles= 8.96/22.4 =0.4 mole
(ii) Weight in grams

1 mole weight = 17

so 0.4 mole weight = 0.4 x 17 = 6.8 gm
(iii) number of molecules

1 mole = 6.023 x 10²³

0.4 mole = 0.4 x 6.023 x 10²³

2.4 x 10²²

Exercise-1 Goyal Brothers Prakashan Class-10 Mole Concept and Stoichiometry Ch-5

Page-84

Question 1.State “Gay Lussac’s Law” of combining gas. Support your answer by two examples.

Answer :

This law, formulated by Gay Lussac, states that, “the ratio between the volumes of gaseous reactants and products can be expressed in simple whole numbers.

For example, in the following reaction, the ratio of volumes of hydrogen, chlorine, and hydrogen chloride is 1:1:2 (a simple ratio):

Question 2.State “Avogadro’s Law” Explain the law for reaction between nitrogen and hydrogen, when they react to form ammonia gas.

Answer :

Avogadro’s lawstates that equalvolumesof different gases contain an equal number ofmolecules.This applies when the temperature andpressurestay the same.

Avogadro’s law can be used to calculate the volumes of gases involved in reactions.

Question 3.What do you understand by the term “atomic mass unit”?

Answer :

a unit of mass used to express atomic and molecular weights, equal to one twelfth of the mass of an atom of carbon-12.

Question 4.Define relative atomic mass of an element.

Answer :

Arelative atomic mass(also calledatomic weight; symbol:A_r) is a measure of how heavyatomsare. It is theratioof the average mass per atom of anelementfrom a given sample to 1/12 the mass of acarbon-12atom.^[1][2]In other words, a relative atomic mass tells you the number of times an average atom of an element from a given sample is heavier than one-twelfth of an atom of carbon-12

Question 5.What is gram atomic mass?” How does gram atomic mass differ from relative atomic mass of an element ?

Answer :

Gram atomic mass(weight) of an element is the mass of Avogadro number ( 6.023 x 10²³) of atoms of that elements in grams.”

Ex –

mass of 1 atom of oxygen = 16 amu = 16 x 1.66 x 10^-24gm

Mass of 6.023 x 10²³atoms of oxygen= 16 x 6.023 x 10²³x 1.66 x 10^-24= 16 gm.

Hence, Gram atomic weight of oxygen = 16 gm.

Question 6.Define relative molecular mass of a substance. How does it differ from gram molecular mass.

Answer :

Expression ofmolar massisgramspermole. It can also be expressed as kilogram permole.Molecular massis expressed in atomicmassunits

Moleculesare substances formed due to combinations of the same or different atoms in different ratios. Thesemoleculesattached to each other to formcompounds. Besides, the molar mass and molecular weight are two physical properties of substances.

Question 7.Define gram molecular volume. State its experimental value.

Answer :

gram–molecular volume, thevolumeoccupied by onegram–molecularweight of any substance at STP. Its value at STP is 22.4 liter / 22400 ml/ 22.4 dm3 / 22400 cm3

Question 8.Define Avogadro’s number. State its value.

Answer :

Avogadro’s number,numberof unitsinone mole of any substance (definedasitsmolecular weightingrams), equal to 6.02214076 × 10²³.Theunits may be electrons, atoms, ions, or molecules, depending onthenature ofthesubstance andthecharacter ofthereaction (if any)

Question 9.What do you understand by the term “mole”? How many elementary units are in one mole of a substance

Answer :

Amoleis defined as 6.02214076 × 10²³of some chemical unit, be it atoms, molecules, ions, or others. Themoleis a convenient unit to use because of the great number of atoms, molecules, or others in any substance

Question 10.Fill in the blank spaces:

(i) Weight of a substance (in grams) = Number of moles X …

(ii) Volume of a gas at S.T.P=… .. x 22.4 dm³ at S.T.P

Answer :

(i) Weight of a substance (in grams) = Number of moles X …Molecular mass

(ii) Volume of a gas at S.T.P=…No of Mole .. x 22.4 dm³ at S.T.P

Question 11. State four applications of Avogadro’s Law.

Answer :

four applications of Avogadro’s Law

(i) It determinesthemolecule formula of a gas.

(ii) It determines atomicity of gases.

(iii) It explains Gay-Lussac’slawof combining volumes.

(iv) It establishestherelation between molecular weight and vapour density of a gas

Question 12.

(i) What do you understand by the term atomicity?

(ii) What is the atomicity of the following elements (1) helium (2) nitrogen (3) ozone (4) phosphorus (5) sulphur?

Answer :

(i)Atomicityis defined as the total number of atoms that constitute a molecule. For example, each molecule of oxygen (O₂) is composed of two oxygen atoms. Soatomicityof oxygen is 2.In older contexts, theterm atomicityis sometimes used in the same sense as valency

(ii) the atomicity of the following elements

(1) helium–1

(2) nitrogen –2

(3) ozone—3

(4) phosphorus—5

(5) sulphur—8

Question 13.

