Class 10 SELINA Solutions Chemistry Chapter 5 - Mole Concept And Stoichiometry

(a)2CO + O₂2CO₂

     2 V    1 V     2 V

2 V of CO requires = 1V of O₂

so, 100 litres of CO requires = 50 litres of O₂

(b) 2H₂ + O₂   2H₂O

       2 V         1V            2V

 From the equation, 2V of hydrogen reacts with 1V of oxygen

so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³

Hence, the unreacted oxygen is 150 - 100 = 50cm³ of oxygen.

This experiment supports Gay lussac's law of combining volumes. 

Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 - 58 = 48cc 

According to Gay lussac's law, the volumes of gases reacting should be in a simple ratio.

CH₄+2O₂CO₂ + 2H₂O

1 V2 V

24 cc48 cc

i.e. methane and oxygen react in a 1:2 ratio.

2C₂H₂+ 5O₂4CO₂ + 2H₂O (l)

2 V5 V4 V

From equation, 2 V of C₂H₂ requires = 5 V of O₂

So, for 400ml C₂H₂ , O₂ required = 400 5/2 =1000 ml

Similarly, 2 V of C₂H₂ gives = 4 V of CO₂

So, 400ml of C₂H₂ gives CO₂ = 400 4/2 = 800ml

Balanced chemical equation:

(i)At STP, 1 mole gas occupies 22.4 L.

As 1 mole H₂S gas produces 2 moles HCl gas,

22.4 L H₂S gas produces 22.4 × 2 = 44.8 L HCl gas.

Hence, 112 cm³ H₂S gas will produce 112 × 2 = 224 cm³ HCl gas.

(ii) 1 mole H₂S gas consumes 1 mole Cl₂ gas.

This means 22.4 L H₂S gas consumes 22.4 L Cl₂ gas at STP.

Hence, 112 cm³ H₂S gas consumes 112 cm³ Cl₂ gas.

120 cm³ - 112 cm³ = 8 cm³ Cl₂ gas remains unreacted.

Thus, the composition of the resulting mixture is 224 cm³HCl gas + 8 cm³ Cl₂ gas.

From the equation, 2V of ethane reacts with 7V oxygen.

So, 300 cc of ethane reacts with

Hence, unused O₂ = 1250 - 1050 = 200 cc

From 2V of ethane, 4V of CO₂ is produced.

So, 300 cc of ethane will produce

  C₂H₄+3O₂2CO₂ + 2H₂O

  1V      3V

11litre   33 litre

CH₄ + 2Cl₂ CH₂Cl₂ +2HCl

1 V2 V1 V2 V 

From equation, 1V of CH₄ gives = 2 V HCl

so, 40 ml of methane gives = 80 ml HCl

For 1V of methane= 2V of Cl₂ required

So, for 40ml of methane = 40 2 = 80 ml of Cl₂

C₃H₈ + 5O₂ 3CO₂ + 4H₂O

1 V5 V3 V

From equation, 5 V of O₂ required = 1V of propane

so, 100 cm³ of O₂will require = 20 cm³ ofpropane

2NO + O₂2NO₂

2 V1 V2 V

From equation, 1V of O₂ reacts with = 2 V of NO

200cm³ oxygen will react with = 200 2 =400 cm³ NO

Hence, remaining NO is 450 - 400 = 50 cm³

NO₂ produced = 400cm³ because 1V oxygen gives 2 V NO₂

Total mixture = 400 + 50 = 450 cm³

 i. 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.

 ii. When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leving behind only 2 litres of hydrogen gas.

 iii. Therefore, the volume of the residual gas will be 2 litres.

4NH₃ + 5O₂4NO + 6H₂O

4 V5 V4 V

9 litres of reactants gives 4 litres of NO

So, 27 litres of reactants will give = 27 4/9 = 12 litres of NO

H₂ + Cl₂ 2HCl

1V1V2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had composition: 16 cm³ hydrogen and 16 cm³ chlorine.

Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

CH₄ + 2O₂ CO₂ + 2H₂O

1 V2 V1 V

2C₂H₂ + 5O₂ 4CO₂ + 2H₂O

2 V5 V4 V

From the equations, we can see that

1V CH₄requires oxygen =2 V O₂

So, 10cm³ CH₄ will require =20 cm³ O₂

Similarly 2 V C₂H₂ requires = 5 V O₂

So, 10 cm³ C₂H₂ will require= 25 cm³ O₂

Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

C₃H₈ + 5O₂3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂ 8CO₂ + 10H₂O

60 ml of propane (C₃H₈) gives 3 60 = 180 ml CO₂

40 ml of butane (C₄H₁₀) gives = 8 40/2 = 160 ml of CO₂

Total carbon dioxide produced = 340 ml

So, when 10 litres of the mixture is burnt = 34 litres of CO₂ is produced.

2C₂H₂(g) + 5O₂(g)4CO₂(g)+ 2H₂O(g) 

4 V CO₂ is collected with 2 V C₂H₂

So, 200cm³ CO₂ will be collected with = 100cm³ C₂H₂ 

Similarly, 4V of CO₂ is produced by 5 V of O₂

So, 200cm³ CO² will be produced by = 250 ml of O₂

According to Avogadro's law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.

So,

(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.

(b) 1 litre of SO₂ contains the least number of molecules since it has the smallest volume.

(i) According to Avogadro's law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

As 150 cc of gas A contains X molecules, 150 cc of gas B also contains X molecules.

So, 75 cc of B will contain X/2 molecules.

(ii) The problem is based on Avogadro's law. 

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444

(b) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5

(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)5 x 18 = 249.5

(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132

(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82

(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5

(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

(a) No. of molecules in 73 g HCl = 6.023 x10²³ x 73/36.5(mol.

mass of HCl)

= 12.04 x 10²³

(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) x 0.5=16 g

(c) No. of molecules in 1.8 g H₂O = 6.023 x 10²³ x 1.8/18

= 6.023 x 10²²

(d) No. of moles in 10g of CaCO₃ = 10/100(mol. Mass CaCO₃)

= 0.1 mole

(e) Weight of 0.2 mole H₂ gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO₂ = 6.023 x 10²³x 3.2/64

= 3.023 x 10²²

Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28

So, the weight of 1 mole of CO₂ is more than the other three.

