Most expected three mark questions in Basic Concepts of Chemistry and Chemical Calculations - III

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Questions:

Generic 3

52 x 3 = 156

Balance the following reaction:
Sb³⁺+ Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) Sb⁵⁺+ Mn²⁺

Balance the following reaction
S₂O₃^2- + I₂\(\rightarrow \) S₂O₄^2- + I^-

Balance the following reaction:
C₂O₄²-+ Cr₂O₇^2-\(\rightarrow \)Cr³⁺+ CO₂

0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.

Balance the following reaction:
S₂O₃^2-+ I²\(\rightarrow\)S₂O₆^2-+ I^-

Balance the following reaction:
MnO₄^- + Fe²⁺ \(\rightarrow\) Mn²⁺ + Fe³⁺

Balance the following reaction:
MnO₄^2-\(\rightarrow\) MnO₄^2- + MnO₂

Balance the following reaction:
MnO^-₄ + Sn²⁺\(\rightarrow\) Mn²⁺ + Sn⁴⁺

Balance the following reaction:
Zn + HNO₃ \(\rightarrow\) Zn (NO₃)₂ + NH₄NO₃ +H₂O

10.

Balance the following reaction:
CuO+ NH₃ \(\rightarrow\)Cu + N₂+ H₂O

11.

Balance the following reaction:
H₂C₂O₄+KMnO₄ + H₂SO₄ \(\rightarrow\) H₂SO₄ + MnSO₄ + CO₂ + H₂O

12.

Balance the following reaction:
P+5HNO₃ \(\rightarrow\) H₃PO₄ + 5NO₂ + H₂O

13.

Balance the following reaction:
P+ HNO₃\(\rightarrow\) H₃PO₄ + NO₂ + H₂O

14.

What will be oxidation number of sulphur in\({ S }_{ 2 }{ O }_{ 8 }^{ 2- }\) ions and \({ S }_{ 2 }{ O }_{ 6 }^{ 2- }\) ion?

15.

Balance the following reaction:
Cu + HNO₃\(\rightarrow\)Cu(NO₃)₂ + NO₂ + H₂O

16.

Balance the following redox reaction:
K₂Cr₂O₇ + KCI + H₂SO₄ \(\rightarrow\) KHSO₄ + CrO₂CI₂ + H₂O

17.

Balance the following reaction:
KMnO₄ + Na₂SO₃\(\rightarrow\)MnO₂ + Na₂SO₄ + KOH (Alkaline medium)

18.

Balance the following reaction:
K₂Cr₂O₇ + KI + H₂SO₄ \(\rightarrow\) K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O

19.

Identify the type of redox reaction taking place in the following
\(\overset { 0 }{ { Cl }_{ 2(g) } } +{ 2OH^{ - }_{ (aq) } }\longrightarrow \overset { -1 }{ Cl } { O }_{ (aq) }^{ - }+\overset { -1 }{ { Cl }^{ - } } _{ (aq) }+{ H }_{ 2 }{ O }_{ (l) }\)

20.

Identify the type of redox reaction taking place in the following
\({ Br }_{ 2(1) }+2I_{ (aq) }\longrightarrow 2{ Br }_{ (aq) }^{ - }+{ I }_{ 2(s) }\)

21.

Identify the type of redox reaction taking place in the following
\(\overset { 0 }{ { Ca }_{ (s) } } +\overset { +1 }{ 2 } \overset { -2 }{ { H }_{ 2 }O_{ (l) } } \longrightarrow \overset { +2 }{ Ca } \overset { -2+1 }{ { (0H) }_{ 2(aq) } } +\overset { 0 }{ H } _{ 2(g) }\)

22.

Identify the type of redox reaction taking place in the following
\(\overset { +1 }{ 2K } \overset { +5 }{ Cl } \overset { -2 }{ O_{ 3(s) } } \longrightarrow 2\overset { +1 }{ K } \overset { -1 }{ { Cl }_{ (s) } } +3\overset { 0 }{ { O }_{ 2 }(g) } \)

23.