(i) What do you understand by the term vapour density?

(ii) How is vapour density of a substance related to its gram molecular mass?

Answer :

(i)Vapour densityis thedensityof avapourin relation to that of hydrogen. It may be defined as mass of a certain volume of a substance divided by mass of same volume of hydrogen.

(ii) The vapour density of a substance related to its gram molecular mass?

molar mass= 2 ×vapour density

Part-2 Stoichiometry Goyal Brothers Prakashan ICSE Class-10

Numerical Based on Percentage Composition

Practice Problems-1 Page- 85

Answer

(1)

Pb + (N)₂+ (O)₆
207 + 2 x 14 + 6 x 16 = 331.
So, the molecular mass of Pb(NO₃)₂= 331.
331 by weight of Pb(NO₃)₂contain 96 parts by weight of oxygen .
100 parts will contain = 96 x 100 / 331 = 29 ‰
So, the percentage composition of oxygen in lead nitrate is 29‰

(2)

Atomic mass of calcium (Ca) = 40

Atomic mass of phosphorus (P) = 31.

Atomic mass of oxygen (O) = 16.

Now, molecular mass of calcium phosphate = 40 (3) + (31+64)2 = 120 + 95(2) = 120+190 = 310 g/mol.

Percentage of Ca in Ca3(PO4)2 = 40(3)/310 ×100 = 120/310 ×100 = 38.7….%

(3)

(4)

The formula mass of ferrous sulphate
At.massofFe+At.,massofS+At.massofoxygen+7×Mol.massofH2​O
56.0+32.0+4×16.0+7×18.0=278.0
So % of water of crystallisation=278126​×100=45.32 %

Practice Problems-2 Page- 86

Answer

(1)

50 x 60/100 =30 gm

Molecular mass of KClO3 = 39+35.5+ 16 x 3

=122.5

% of O2 = (48 x 100 /122.5)

………….39.18……..

now

39.18 x 60/100 = 23.5%

(2)

Molecular weight of Al2​O3​=102g

Molecular weight ofAl=27g

Weight ofAlinAl2​O3​=2×27=54g

As given that the given sample ofAl2​O3​is90%pure, i.e.,10Kgof sample has9Kgof pureAl2​O3​.

Therefore,

Amount ofAlin102gofAl2​O3​=54g

Amount ofAlin9000gofAl2​O3​=10254​×9000=4764.7g

Total weight of sample=10Kg=10000g

∴percentage ofAlin given sample=100004764.7​×100=47.64%

Practice Problems-3 Page- 87

Answer

(1)

The mass ofNis28g.

Percent composition ofN is = 28*100/80 =35%

The mass ofNis42g.

Percent composition ofN is = 42*100/149 =28.19%

hence ammonium nitrate is better

(2)

GivenUrea= NH2- CO- NH2 And we know that Total mass ofurea= 60 Mass ofnitrogen= 14 Two molecules ofnitrogen= 2*14= 28Percentage of nitrogen= Mass ofNitrogen/(Urea) x 100 = 28/60 x 100 = 46.6% HencePercentage of nitrogen in ureais 46%

You can write thepercentcomposition ofnitrogen in ammonium sulfateby dividing the molar mass of the twonitrogenatoms by the molar mass of the compound, and multiplying the result by 100. This means that every 100g ofammonium sulfatecontain 21.2 g ofnitrogen

hence urea is better

Numerical Problems Based on Empirical Formula and Molecular Formula

(Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Practice Problems-1 Page- 89

Answer

(1)

Steps to calculate empirical formula-

(2)

(4)

mole value of Na= 16.10/23=0.7

mole value of C= 4.20/12=0.35

mole value of O= 16.77/16= 1.08

simple mole ratio of Na= 0.7/0.35=2

simple molar ratio of C=0.35/0.35=1

simple mole ratio of O= 1.08/0.35= 3

so, empirical formula is

Na₂CO₃10H₂O

Practice Problem-2 Page-90

Answer

(1)

mole value of C= 26.70 /12 = 2.21

mole value of H= 2.20/1= 2.22

mole value of O= 71.10/16= 4.44

(a) its empirical formula = CHO₂

(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D × 2 = 90

so, molecular formula = C₂H₂O₄

₍₂₎

(3)

Practice Problem-3 Page-91

Answer

(1)

VD= Empirical formula

VD= 12+1 x2+ 16=30

Molecular Formula =2 * Empirical formula

= 2( CH₂O)

= C₂H₄O₂

(2)

EF = 2 VD =Molecular Formula

12*2 + 1*6 +16 = 46

Empirical formula is same to Molecular Formula

C₂H₆O

Exercise 2 Goyal Brothers Class-10 Mole Concept and Stoichiometry Ch-5

Page-91

Question 1.What do you understand by the term “molecular formula”?