4g of NH₃ having minimum molecular mass contain maximum molecules.

a) No. of particles in s1 mole = 6.023 x 10²³

So, particles in 0.1 mole = 6.023 x 10 ²³ x 0.1 = 6.023 x 10²²

b)1 mole of H₂SO₄ contains =2 x 6.023 x 10²³

So, 0.1 mole of H₂SO₄ contains =2 x 6.023 x 10²³ x0.1

= 1.2x10²³ atoms of hydrogen

c)111g CaCl₂ contains = 6.023 x 10²³ molecules

So, 1000 g contains = 5.42 x 10²⁴molecules

(a) 1 mole of aluminium has mass = 27 g

So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g

(b) 0.1 mole of HCl has mass = 0.1 x 36.5(mass of 1 mole)

= 3.65 g

(c) 0.2 mole of H₂O has mass = 0.2 x 18 = 3.6 g

(d) 0.1 mole of CO₂ has mass = 0.1 x 44 = 4.4 g

(a) 5.6 litres of gas at STP has mass = 12 g

So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6

= 48g(molar mass)

(b)1 mole of SO₂ has volume = 22.4 litres

So, 2 moles will have = 22.4 x 2 = 44.8 litre

(a) 1 moleof CO₂ contains O₂ = 32g

So, CO₂ having 8 gm of O₂ has no. of moles = 8/32 = 0.25 moles

(b) 16 g of methane has no. of moles = 1

So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

(a) 6.023 x 10 ²³ atoms of oxygen has mass = 16 g

So, 1 atom has mass = 16/6.023 x 10²³ = 2.656 x 10^-23 g

(b) 1 atom of Hydrogen has mass = 1/6.023 x 10²³ = 1.666 x 10^-24

(c) 1 molecule of NH₃ has mass = 17/6.023 x10²³ = 2.82 x 10^-23 g

(d) 1 atom of silver has mass = 12 × 10²² /6.023 x 10²³ = 12/60 = 1/5 = 0.2 g

(e) 1 molecule of O₂ has mass = 32/6.023 x 10²³ = 5.314 x 10^-23 g

(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

(a) 0.1 mole of CaCO₃ has mass =100(molar mass) x 0.1=10 g

(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 x 0.1 = 32.2 g

(c) 0.1 mole of CaCl₂ has mass = 111 x 0.1 = 11.1g

(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

(a) 1molecule of Na₂CO₃.10H₂O contains oxygen atoms = 13

So, 6.023 x10²³ molecules (1mole) has atoms=13 x 6.023 x 10²³

So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10²³ =7.8x10²³

(b) Given Na = 4.6 gm

At. mass = 23

No. of gram atoms of Na=    Mass of Na   

                           = 4.6

                              23

                         = 0.2 gms

(c) 32 g of oxygen gas = 1 mole
1 gram of oxygen gas = 1/32 mole
Given that 12 g of oxygen gas
No: of moles = given mass / molar mass
= 12/32 = 0.375 mole          

3.2 g of S has number of atoms = 6.023 x10²³ x 3.2 /32

= 0.6023 x 10²³

So, 0.6023 x 10²³ atoms of Ca has mass=40 x0.6023x10²³/6.023

x10²³

= 4g

(a) No. of atoms = 52 x 6.023 x10²³ = 3.131 x 10²⁵

(b) 4 amu = 1 atom of He

so, 52 amu = 13 atoms of He

(c) 4 g of He has atoms =6.023 x10²³

So, 52 g will have = 6.023 x 10²³ x 52/4 = 7.828 x10²⁴ atoms

Molecular mass of Na₂CO₃ = 106 g

106 g has 2 x 6.023 x10²³ atoms of Na

So, 5.3g will have = 2 x 6.023 x10²³x 5.3/106=6.022 x10²² atoms

Number of atoms of C = 6.023 x10²³ x 5.3/106 = 3.01 x 10²² atoms

And atoms of O = 3 x 6.023 x 10²³ x 5.3/106= 9.03 x10²² atoms

(a) 60 g urea has mass of nitrogen(N₂) = 28 g

So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg

(b) 64 g has volume = 22.4 litre

So, 320 g will have volume = 22.4 x 320/64=112 litres

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.

(b) Vapour density of Chlorine atom is 35.5.

22400 cm³ of CO has mass = 28 g

So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

18 g of water has number of molecules = 6.023 x 10²³

So, 0.09 g of water will have no. of molecules = 6.023 x 10²³ x 0.09/18 = 3.01 x 10²¹molecules

(a) No. of moles in 256 g S₈ = 1 mole

So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 10²³ = 1.2 x 10²² molecules

No. of atoms in 1 molecule of S= 8

So, no. of atoms in 1.2 x 10²² molecules = 1.2 x 10²² x 8

= 9.635x 10²² molecules

Atomic mass of phosphorus P = 30.97 g

Hence, molar mass of P₄ = 123.88 g

If phosphorus is considered as P₄ molecules,

then 1 mole P₄ ≡ 123.88 g 

Therefore, 100 g of P₄ = 0.807 g

(a) 308 cm³ of chlorine weighs = 0.979 g

So, 22400 cm³ will weigh = gram molecular mass

= 0.979 x 22400/308 =71.2 g

(b) 2 g(molar mass) H₂ at 1 atm has volume = 22.4 litres

So, 4 g H₂ at 1 atm will have volume = 44.8 litres

Now, at 1 atm(P₁) 4 g H₂ has volume (V₁) = 44.8 litres

So, at 4 atm(P₂) the volume(V₂) will be =

(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)

So, mass of oxygen in 2.2 litres = 2.2 x 32/22.4=3.14 g

No. of atoms in 12 g C = 6.023 x10²³

So, no. of carbon atoms in 10^-12 g = 10^-12 x 6.023 x10²³/12

= 5.019 x 10¹⁰ atoms

Given:

P= 1140 mm Hg

Density = D = 2.4 g / L

T = 273 ⁰C = 273+273 =   546 K

M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L

Hence we have to find out the volume of the unknown gas at STP.

First, apply Charle’s law.

 We have to find out the volume of one litre of unknown gas at standard temperature 273 K.

V₁= 1 L  T₁ = 546 K

V₂=?       T₂ = 273 K

V₁/T₁ = V₂/ T₂

V₂ = (V₁ x T₂)/T₁

      = (1 L x 273 K)/546 K

      = 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.

 P ₁ = 1140 mm Hg  V₁ = 0.5 L

P₂ = 760 mm Hg  V₂ = ?

P₁ x V₁ = P₂ x V₂

V₂ = (P₁ x V₁)/P₂

      = (1140 mm Hg x 0.5 L)/760 mm Hg

      = 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP

X moles = 0.75 L / 22.4 L

                =  0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles    = 2.4 g / M

M = 2.4 g / 0.0335 moles

M= 71.6 g / mole

Hence, the gram molecular mass of the unknown gas is 71.6 g

1000 g of sugar costs = Rs. 40

So, 342g(molar mass) of sugar will cost=342x40/1000=Rs. 13.68

(a) CH 
(b) C₂H₆O 
(c) CH 
(d) CH₂O

(a) Molecular mass of Ca(H₂PO₄)₂ = 234

So, % of P = 2 31 100/234 = 26.5%

(b) Molecular mass of Ca₃(PO₄)₂ = 310

% of P = 2 31 100/310 = 20%

Molecular mass of KClO₃ = 122.5 g

% of K = 39 /122.5 = 31.8%

% of Cl = 35.5/122.5 = 28.98%

% of O = 3 16/122.5 = 39.18%

Element%At. massAtomic ratioSimple ratio

Pb62.52071

N8.5142

O29.0166

So, Pb(NO₃)₂is the empirical formula.