Identify the type of redox reaction taking place in the following
\(\overset { +5 }{ { Y }_{ 2 } } \overset { -2 }{ { O }_{ 5(s) } } +5\overset { 0 }{ { Ca }_{ (s) } } \longrightarrow \overset { 0 }{ 2 } { V }_{ (s) }+\overset { +2 }{ 5 } \overset { -2 }{ { CaO }_{ (s) } } \)

24.

Identify the type of redox reaction taking place in the following
\(3{ M }^{ 0 }g_{ (s) }+{ N }_{ 2(g) }^{ 0 }\longrightarrow \overset { +2 }{ { Mg }_{ 3 } } \overset { -3 }{ N_{ 2(s) } } \)

25.

Determine the empirical formula of a compound containing K = 24.75%, Mn = 34.77% and rest is oxygen

26.

A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula.

27.

Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eyewash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample ?

28.

Calculate the equivalent mass of barium hydroxide

29.

Decide whether each of the following reaction involves oxidation reduction reaction or not. If it does , identify which species is oxidised and which gets oxidised ?
\({ CH }_{ 3 }-\overset { \underset { || }{ O } }{ C } -OH+{ CH }_{ 3 }{ NH }_{ 2 }\rightleftharpoons { CH }_{ 3 }-\overset { \underset { || }{ O } }{ C } -O^{ - }+{ CH }_{ 3 }{ NH }_{ 3 }^{ + }\)

30.

Decide whether each of the following reaction involves oxidation reduction reaction or not. If it does ,identify which species is oxidised and which gets oxidised?
\({ CH }_{ 3 }{ CH }_{ 2 }OH\overset { { H }_{ 2 }{ SO }_{ 4 } }{ \longrightarrow } { CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }O\)

31.

Decide whether each of the following reaction involves oxidation reduction reaction or not. If it does,identify which species is oxidised and which gets oxidised ?
\({ 4CH }_{ 3 }-\overset { \overset { O }{ || } }{ C } -{ CH }_{ 3 }+LiAl{ H }_{ 4 }={ 4H }_{ 2 }O\rightleftharpoons { 4CH }_{ 3 }-\overset { \overset { OH }{ || } }{ CH } -{ CH }_{ 3 }+LiOH+Al({ OH })_{ 3 }\)

32.

0.456 g of a metal gives 0.606g of its chloride Calculate its equivalent mass

33.

Calculate the equivalent mass of the following : Sodium

34.

Calculate the equivalent mass of the following : nitrate ion (NO₃^-)

35.

Calculate the equivalent mass of the following : Zn

36.

Calculate the molar volume of the following 3.0115 \(\times\)10²³ molecules of SO₂ gas

37.

Calculate the molar volume of the following 460g of formic acid

38.

Calculate the molar volume of the following 5 moles of methane

39.

Calculate the molar volume of the following 88 g of CO₂

40.

Calculate the equivalent mass of bicarbonate ion.

41.

Calculate the weight of 0.2 mole of sodium carbonate.

42.

How many moles of glucose are present in 720 g of glucose

43.

The density of CO₂ = 1.977 kgm-3 at STP. Calculate I the molecular mass of CO₂

44.

Calculate the number of moles present in the following 19.5g of potassium

45.

Calculate the number of moles present in the following 90 g of magnesium oxide

46.

Calculate the number of moles present in the following 46g of ethanol

47.

Calculate the number of moles present in the following 120g of sodium hydroxide

48.

Calculate the number of moles present in the following 50g of calcium chloride

49.

Find the molecular mass of FeSO₄7H₂O.

50.

Calculate number of moles of carbon atoms in three .moles of ethane.

51.

The approximate production of Na₂C0₃ per month is 424 x 106g while that of methyl alcohol is 320 x 10₆g. Which is produced more in terms of moles?

52.

Calculate the number of atoms/molecules present in the following 1Kg of acetic acid

Generic 5

2 x 5 = 10

53.