Answer :

achemical formulathat indicates the kinds of atoms and the number of each kind in amoleculeof a compound.

Question 2.What do you understand by the term “empirical formula”?

Answer :

In chemistry, theempirical formulaof a chemical compound is the simplest positive integer ratio of atoms present in a compound. .Anempirical formulamakes no mention of the arrangement or number of atoms

Question 3.What is the empirical formula of the following chemical compounds:
(i) C6H6 (ii) C6H12O6 (iii) CH3COOH, (iv) C2H5OH, (v) C2H4 (vi) C2H2

Answer :

the empirical formula of the following chemical compounds:
(i) C6H6 —CH

(ii) C6H12O6—-CH2O

(iii) CH3COOH,—–CH2O

(iv) C2H5OH,—-C2H6O

(v) C2H4—– CH2

(vi) C2H2——CH

Question 4.What is the difference between molecular formula and empirical formula?

Answer :

Molecular formulastell you how many atoms of each element arein acompound, andempirical formulastell you the simplest or most reduced ratio of elementsin acompound. If a compound’smolecular formulacannot be reduced any more, then theempirical formulais the same as themolecular formula

Question 5.The molecular weight of a chemical compound is six times its empirical formula weight. If the empirical formula of the compound is CH2, find its molecular formula.

Answer :

empirical mass = CH2 = 12+ 2×1 =14

molecular mass= 6 time of empirical mass

molecular mass= 6 x 14

molecular mass = 84

n = molecular mass/ empirical mass

n= 84/14

n=6

Molecular formula = n (empirical formulas)

Molecular formula = n (CH2)

Molecular formula = 6 (CH2)

Molecular formula = C6H12

A. Problems on Weight – Weight Relationships

( Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Practice Problem-1 Page-92

Answer

(1)

molecular mass of ammonium nitrate =80 g

mass of nitrogen in ammonium nitrate= 28

when ammonium nitrate 80 g then nitrogen is= 28

when ammonium nitrate 1 g then nitrogen is= 28/80

when ammonium nitrate 100*1000 g then nitrogen is= (28*100*1000)/80

=35 000 g = 35 kg

(2)

molecular mass of ferric oxide(Fe2O3) =160 g

mass of iron in ferric oxide(Fe2O3)=56*2 = 112

when ferric oxide(Fe2O3) 160 g then iron mass= 112

when ferric oxide(Fe2O3) is 1 g then iron mass= ( 112) /160

when ferric oxide(Fe2O3) is 60*1000*1000 g then iron mass= (112*60*1000*1000)/160

=5600000 g = 5600 quintol= 56 tonn

Practice Problem-2 Page-93

Answer of Practice Problem-2 Page-93

(1)

Now calculate weight of potassium chloride formed

( 2)

(i)

(ii)

Practice Problem-3 Page-93

Answer

(1)

Pure MgCO3 mass= 500 x 60/100

=320 gm

Mole of MgCO3 =Mass / MM

= 320/84

=3.81 mole

1 mole MgCO3 form = 1 mole CO2

3.81 mole = 3.81 mole CO2

hence

1 mole CO2 = 44 gm

3.81 mole CO2 =3.81 x 44

= 167.62 gm

(2)

Practice Problem-4 Page-94

Answer

(1)

( 2 )

B- Weight- Volume Relationship

( Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5)

Practice Problem-1 Page-94 , 95

Answer

(2)

C. Chemical Calculation Based on Mole Concept

( (Goyal Brothers Class-10 Mole Concept and Stoichiometry for ICSE Chemistry Ch-5 )

Practice Problem-1 Page -95

Answer

(2)

Objective Type Questions

(Goyal Brothers Class-10 Mole Concept and Stoichiometry )

I. Multiple Choice Questions

Choose the correct answer from the options given below :

1. ……………..
2. …………….
3. …………..
4. ………..
5. ……………………
6. …………
7. …………
8. ……………..
9. ………..
10. ………………………
11. ………………………….
12. …………………….
13. ………………….
14. ……………….
15. …………………..
16. ……………….
17. ………..
18. ……………….
19. ……………………….
20. …………………….

II Fill in the blanks space with the choices given in brackets:

1. …………
2. ………….
3. ……….
4. …………
5. ………..
6. ………….
7. …………
8. …………
9. …………………
10. …………………

III Choose from the following list, as what matches the descriptions given blew.

1. …………………
2. ……………
3. ………….
4. …………
5. ………
6. ……….
7. ………
8. ……….
9. ………
10. ……………..

–: End of Mole Concept and Stoichiometry Solutions Goyal Brothers :–

Return of :Chemistry Class-10GoyalBrothers Prakashan

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