In Fe₂O₃ , Fe = 56 and O = 16

Molecular mass of Fe₂O₃ = 2 56 + 3 16 = 160 g

Iron present in 80% of Fe₂O₃ =

So, mass of iron in 100 g of ore = 56 g

mass of Fe in 10000 g of ore = 56 10000/100

= 5.6 kg

For acetylene , molecular mass = 2 V.D = 2 13 = 26 g

The empirical mass = 12(C) + 1(H) = 13 g

n =

Molecular formula of acetylene= 2 Empirical formula =C₂H₂

Similarly, for benzene molecular mass= 2 V.D = 2 39 = 78

n = 78/13=6

So, the molecular formula = C₆H₆

Element%at. massatomic ratiosimple ratio

C54.54122

H9.0914

O36.36161

(a) So, its empirical formula= C₂H₄O

(b) empirical formula mass = 44

Since, vapour density = 44

So, molecular mass = 2 V.D = 88

Or n = 2

so, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Element%at. massatomic ratiosimple ratio

C26.59121

H2.2211

O71.19162

(a) its empirical formula = CHO₂

(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D 2 = 90

so, molecular formula = C₂H₂O₄

Element%at. massatomic ratiosimple ratio

Cl71.6535.51

H4.0712

C24.28121

(a) its empirical formula = CH₂Cl

(b) empirical formula mass = 49.5

Since, molecular mass = 98.96

so, molecular formula = (CH₂Cl)₂ = C₂H₄Cl₂

(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1

(b) ElementGiven massAt. massGram atomRatio

C4.8120.412

H1112.55

So, the empirical formula = C₂H₅

(c) Empirical formula mass = 29

Molecular mass = V.D 2 = 29 2 = 58

So, molecular formula = C₄H₁₀

Since, g atom of Si = given mass/mol. Mass

so, given mass = 0.2 28 = 5.6 g

ElementmassAt. massGram atomRatio

Si5.6280.21

Cl21.335.53

Empirical formula = SiCl₃

% of carbon = 82.76%

% of hydrogen = 100 - 82.76 = 17.24%

Empirical formula = C₂H₅

Empirical formula weight = 2 x 12 + 1 x 5 = 24 + 5 = 29

Vapour Density = 29

Relative molecular mass = 29 x 2 = 58

N =  

Molecular formula = n x empirical formula

  = 2 x C₂H₅

 = C₄H₁₀

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg

(b) G atoms of nitrogen = 7/14= 0.5 or 1/2 g- atoms of N

(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3

So, the formula is Mg₃ N₂

Barium chloride = BaCl₂.x H₂O

Ba + 2Cl + x[H₂ + O]

=137+ 235.5 + x [2+16]

=[208 + 18x] containswater = 14.8% water in BaCl₂.x H₂O

=[208 + 18 x] 14.8/100 = 18x

=[104 + 9x] 2148=18000x

=[104+9x] 37=250x

=3848 + 333x =2250x

1917x =3848

x = 2molecules of water

Molar mass of urea; CON₂H₄ = 60 g

So, % of Nitrogen = 28 100/60 = 46.66%

Element%At. massAtomic ratioSimple ratio

C42.1123.51

H6.4816.482

O51.42163.21

The empirical formula is CH₂O

Since the compound has 12 atoms of carbon, so the formula is

C₁₂ H₂₄ O_12.

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A₂B₄.

(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D

Empirical formula weight = V.D/3

So, n = molecular mass/ Empirical formula weight = 6

Hence, the molecular formula is A₆B₆

(c) 

Given:

Wt. of the compound: 10.47g

Wt. of metal A: 6.25g

Wt. of non-metal B: 10.47 – 6.25 = 4.22g

Hence, the empirical formula is AB₄

Atomic ratio of N = 87.5/14 =6.25

Atomic ratio of H= 12.5/1 = 12.5

This gives us the simplest ratio as 1:2

So, the molecular formula is NH₂

Element%at. massatomic ratiosimple ratio

Zn22.65650.3481

H4.8814.8814

S11.15320.3481

O61.32163.8311

Empirical formula of the given compound =ZnSH₁₄O₁₁

Empiricala formula mass = 65.37+32+141+11+16=287.37

Molecular mass = 287

n= Molecular mass/Empirical formula mass = 287/287=1

Molecular formula = ZnSO₁₁H₁₄

=ZnSO₄.7H₂O

(a) 100 g of CaCO₃ produces = 164 g of Ca(NO₃)₂

So, 15 g CaCO₃ will produce = 164 15/100 = 24.6 g Ca(NO₃)₂

(b) 1 V of CaCO₃ produces 1 V of CO₂

100 g of CaCO₃ has volume = 22.4 litres

So, 15 g will have volume = 22.415/100 = 3.36 litres CO₂

2NH₃ + H₂SO₄ (NH₄)₂SO₄

66 g

(a) 2NH₃ + H₂SO₄ (NH₄)₂SO₄

34 g98 g132 g

For 132 g (NH₄)₂SO₄ = 34 g of NH₃ is required

So, for 66 g (NH₄)₂SO₄ = 66 32/132 = 17 g of NH₃ is required

(b) 17g of NH₃ requires volume = 22.4 litres

(c) Mass of acid required, for producing 132g (NH₄)₂SO₄ = 98g

So, Mass of acid required, for 66g (NH₄)₂SO₄ = 66 98/132 = 49g

(a) Molecular mass of Pb₃O₄ = 3 207.2 + 4 16 = 685 g

685 g of Pb₃O₄ gives = 834 g of PbCl₂

Hence, 6.85 g of Pb₃O₄ will give = 6.85 834/685 = 8.34 g

(b) 685g of Pb₃O₄ gives = 71g of Cl₂

Hence, 6.85 g of Pb₃O₄ will give = 6.85 71/685 = 0.71 g Cl₂

(c) 1 V Pb₃O₄produces 1 V Cl₂

685g of Pb₃O₄has volume = 22.4 litres = volume of Cl₂ produced

So, 6.85 Pb₃O₄ will produce = 6.85 22.4/685 = 0.224 litres of Cl₂

Molecular mass of KNO₃ = 101 g

63 g of HNO₃ is formed by = 101 g of KNO₃

So, 126000 g of HNO₃ is formed by = 126000 101/63 = 202 kg

Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃

So, smaller mass of NaNO₃ is required.

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

100g      73g                              2 L

(a)V₁ = 2 litres     

V₂= ?

T₁ = (273 + 27) = 300K

T₂ = 273K

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁ =

Now at STP 22.4 litres of CO₂ are produced using CaCO₃ = 100g

So,  litres are produced by = (2 × 273 × 100) / (300 × 22.4) = 8.125 g

(b)22.4 litres are CO₂ are prepared from acid =73 g

litres are prepared from = (2 × 273 × 73) / (300 × 22.4) =5.9 g 

2H₂O2H₂ + O₂

2 V2 V1 V

2 moles of H₂O gives = 1 mole of O₂

So, 1 mole of H₂O will give = 0.5 moles of O₂

so, mass of O₂ = no. of moles x molecular mass

= 0.5 32 = 16 g of O₂

and 1 mole of O₂ occupies volume =22.4 litre

so, 0.5 moles will occupy = 22.4 0.5 = 11.2 litres at S.T.P.