The balanced equation for a reaction is given below 2x + 3y \(\rightarrow\) 41 + m
When 8 moles of x react with 15 moles of y, then
i) Which is the limiting reagent ?
ii) Calculate the amount of products formed.
iii) Calculate the amount of excess reactant left at the end of the reaction

54.

Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data.
x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2, 1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.
i) The atomic masses of the element x,y,z.
ii) Empirical formula of the compound and
iii) Molecular formula of the compound.

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Answers Key:

Generic 3

52 x 3 = 156

Equalise the increase/decrease in Oxidation number by multiplying with suitable numbers.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) Sb⁵⁺+ Mn²⁺
Balance all other atoms except O and H
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺
Balance Oxygen atom by adding H₂O on the side falling short of oxygen.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺+8H₂O
Balance hydrogen atom by adding H+ on the side falling short of hydrogen atoms.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺+8H₂O + 16H⁺

S₂O₃^2- + I₂\(\rightarrow \) S₂O₄^2- + I^-
S₂O₃^2- + I₂\(\rightarrow \) S₂O₄^2- + 2I^-
S₂O₃^2- + I₂+ H₂O\(\rightarrow \) S₂O₄^2- + 2I^-
S₂O₃^2- + I₂+ H₂O\(\rightarrow \) S₂O₄^2- + 2I^-+ 2H⁺

3C₂O₄²-+ Cr₂O₇^2-\(\rightarrow \)2Cr³⁺+ 6CO₂+ 7H₂O
3C₂O₄^2-+ Cr₂O₇^2-+ 14H⁺\(\rightarrow \)2Cr³⁺+ 6CO₂+ 7H₂O

Mass of the metal = 0.456 g
Mass of the metal chloride = 0.606 g
0.456 g of the metal combines with 0.15 g of chlorine.
Mass of the metal that combines with 35.5 g of chlorine is \(\frac{0.456}{0.15}\) x 35.5
= 107.92 g eq^-1.

S₂O₃^2-+I²\(\rightarrow\)S₂O₆^2-+I^-

Equalise the increase/decrease in O.N by multiplying the S species by 1 and I species by 3.
S₂O₃^2-+I²\(\rightarrow\)S₂O₆^2-+3I^-
Balance all other atoms except 0 and H
S₂O₃^2-+3I²\(\rightarrow\)S₂O₆^2-+6I^-
Balance 0 atom by adding H₂0 on the side falling short of Oxygen.
S₂O₃^2-+3I²+3H₂O\(\rightarrow\)S₂O₆^2-+6I^-
Balance H atom by adding H+ ion on the side falling short of hydrogen.
S₂O₃^2-+3I²+3H₂O\(\rightarrow\)S₂O₆^2-+6I^- + 6H⁺
Add the equal number of OH- ion on both side since the medium is alkaline
S₂O₃^2-+3I²+3H₂O + 6OH^-\(\rightarrow\)S₂O₆^2-+6I^- + 6H⁺ + 60H^-
S₂O₃^2-+3I²+3H₂O + 6OH^-\(\rightarrow\)S₂O₆^2-+6I^- + 6H₂

MnO₄^- + Fe²⁺ \(\rightarrow\) Mn²⁺ + Fe³⁺

Equalise the increase / decrease in Oxidation number by multiplying Mn species by 1 and Fe species by 5.
MnO₄^- + 5Fe²⁺ \(\rightarrow\) Mn²⁺ + 5Fe³⁺
Balance all other atoms except O and H
MnO₄^- + 5Fe²⁺ \(\rightarrow\) Mn²⁺ + 5Fe³⁺
Balance O atom by adding H₂O on the side falling short of hydrogen and equal number OH^- on the opposite side.
MnO₄^- + 5Fe²⁺ \(\rightarrow\) Mn²⁺ + 5Fe³⁺ + 4H₂O + 8OH^-
MnO₄^- + 5Fe²⁺ \(\rightarrow\) Mn²⁺ + 5Fe³⁺ + 4H₂O + 8OH^-