2Na₂O₂ + 2H₂O4NaOH + O₂

2 V4 V1 V

(a) Mol. Mass of Na₂O₂ = 2 23 + 2 16 = 78 g

Mass of 2Na₂O₂= 156 g

156 g Na₂O₂ gives = 160 g of NaOH (4 40 g)

So, 1.56 Na₂O₂ will give = 160 1.56/156 = 1.6 g

(b) 156 g Na₂O₂ gives = 22.4 litres of oxygen

So, 1.56 g will give = 22.4 1.56/156 = 0.224 litres

= 224 cm³

(c)156 g Na₂O₂ gives = 32 g O₂

So, 1.56 g Na₂O₂ will give = 32 1.56/156

= 32/100 = 0.32 g

2NH₄Cl + Ca(OH)₂CaCl₂+2H₂O + 2NH₃

2 V1 V1 V2 V

Mol. Mass of 2NH₄Cl = 2[14 + (1 4) + 35.5] = 2[53.5] = 107 g

(a) 107 g NH₄Cl gives = 34 g NH₃

So, 21.4 g NH₄Cl will give = 21.4 34/107 = 6.8 g NH₃

(b) The volume of 17 g NH₃ is 22.4 litre

So, volume of 6.8 g will be = 6.8 22.4/17 = 8.96 litre

MnO₂ + 4HCl MnCl₂ + 2H₂O +Cl₂

1 V4 V1 V1 V

(a) 1 mole of MnO₂ weighs = 87 g (mol. Mass)

So, 0.02 mole will weigh = 87 0.02 = 1.74 g MnO₂

(b) 1 mole MnO₂ gives = 1 mole of MnCl₂

So, 0.02 mole MnO₂will give =0.02 mole of MnCl₂

(c) 1 mole MnCl₂ weighs = 126 g(mol mass)

So, 0.02 mole MnCl₂ will weigh = 126 0.02 g = 2.52 g

(d) 0.02 mole MnO₂will form =0.02 mole of Cl₂

(e) 1 mole of Cl₂ weighs = 35.5 g

So, 0.02 mole will weigh = 71 0.02 = 1.42 g of Cl₂

(f) 1 mole of chlorine gas has volume = 22.4 litres

So, 0.02 mole will have volume = 22.4 0.02 = 0.448 litre

(g) 1 mole MnO₂requires HCl = 4 mole

So, 0.02 mole MnO₂ will require =4 0.02 = 0.08 mole

(h) For 1 mole MnO₂ , acid required = 4 mole of HCl

So, for 0.02 mole, acid required = 4 0.02 =0.08 mole

Mass of HCl = 0.08 x 36.5 = 2.92 g

N₂ + 3H₂ 2NH₃

28g6g34g

28g of nitrogen requires hydrogen = 6g

2000g of nitrogen requires hydrogen = 6/28 2000=3000/7g

So mass of hydrogen left unreacted =1000-3000/7=571.4g of H₂

(b)28g of nitrogen forms NH₃ = 34g

2000g of N₂ forms NH₃

= 34/28 2000

=2428.6g

(a) Weight of 1 g atom N = 14 g

So, weight of 2 g atom of N = 28 g

(b) 6.023 x10²³ atoms of C weigh = 12 g

So, 3 x10²⁵ atoms will weigh =

(c) 1 mole of sulphur weighs = 32 g

(d) 7 g of silver

 So, 7 grams of silver weighs least.

Option C is correct.

40 g of NaOH contains 6.023 x 10²³ molecules

So, 4 g of NaOH contains = 6.02 x10²³ x 4/40

= 6.02 x10²² molecules

The number of molecules in 18 g of ammonia= 6.02 x10²³

So, no. of molecules in 4.25 g of ammonia = 6.02 x10²³ x 4.25/18

= 1.5 x 10²³

Mass of gas X =10g

Mass of hydrogen gas= 2

Relative vapour density

 ===5

 Relative molecular mass of the gas= 2×relative vapour  

 density = 2×5

 =10

Correct option: (c) molecular mass

Solution:

The relative molecular mass of a gas or vapour is twice its vapour density.

2 × Rel. V. D.= Rel. molecular mass of a gas or vapour

Correct option: (a) C₃H₆O₃

Solution:

Molecular formula = n × Empirical formula

Where, n = an integer

In the given case, if n = 3, then

C₃H₆O₃ = 3 × CH₂O

Correct option: (b) 29

Solution:

2 × Rel. V. D.= Rel. molecular mass of a gas or vapour

2 × Rel. V. D.= 58

Rel. V. D. = 58 / 2

Rel. V. D. = 29

Correct option: (d) P₃Q₆

Solution:

Molecular formula = n × Empirical formula

Where, n = an integer

Also,

Molecular formula mass = n × Empirical formula mass

30 = n × 10

n = 3

Molecular formula = n × Empirical formula

Molecular formula = 3 × PQ₂

Molecular formula = P₃Q₆

Correct option: (c) 1 : 1

Solution:

Molecular mass of hydrogen is 1 g = 1 mole of hydrogen = 6.023 × 10²³ molecules of hydrogen

So, 2 g of hydrogen = 2 moles of hydrogen = 2 × 6.023 × 10²³ molecules of hydrogen

Molecular mass of oxygen is 16 g = 1 mole of oxygen = 6.023 × 10²³ molecules of oxygen

So, 32 g of oxygen = 2 moles of oxygen = 2 × 6.023 × 10²³ molecules of oxygen

Therefore,

Ratio of molecules in 2 g of hydrogen to molecules in 32 g of oxygen is 1:1. 

(a) Moles:1 mole + 2 mole 1 mole + 2 mole

(b) Grams: 42g + 36g74g+ 4 g

(c) Molecules = 6.02 10²³ + 12.046 10²³ 6.02 10²³+ 12.046 10²³

(a) One mole of chlorine contains 6.023 x 10²³ atoms of chlorine.

(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.

(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.

(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

According to Avogadros law:

Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.

So, 1 mole of each gas contains = 6.02 10²³ molecules

Mol. Mass of H₂ (2),O₂(32) ,CO₂(44),SO₂(64),Cl₂(71)

(1)Now 2 g of hydrogen contains molecules =6.02 10²³

So, 8g of hydrogen contains molecules = 8/2 6.02 10²³

=4 6.02 10²³ = 4M molecules

(2)32g of oxygen contains molecules = 8/32 6.02 10²³=M/4

(3)44g of carbon dioxide contains molecules = 8/44 6.02 10²³=2M/11

(4)64g of sulphur dioxide contains molecules =6.02 10²³

So, 8g of sulphur dioxide molecules = 8/64 6.02 10²³= M/8

(5)71 g of chlorine contains molecules =6.02 10²³

So, 8g of chlorine molecules = 8/72 6.02 10²³ = 8M/71

Since 8M/71<M/8<2M/11<M/4<4M

Thus Cl₂<SO₂<CO₂<O₂<H₂

(i)Least number of molecules in Cl₂

(ii)Most number of molecules in H₂

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm³.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

(d) The relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10²³ atoms.

(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.

(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

(a) Gay-Lussac's law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro's law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Now volume of hydrogen gas =volume of helium gas

n molecules of hydrogen =n molecules of helium gas

nH₂=nHe

1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium

Therefore 2H=He

Therefore atoms in hydrogen is double the atoms of helium.

(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.

The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Information conveyed by H₂O

(1) That H₂O contains 2 volumes of hydrogen and 1 volume of oxygen.

(2) That ratio by weight of hydrogen and oxygen is 1:8.

(3) That molecular weight of H₂O is 18g.

(a) stoichiometry measures quantitative relationships and is used to determine the amount of products/reactants that are produced/needed in a given reaction.

(b) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.

(c) N₂ means 1 molecule of nitrogen and 2N means two atoms of nitrogen.

N₂ can exist independently but 2N cannot exist independently.

(a) Applications of Avogadro's Law :

(1) It explains Gay-Lussac's law.

(2) It determines atomicity of the gases.

(3) It determines the molecular formula of a gas.

(4) It determines the relation between molecular mass and vapour density.

(5) It gives the relationship between gram molecular mass and gram molecular volume.

(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac's Law says.

H₂+Cl₂?2HCl

1V1V2V(By Gay-Lussacs law)

n moleculesnmolecules2n molecules(By Avogadros law)

a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.

b)The value of avogadro's number is 6.023 10²³

c) The molar volume of a gas at STP is 22.4 dm³ at STP

From equation: 2H₂+ O₂2H₂O

1 mole of Oxygen gives = 2 moles of steam

so, 0.5 mole oxygen will give = 2 0.5 = 1mole of steam

3Cu + 8HNO₃ 3Cu (NO₃)₂+ 4H₂O + 2NO

1 V8 V3 V2 V

Mol. Mass of 8HNO₃ = 8 63 = 504 g

(a) For 504 g HNO₃, Cu required is = 192 g

So, for 63g HNO₃Cu required = 192 63/504 = 24g

(b) 504 g of HNO₃ gives = 2 22.4 litre volume of NO

So, 63g of HNO₃ gives =2 22.4 63/504 =5.6 litre of NO

(a) 28g of nitrogen = 1mole

So, 7g of nitrogen = 1/28 7= 0.25 moless

(b) Volume of 71 g of Cl2 at STP =22.4 litres

Volume of 7.1 g chlorine =22.4 7.1/71=2.24 litre

(c) 22400cm³ volume have mass =28 g of CO(molar mass)

So, 56cm³ volume will have mass =28 56/22400= 0.07 g

% of N in NaNO₃=

% of N in (NH₄)₂SO₄ =

% of N in CO(NH₂)₂ =

So, highest percentage of N is in urea.

2H₂O2H₂+O₂

2 V2 V1 V

(a) From equation, 2 V of water gives 2 V of H₂ and 1 V of O₂

where 2 V = 2500 cm³

so, volume of O₂ liberated = 2V/V = 1250 cm³

(b)

P₁ = 1 Atm.

P₂ = 2.5 Atm.

V₁ = 2500

V₂ = ?

P₁ V₁ = P₂ V₂

1 x 2500 = 2.5 x V₂

V₂ = 2500 x 10/25

V₂ = 1000 cm³

(c)

(If volume of H₂ = 1000 cm³, temp. is T₁, V₂ = 2500 and New temp. T₂

V₁ / T₁ = V₂ / T₂

1000/T₁ = 2500/T₂

or

T₂/T₁= 2500/1000

= 2.5

T₂ = 2.5 T₁

It must be 2.5 times of original temperature.

Molecular mass of urea=12 + 16+2(14+2) =60g

60g of urea contains nitrogen =28g

So, in 50g of urea, nitrogen present =23.33 g

50 kg of urea contains nitrogen=23.33kg

% of hydrogen = 20%

% of carbon = 100 - 20 = 80%

Empirical formula = CH₃

Empirical formula weight = 1 x 12 + 1 x 3 = 12 + 3 = 15

Vapour Density = 15

Relative molecular mass = 15 x 2 = 30

N = 

Molecular formula = n x empirical formula

 = 2 x CH₃

 = C₂H₆

22400cm³ CO₂ has mass = 44g

so, 224 cm³ CO₂ will have mass= 0.44 g

Now since CO₂ is being formed and X is a hydrocarbon so it contains C and H.

In 0.44g CO₂, mass of carbon=0.44-0.32=0.12g=0.01g atom

So, mass of Hydrogen in X = 0.145-0.12 = 0.025g

= 0.025g atom

Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5

i.e. the formula of hydrocarbon is C₂H₅

(a) C and H

(b) Copper (II) oxide was used for reduction of the hydrocarbon.

(c)

(i) no. of moles of CO₂= 0.44/44 = 0.01 moles

(ii) mass of C = 0.12 g

(iii) mass of H = 0.025 g

(iv) The empirical formula of X = C₂H₅

Mass of X in the given compound =24g

Mass of oxygen in the given compound =64g

So total mass of the compound =24+64=88g

% of X in the compound = 24/88100 = 27.3%

% of oxygen in the compound=64/88 100 =72.7%

Element%At. MassAtomic ratioSimplest ratio

X27.31227.3/12=2.271

O72.71672.2/16=4.542

So simplest formula = XO₂

(a) V.D =

(b) Molecular mass = 17(V.D) x 2= 34g

(a) CO₂+C2CO

1 V1 V2 V

12 g of C gives =44.8 litre volume of CO

So, 3 g of C will give = 11.2 litre of CO

(b) 2CO + O₂2CO₂

2 V1 V2 V

(i) 2 V CO requires oxygen = 1 V

so, 24 cm³ CO will require = 24/2 =12 cm³

(ii) 2 x 22400 cm³ CO gives = 2 x 22400 cm³ CO₂

so, 24cm³ CO will give = 24 cm³ CO₂

Molecular weight of  =

 =328g

 Molecular weight of CaO =2(40+16)

 =112g

a. 328g of Ca(NO₃)₂ liberates 4 moles of NO₂

328g of Ca(NO₃)₂ liberates L of NO₂  82g will liberate 

 =22.4dm³ of NO₂

b. 328 g of calcium nitrate gives 112g of CaO

82 g will give 

 =28 g of CaO

2C₈H₁₈ + 25O₂16CO₂ + 18H₂O

2 V25 V16 V18 V

(i) 2 moles of octane gives = 16 moles of CO₂

so, 1 mole octane will give = 8 moles of CO₂

(ii) 1 mole CO₂ occupies volume = 22.4 litre

so, 8 moles will occupy volume = 8 22.4 = 179.2 litre

(iii) 1 mole CO₂ has mass = 44 g

so, 16 moles will have mass = 44 16 = 704 g

(iv) Empirical formula is C₄H₉.

The relative atomic mass of Cl = (35 3 + 1 37)/4=35.5 amu

Mass of silicon in the given compound =5.6g

Mass of the chlorine in the given compound=21.3g

Total mass of the compound=5.6g+21.3g=26.9g

% of silicon in the compound = 56/26.9 100 = 20.82%

% of chlorine in the compound = 21.2/26.9 100 = 79.18%

Element%At. MassAt. RatioSimplest ratio

Si20.822820.82/28=0.741

Cl79.18 35.579.18/35.5=2.233

So the empirical formula of the given compound =SiCl₃

% compositionAtomic ratioSimple ratio

P=38.27%38.27/31 =1.231

H= 2.47%2.47/1 = 2.472

O= 59.26%59.26/16 = 3.703

So, empirical formula is PH₂O₃ or H₂PO₃

Empirical formula mass = 31+ 2 1 + 3 16 = 81

The molecular formula is = H₄P₂O₆, because n = 162/81=2

a) V₁ = 10 litres V₂=?