MnO₄^2-\(\rightarrow\) MnO₄^2- + MnO₂
MnO₄^2-\(\rightarrow\) MnO₄^2- + MnO₂ +e^- (Oxidation)....(1)
+6 +7
MnO₄^2-+2e^-\(\rightarrow\) MnO₂ (Reduction) ...(2)
+6 +4
Multiply equation (1) by (2)
2 MnO₄^2-\(\rightarrow\) MnO₄^2- + 2e^-
Add equation (2) and (3)
3 MnO₄^2-^-\(\rightarrow\) MnO₂ +2MnO₄-
Balance O atoms by adding H₂O on the side falling short of oxygen atoms
3 MnO₄^2-\(\rightarrow\) MnO₂ +2MnO₄- + 2H₂O
Balance H atoms by adding H+ on the side falling short of hydrogen atoms
3MnO₄^2-+4H⁺\(\rightarrow\) MnO₂ +2MnO₄- + 2H₂O

MnO^-₄ + Sn²⁺\(\rightarrow\) Mn²⁺ + Sn⁴⁺

Equalise the increase / decrease in O.N by multiplying the oxidant and reductant by suitable numbers.
2MnO^-₄ + 5Sn²⁺\(\rightarrow\) 2Mn²⁺ + 5Sn⁴⁺
Balance all other atoms except O and H
2MnO^-₄ + 5Sn²⁺\(\rightarrow\) 2Mn²⁺ + 5Sn⁴⁺
Balance O atom by adding water on the the side falling short of oxygen atoms.
2MnO^-₄ + 5Sn²⁺\(\rightarrow\) 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O
Balance H atom by adding H+ on the side falling short of hydrogen atoms.
2MnO^-₄ + 5Sn²⁺ 16H⁺\(\rightarrow\) 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O

Zn + HNO₃ \(\rightarrow\) Zn (NO₃)₂ + NH₄NO₃ +H₂O
Zn + HNO₃ \(\rightarrow\) Zn (NO₃)₂ + NH₄NO₃ +H₂O
Zn^o \(\rightarrow\)Zn²⁺ + 2e^-
N⁺⁵ + 8e^- \(\rightarrow\) N^-3
Multiply Equation (1) by 4 to balance the electrons
4Zn^o \(\rightarrow\)4Zn²⁺ + 2e^-
Add equation (3) and (2)
4Zn^o \(\rightarrow\)4Zn²⁺ + ~~8e^-~~
N⁵⁺ + ~~8e¹~~\(\rightarrow\) N^3-
4Zn^o +N⁵⁺\(\rightarrow\)4Zn²⁺ + N^3-
Overall equation
4Zn + 10HNO₃\(\rightarrow\) 4Zn (NO₃)₂ + NH₄NO₃ + H₂O
Balance all the atoms except O and H
4Zn + 10HNO₃\(\rightarrow\) 4Zn (NO₃)₂ + NH₄NO₃ + H₂O
Balance oxygen atom by adding H₂O on the side falling short of oxygen atom.
4Zn + 10HNO₃\(\rightarrow\) 4Zn (NO₃)₂ + NH₄NO₃ +3H₂O

10.

CuO+ NH₃ \(\rightarrow\)Cu + N₂+ H₂O
CuO+ NH₃ \(\rightarrow\)Cu + N₂+ H₂O
Cu + 2e^-\(\rightarrow\)Cu (Reduction) ...(1)
N^-3 \(\rightarrow\)N⁰₂ + 3e^-(oxidation)..(2)
Multiply Equation (2) by 2 to balance nitrogen atom
2N^-3 \(\rightarrow\) N⁰₂ + 6e^- ...(3)
Multiply equation (1) by 3 to balance the number of electrons.
3Cu²⁺ + 6e^- \(\rightarrow\) 3Cu ...(4)
Add equation (3) and (4)
3Cu²⁺ + 2N^-3 \(\rightarrow\) 3Cu + N₂
Over all balanced equation
3CuO + 2NH₃\(\rightarrow\) 3Cu + N₂ + H₂O
Balance O atom by adding H₂O on the side falling short of it.
3CuO + 2NH₃ \(\rightarrow\) 3Cu + N₂ + 3H₂O

11.