T₁= 27+ 273 = 300KT₂=273K

P₁=700 mmP₂ = 760 mm

Using the gas equation

b)

(a) Molecular mass of CO₂ = 12+ 2x16 = 44 g

So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22

V.D =

So, mass of CO₂ = 22 kg

(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X

(a) The volume occupied by 1 mole of chlorine = 22.4 litre

(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.

(c) V₁/V₂ = T₁/T₂

22.4/V₂ =273/546

V₂ = 44.8 litres

(d) Mass of 1 mole Cl₂ gas =35.5 x 2 =71 g

(a) Total molar mass of hydrated CaSO₄.xH₂O = 136+18x

Since 21% is water of crystallization, so

So, x = 2 i.e. water of crystallization is 2.

(b) For 18 g water, vol. of hydrogen needed = 22.4 litre

So, for 1.8 g, vol. of H₂ needed= 1.8 x 22.4/18 = 2.24 litre

Now 2 vols. of water = 1 vol. of oxygen

1 vol. of water =1/2 vol. of O₂ =22.4/2=11.2 lit.

18 g of water = 11.2 lit. of O₂

1.8 g of water = 11.2/18 18/10=1.12 lit.

(c) 32g of dry oxygen at STP = 22400cc

2g will occupy = 224002/32=1400cc

P₁=760mmP₂ =740mm

V₁=1400ccV₂ =?

T₁ =273 K, T₂ = 27 +73 = 300K

(d) P₁= 750mmP₂=760mm

V₁= 44lit.V₂=?

T₁= 298KT₂=273K

(e) Since 143.5g of AgCl is produced from =58.5 g of NaCl

so, 1.435 g of AgCl is formed by =0.585 g of NaCl

% of NaCl =0.585 x100 = 58.5%

a.
 

 For CO₂ 12+32 

 i. Molecular mass of sulphuric acid = 2(2+32+64)

  = 196

   196 g of suphuric acid oxidized 12g of Carbon

   49 g of suphuric acid will  

  =3 g

 ii. 196 g of sulphuric acid gives 2(22.4)

 =44.8L

  49 g og sulphuric acid will give  

  =11.2 L of SO₂

b. 

 i.  

Empirical formula = CHCl₂

ii. 

Empirical formula = CHCl₂

Empirical formula weight = 1 x 12 + 1 x 1+(2 x35.5)

= 12 + 1+70

= 83 

Relative molecular mass = 168

N =  2.02≈2

Molecular formula = n x empirical formula

= 2 x CHCl₂

= C₂H₂Cl₄

a. Relative molecular mass of [Mg (NO₃) 6H₂O]

 =24+14+(3 x16)+(6 x18)=194

   Since, 194g of [Mg (NO₃) 6H₂O] contains 144g of oxygen

 100g of [Mg (NO₃) 6H₂O] contains  of oxygen = 74.22%  

b. Relative molecular mass of Na₂B₄O₇.10H₂O

 (23 × 2) + (4 × 11) + (7 × 16) + 10(18) = 382

 Since 382g of Na₂B₄O₇.10H₂O contains 44g of boron

 100g Na₂B₄O₇.10H₂O of contains of boron

   =11.5%

c. Relative molecular mass of Ca(H₂PO₄)₂

  = 40 + 2(2 + 31 + 64) = 234

 Since,234g of Ca(H₂PO₄)₂ contains 62g of phosphorus 

  100g of Ca(H₂PO₄)₂ contains

 =26.5%

The given equation is,

2NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂

Molecular weight of NaOH, sodium hydroxide = 23 + 16 + 1 = 40

Molecular weight of Cu(OH)₂ ,

Copper hydroxide = 64 + 16 + 1 + 16 + 1 = 98

Now, 40 g of NaOH is used to precipitate 98 g of Cu(OH)₂

Hence, 200 g of NaOH will be used to precipitate (98/40)×200 g of Cu(OH)₂ = 490 g of Cu(OH)₂

So, 490 g of copper hydroxide would be prepared using 200 g of sodium hydroxide. 

Solid ammonium dichromate decomposes as:

(a)Molecular mass of ammonium dichromate

 = 2(14+4)+104+112

 = 252 g

 Number of moles=

 =

 =0.25moles

(b)

 252 g of ammonium dichromate gives 22.4 dm³ of N₂

 63 g of ammonium dichromate gives 

  =5.6 L

 = 0.25 moles  

(c)

 252 g of ammonium dichromate gives 22.4 dm³ of N₂

 63 g of ammonium dichromate gives 

 = 5.6 L

(d)

 Number of moles=

 =

 =0.25moles

 0.25 moles of ammonium dichromate gives

 0.25 moles of N₂=7 g

 1 mole of H₂O =18 g

Therefore, total loss of mass=7+18

 =25 g

(e)

 252 g of ammonium dichromate gives 152 g of CrO₃

 63 g of ammonium dichromate gives 

  =38 g

2H₂S + 3O₂2H₂O + 2SO₂

2 V3 V2 V

128 g of SO₂gives = 2 22.4 litres volume

So, 12.8 g of SO₂ gives = 2 22.4 12.8/128

= 4.48 litre volume

Or one can say 4.48 litres of hydrogen sulphide.

2 22.4 litre H₂S requires oxygen = 3 22.4 litre

So, 4.48 litres H₂S will require = 6.72 litre of oxygen

From equation, 2NH₃ + 2½ 2 O₂2NO + 3H₂O

When 60 g NO is formed, mass of steam produced = 54 g

So, 1.5 g NO is formed, mass of steam produced = 54 1.5/60

=1.35 g

In 1 hectare of soil, N₂ removed = 20 kg

So, in 10 hectare N₂ removed = 200 kg

The molecular mass of Ca(NO₃)₂ =164

Now, 28 g N₂ present in fertilizer = 164 g Ca(NO₃)₂

So, 200000 g of N₂ is present in = 164 200000/28

= 1171.42 kg

(a) 1 mole of phosphorus atom = 31 g of phosphorus

31 g of P =1 mole of P

6.2g of P = =0.2 mole of P

(b) 31 g P reacts with HNO₃ = 315 g

so, 6.2 g P will react with HNO₃ = 315 6.2/31 = 63 g

(c)

Moles of steam formed from 31g phosphorus = 18g/18g = 1mol

Moles of steam formed from 6.2 g phosphorus = 1mol/31g6.2=0.2 mol

Volume of steam produced at STP =0.2 22.4 l/MOL=4.48 litre

Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled

So,Volume of steam produced at 760mm Hg and 273⁰C=4.48 2=8.96litre

112cm³ of gaseous fluoride has mass = 0.63 g

so, 22400cm³ will have mass = 0.63 22400/112

= 126 g

The molecular mass = At mass P + At. mass of F

126= 31 + At. Mass of F

So, At. Mass of F = 95 g

But, at. mass of F = 19 so 95/19 = 5

Hence, there are 5 atoms of F so the molecular formula = PF₅

Na₂CO₃.10H₂O Na₂CO₃ + 10H₂O

286 g106 g

So, for 57.2 g Na₂CO₃.10H₂O = 106 57.2/286 = 21.2 g Na₂CO₃

Simple ratio of M = 34.5/56 = 0.616 = 1

Simple ratio of Cl = 65.5/35.5 = 1.845 = 3

Empirical formula = MCl₃

Empirical formula mass = 162.5, Molecular mass = 2 V.D = 325

So, n = 2

So, molecular formula = M₂Cl₆

(i) Element%atomic massatomic ratiosimple ratio

C4.8121

Br95.2803

So, empirical formula is CBr₃

(ii) Empirical formula mass = 12 + 3 80 = 252 g

molecular formula mass = 2 252(V.D) = 504 g

n= 504/252 = 2

so, molecular formula = C₂Br₆

4N₂O + CH₄ CO₂ + 2H₂O + 4N₂

4 V1 V1 V2 V4 V

2 x 22400 litre steam is produced by N₂O = 4 x 22400 cm³

So, 150 cm³ steam will be produced by= 4 22400 150/2 x 22400

= 300 cm³N₂O

(a) Volume of O₂ = V

Since O₂ and N₂ have same no. of molecules = x

so, the volume of N₂ = V

(b) 3x molecules means 3V volume of CO

(c) 32 g oxygen is contained in = 44 g of CO₂

So, 8 g oxygen is contained in = 44 x 8/32 = 11 g

(d) Avogadro's law is used in the above questions.

simple ratio of Na = 42.1/23 = 1.83 = 3

simple ratio of P = 18.9/31 = 0.609 = 1

simple ratio of O = 39/16 = 2.43 = 4

So, the empirical formula is Na₃PO₄

CH₄ + 2O₂ CO₂ + 2H₂O

1 V2 V1 V2 V

From equation:

22.4 litres of methane requires oxygen = 44.8 litres O₂

2H₂ + O₂ 2H₂O

2 V1 V2 V

From equation,

44.8 litres hydrogen requires oxygen = 22.4 litres O₂

So, 11.2 litres will require = 22.4 x 11.2/44.8 = 5.6 litres

Total volume = 44.8 + 5.6 = 50.4 litres

Gram atoms of Pb = 6.21/207=0.03 = 1

Gram atoms of Cl = 4.26/35.5 = 0.12 = 4

So, the empirical formula = PbCl₄

Na₂SO₄+BaCl₂BaSO₄+2NaCl

Molecular mass of BaSO₄ = 233 g

Now, 233 g of BaSO₄ is produced by Na₂SO₄ = 142 g

So, 6.99 g BaSO₄ will be produced by = 6.99 142/233 = 4.26

The percentage of Na₂SO₄ in original mixture = 4.26 100/10

= 42.6%

(a) 1 litre of oxygen has mass = 1.32 g

So, 24 litres (molar vol. at room temp.) will have mass = 1.32 x 24

= 31.6 or 32 g

(b) 2KMnO₄K₂MnO₄ + MnO₂ + O₂

316 g of KMnO₄ gives oxygen = 24 litres

So, 15.8 g of KMnO₄ will give = 24 316/15.8 = 1.2 litres

(a)

(i) The no. of moles of SO₂ = 3.2/64 = 0.05 moles

(ii) In 1 mole of SO₂, no. of molecules present = 6.02 10²³

So, in 0.05 moles, no. of molecules = 6.0210²³ 0.05

= 3.0 10²²

(iii) The volume occupied by 64 g of SO₂ = 22.4 dm³

3.2 g of SO₂ will be occupied by volume= 22.4 3.2/64 =1.12 dm³

(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.

V₁/V₂ = n₁/n₂

V₁/V₂ = n₁/2n₁

So, V₂ = 2V₁

(iii) Gay lussac's law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm³, so the number of molecules = 6 x 10²³ because according to mole concept 22.4 litre volume at STP has = 6 x 10 ²³ molecules

(v) No. of moles of D = 1 because volume is 22.4 litre

so, mass of N₂O = 1 44 = 44 g

(a) NaCl+NH₃+ CO₂ + H₂ONaHCO₃+NH₄Cl

2NaHCO₃ Na₂CO₃+H₂O + CO₂

From equation:

106 g of Na₂CO₃ is produced by = 168 g of NaHCO₃

So, 21.2 g of Na₂CO₃ will be produced by = 168 21.2/106

= 33.6 g of NaHCO₃

(b) For 84 g of NaHCO₃, requiredvolume of CO₂ = 22.4 litre

So, for 33.6 g of NaHCO₃, required volume of CO₂ = 22.4 x 33.6/84

= 8.96 litre

(a) NH₄NO₃N₂O+2H₂O

1mole1mole2mole

1 V1 V2 V

44.8 litres of water produced by = 22.4 litres of NH₄NO₃

So, 8.96 litres will be produced by = 22.4 x 8.96/44.8

= 4.48 litres of NH₄NO₃

So, 4.48 litres of N₂O is produced.

(i) 44.8 litre H₂O is produced by = 80 g of NH₄NO₃

So, 8.96 litre H₂O will be produced by = 80 x 8.96/44.8

= 16g NH₄NO₃

(iii) % of O in NH₄NO₃ = 3x16/80 = 60%

Molecular mass of =(240) g

Molecular mass of  = 2 × 22.4 = 44.8dm³

Molecular mass of  = (192)g

240 g of CuO requires 44.8 dm³ of NH₃

120g of CuO will require  

 =22.4dm³

(a) The molecular mass of ethylene(C₂H₄) is 28 g

No. of moles = 1.4/28 = 0.05 moles

No. of molecules = 6.023 x10²³ x 0.05 = 3 x 10²² molecules

Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D

S0, V.D = 28/2 = 14

(a) Molecular mass of Na₃AlF₆ = 210

So, Percentage of Na = 3x23x100/210 = 32.85%

(b) 2CO + O₂2CO₂

2 V1 V2 V

1 mole of O₂ has volume = 22400 ml

Volume of oxygen used by 2 x 22400 ml CO = 22400 ml

So, Vol. of O₂ used by 560 ml CO =22400 x 560/(2 x 22400)

= 280 ml

So, Volume of CO₂ formed is 560 ml.

(a) 

 i. The combustion reaction

 According to Gay-Lussac's law,

 2 volume of acetylene requires 5 volume of oxygen to burn it

 1 volume of acetylene requires 2.5 volume of oxygen to burn it

 200cm³ requires 2.5×200=500 cm³ of oxygen

  2 volume of acetylene on combustion gives 4CO₂ 

  1 volume of acetylene on combustion gives 2CO₂ 

 200cc of acetylene on combustion will give  200×2=400cc of CO₂

 (b) Hydrogen = 12.5%

∴ Nitrogen= 100-12.5= 87.5%

 The Empirical formula of the compound is NH₂

 Empirical formula weight =14+2=16

 Relative molecular mass =37

 N = 2.3≈2

 Molecular formula = n x empirical formula  = 2 x NH₂

  =N₂H₄

(c)  

 Molecules of nitrogen gas in a cylinder = 24 x 10²⁴

 Avogadro's number = 6 x 10²³

(i) Mass of nitrogen in a cylinder  =

 =1120g

(ii) Volume of nitrogen at stp

Volume of 28 g of N₂  = 22.4dm³ 

Volume of 1120g of N₂ =  dm³

   =896 dm³

(d) NaCl + AgNO₃ ⟶ AgCl + NaNO₃

As, 143 g of AgCl is obtained from = 58 g of NaCl 

So, 14.3 g of AgCl will be obtained from = 58 × 14.3 / 143 = 5.8 g of NaCl 

Weight of commercial NaOH  = 30 g 

Percentage of NaCl in NaOH = 5.8 × 100 / 30 = 19.33 % 

(e) At STP, 100 cm³ of gas weighs = 0.5 g

So, at STP 22400 cm³ of gas will weigh = 0.5 x 22400 / 100 = 112 g

(a) 

 (i) C₃H_8(g) + 5O_2(g) ⟶ 3CO_2(g) + 4H₂O_(g)

2C₄H_10(g) +  13O_2(g) + 8CO_2(g) + 10H₂O_(g)

60 ml of propane (C₃H₈) gives 3 × 60 = 180 ml CO₂

40 ml of butane (C₄H₁₀) gives = 8 × 40 / 2 = 160 ml of CO₂

Total carbon dioxide produced = 340 ml

So, when 10 litres of the mixture is burnt = 34 litres of CO2 is produced.