H₂C₂O₄+KMnO₄ + H₂SO₄ \(\rightarrow\) H₂SO₄ + MnSO₄ + CO₂ + H₂O

Equalise the increase I decrease in O N by multiplying Cu species by 5 and Mn species by 1
5H₂C₂O₄+KMnO₄ + H₂SO₄ \(\rightarrow\) K₂SO₄ + MnSO₄ + 5CO₂ + H₂O
Balance all other atoms except H and O atoms
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
Balance O atom by adding H₂O on the side falling short of oxygen atoms.
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O+ 7H₂O
The balanced equation is
5H₂C₂O₄+2KMnO₄ + 3H₂SO₄ \(\rightarrow\) K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O

12.

P+5HNO₃ \(\rightarrow\) H₃PO₄ + 5NO₂ + H₂O
All atoms are balanced
Balanced Equation is
P+5HNO₃ \(\rightarrow\) H₃PO₄ + 5NO₂ + H₂O

13.

P+ HNO₃\(\rightarrow\) H₃PO₄ + NO₂ + H₂O

Equalise the increase I decrease in O N by multiplying P species by +1 and N species by +5

14.

(i) In \({ S }_{ 2 }{ O }_{ 8 }^{ 2- }\)there is one peroxide bond (-O-O-) therefore, two oxygen atoms having oxidation number -1 (i.e O₂^2-) and for the other six oxygen atoms, the oxidation number is -2
\({ S }_{ 2 }{ O }_{ 8 }^{ 2- }\) = 2x +(-2 \(\times\) 6)
+(-1 \(\times\)2) = -2
2x = +12 ⇒ x = +6

(ii) In \({ S }_{ 2 }{ O }_{ 6 }^{ 2- }\), two S-atoms have oxidation state +5 while another two S-atoms have 0 oxidation state.

15.

Cu + HNO₃\(\rightarrow\)Cu(NO₃)₂ + NO₂ + H₂O

Equalise the increase / decrease in O N by multiplying Cu species by +1 and N species by +2
Cu + 2HNO₃\(\rightarrow\)Cu(NO₃)₂ + 2NO₂ + H₂O
Balance all other atoms except H and O atoms
Cu + 2HNO₃\(\rightarrow\)Cu(NO₃)₂ + 2NO₂ + H₂O
Balance O atom by adding H₂O molecules on the side falling short of oxygen atoms.
Cu + 4HNO₃\(\rightarrow\)Cu(NO₃)₂ + 2NO₂ + H₂O + H₂O
The balanced equation is
Cu + 4HNO₃\(\rightarrow\)Cu(NO₃)₂ + 2NO₂ + 2H₂O

16.

K₂Cr₂O₇ + KCI + H₂SO₄ \(\rightarrow\) KHSO₄ + CrO₂CI₂ + H₂O
It is not a redox reaction.

17.

KMnO₄ + Na₂SO₃\(\rightarrow\)MnO₂ + Na₂SO₄ + KOH (Alkaline medium)

Equalise the increase/decrease in O N by multiplying Mn species by 2 and S species by 3
2KMnO₄ + 3Na₂SO₃ \(\rightarrow\) 2MnO₂ + 3Na₂SO₄ + KOH
Balance all other atoms except H and O
2KMnO₄ + 3Na₂SO₃ \(\rightarrow\) 2MnO₂ + 3Na₂SO₄ + 2KOH
Balance O atoms by adding H₂O molecules on the side falling short of oxygen atom
2KMnO₄ + 3Na₂SO₃ +H₂O\(\rightarrow\) 2MnO₂ + 3Na₂SO₄ + 2KOH

18.