(ii) Molecular mass of NH₄(NO₃) =80

 H=1, N=14, O=16

 % of Nitrogen

 As 80 g of NH4(NO3) contains 28 g of nitrogen

  100 g of of NH₄(NO₃) will contain  

 = 35%

 % of Oxygen

 As,80 g of NH4(NO3) contains 48 g of oxygen

  100 g of of NH₄(NO₃) will contain  

 = 60%

(b) 

 (i) Equation for reaction of calcium carbonate with dilute hydrochloric acid:

 
(ii) Relative molecular mass of calcium carbonate=100

 Mass of 4.5 moles of calcium carbonate

 = No. of moles× Relative molecular mass 

 = 4.5×100

 = 450g

(iii) 

As, 100g of calcium carbonate gives 22.4dm³ of CO₂

 450 g of calcium carbonate will give

 =100.8 L

 
(iv) Molecular mass of calcium carbonate =100

 Relative molecular mass of calcium chloride =111

 As 100 g of calcium carbonate gives 111g of calcium chloride

  450 g of calcium carbonate will give  

 =499.5 g 

 
(v) Molecular mass of HCl=36.5

 Molecular mass of calcium carbonate =100

 As 100 g of calcium carbonate gives (2×36.5)= 73g of HCl

 450 g of calcium carbonate will give

  =328.5g

Number of moles of HCl= 

 =

 = 9 moles

(a) 

 (i) Atomic mass: S = 32 and O = 16

 Molecular mass of SO₂=32+(2×16)

  =64g

 As 64 g of SO₂ = 22.4dm³

 Then, 320 g of SO₂ = 

 =112 L

(ii) Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."

(iii) C₃H₈ + 5O₂→3CO₂ + 4H₂O

 Molar mass of propane = 44

 44 g of propane requires 5 × 22.4 litres of oxygen at STP.

8.8 g of propane requires  = 22.4 litres

(b)  

 (i)  

 Empirical formula = CH₂Br 

n(Empirical formula mass of CH₂Br) = Molecular mass (2 × VD)

n(12 + 2 + 80) = 94 × 2

  n = 2

 Molecular formula = Empirical formula × 2

 = (CH₂Br) × 2

  = C₂H₄Br₂

 
(ii) 
1.  10²² atoms of sulphur

     6.022 × 10²³ atoms of sulphur will have mass = 32 g

     10²² atoms of sulphur will have mass =  

     = 0.533 g

2.  0.1 mole of carbon dioxide

     1 mole of carbon dioxide will have mass = 44 g

     0.1 mole of carbon dioxide will have mass = 4.4 g

(a)  

(i) Number of moles of phosphorus taken =  

   = 0.3 mol

(ii) 1 mole of phosphorus gives 98 gm of phosphoric  acid.

 So, 0.3 mole of phosphorus gives (0.3 × 98) gm of  phosphoric acid

 = 29.4 gm of phosphoric acid

 (iii) 1 mole of phosphorus gives 112 L of NOgas at STP.

  So, 0.3 mole of phosphorus gives (112 × 0.3) L of

 NOgas at STP.

 = 33.6 L of NOgas at STP

(b)  

 (i) According to the equation

3 volumes of hydrogen produce 2 volumes of ammonia

67.2 litres of hydrogen produce  = 44.8 L

3 volumes of hydrogen combine with 1 volume of ammonia.

67.2 litres of hydrogen combine with  =22.4L Nitrogen left = 44.8 - 22.4 = 22.4 litres

(ii) 5.6 dm³ of gas weighs 12 g

 1 dm³ of gas weighs = (12/56) gm

 22.4 dm³ of gas weighs = (12/56 × 22.2) gm = 48g

 Therefore, the relative molecular mass of gas = 48 gm.

 
(iii) Molar mass of Mg (NO₃)₂.6H₂O

 = 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g

 Mass percent of magnesium =  

(a)  

 (i) 

  2 vols. of butane requires O₂ = 13 vols

 90 dm³ of butane will require O₂ = × 90

  = 585 dm³

 
(ii) Molecular mass = 2 × Vapour density

 So, molecular mass of gas = 2 × 8 = 16 g

 As we know, molecular mass or molar mass occupies 22.4 litres.

 That is,

 16 g of gas occupies volume = 22.4 litres

 So, 24 g of gas will occupy volume

 =

(iii) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

So, molecules of nitrogen gas present in the same vessel = X

(b)  

 (i)  

  3 vols. of oxygen require KClO₃ = 2 vols. 

 So, 1 vol. of oxygen will require KClO₃ =

 So, 6.72 litres of oxygen will require KClO₃

 =

 22.4 litres of KClO₃ has mass = 122.5 g

 So, 4.48 litres of KClO₃ will have mass

 = 

(ii) 22.4 litres of oxygen = 1 mole

 So, 6.72 litres of oxygen =

 No. of molecules present in 1 mole of O₂

 = 6.023 × 10²³

 So, no. of molecules present in 0.3 mole of O₂

 = 6.023 × 10²³ × 0.3

  = 1.806 × 10²³

 
(iii) Volume occupied by 1 mole of CO₂ at STP = 22.4  litres 

So, volume occupied by 0.01 mole of CO₂ at STP = 22.4 × 0.01= 0.224 litres

(a)  

 (i) 

 2 moles of C₂H₂=4moles of CO₂

 x dm³ of C₂H₂  =8.4 dm³ of CO₂

 x=

 =4.2 dm³ of C₂H₂  

 
(ii) Empirical formula= X₂Y

  Atomic weight (X)= 10

  Atomic weight (Y)= 5 

  Empirical formula weight = (2 × 10) + 5

 =25

  = 2

 So, molecular formula = X₂Y×2

 = X₄Y₂    

(b)  

 (i) A cylinder contains 68 g of ammonia gas at STP.

  Molecular weight of ammonia = 17 g/mole

  68 g of ammonia gas at STP =?

  1 mole = 22.4 dm³

 ∴ 4 mole = 22.4 × 4 = 89.6 dm³ 

 
(ii) 4 moles of ammonia gas is present in the cylinder.

 
(iii) 1 mole = 6.023 × 10²³ molecules

      4 moles = 24.092 × 10²³ molecules