K₂Cr₂O₇ + KI + H₂S0₄ \(\rightarrow\) K₂SO ₄ + Cr2(SO₄)₃ + I₂ + H₂O

Equalise the increase / decrease in O N by multiplying I species by 1
K₂Cr₂O₇ + 3KI + H₂S0₄ \(\rightarrow\) K₂SO ₄ + Cr2(SO₄)₃ + 3I₂ + H₂O
Balance all other atoms except H and O
K₂Cr₂O₇ + 6KI + 7H₂S0₄ \(\rightarrow\) 4K₂SO ₄ + Cr2(SO₄)₃ + 3I₂ + H₂O
Balance O atom by adding H₂O on the side falling short of oxygen
K₂Cr₂O₇ + 6KI + 7H₂S0₄ \(\rightarrow\) 4K₂SO ₄ + Cr2(SO₄)₃ + 3I₂ + H₂O +6H₂O
So the balanced equation is
K₂Cr₂O₇ + 6KI + 7H₂S0₄ \(\rightarrow\) 4K₂SO ₄ + Cr2(SO₄)₃ + 3I₂ +7H₂O

19.

Disproportionation reaction

20.

Non-metal displacement reaction

21.

Metal displacement reaction

22.

Decomposition reaction

23.

Displacement reaction

24.

Combination reaction

25.

Element	Percentage	Atomic mass	Relative No. of moles	Simple Ratio Moles	Simplest whole numberRatio
K	24.75	39	\(\frac{24.75}{39}=0.63\)	\(\frac{0.63}{0.63}=1\)	1
Mn	34.77	55	\(\frac{34.77}{55}=0.63\)	\(\frac{0.63}{0.63}=1\)	1
O	100-(24.75++34.77)=40.48	16	\(\frac{40.48}{46}=2.53\)	\(\frac{2.53}{0.63}=4\)	4

The empirical formula is KMn0₄

26.

Element	Percentage	Atomic mass	Relative No. of moles	Simple Ratio Moles	Simplest whole numberRatio
X	50	10	\(\frac{50}{10}=5\)	\(\frac{5}{2.5}=2\)	2
Y	50	20	\(\frac{50}{20}=25\)	\(\frac{2.5}{2.5}=1\)	1

Its simplest formula = X₂Y

27.

Formula mass of boric acid H₃BO₃ = 61.834 amu
1 mole of H₃BO₃ = Molar mass of H₃BO₃
= 61.834 g
0.543 mole of H₃BO₃ = 61.834 x 0.543
= 33.57 g of H₃BO₃

28.

Equivalent mass of Ba(OH)₂
Molar mass of Ba(OH)₂ = 171.34 g/mol
Acidity of the Ba(OH)₂ = 2
Equivalent mass of the Ba(OH)₂ = \(\frac { Molar \ mass \ of \ the \ base }{Acidity \ of \ the \ base}\)
\(=\frac{171.34}{2}=85.5\)

29.

It is not a redox reaction because neither hydrogen or oxygen or electron is removed or added

30.

It is not a redox reaction, because neither hydrogen or oxygen or electron is removed or added

31.

Here, H in Li AlH₄ gets oxidised because of the addition of oxygen atom that leads to the formation of OH^-

32.

Mass of chloride = 0.606 - 0.456
= 0.146 g
0.146g of chlorine combines with 0.456 g of metal
∴ 35.5 g of chlorine will combine with
= \(\frac { 35.5\times 0.456 }{ 0.146 } \)
= 110.8 g of metal
∴ equivalent mass of metal = 110.8g equ^-1

33.

Equivalent mass = \(\frac { Atomic\ mass }{ Valency } \)
Equivalent mass of sodium = \(\frac { 23 }{ 1 } \) = 23

34.

Equivalent mass of an ion = \(\frac { Formula\ mass }{ Change\ of\ ion } \)
Equivalent mass of NO₃^- = \(\frac { 62 }{ 1 } \) = 62

35.

Equivalent mass = \(\frac { Atomic\ mass }{ Valency } \)
= \(\frac { 65 }{ 2 } \) = 32.5 q eq^-1

36.

6.023 1023 molecules = 1 mole
3.0115 \(\times\)10²³ molecules = \(\frac { 1 }{ 6.023\times { 10 }^{ 23 } } \times 3.0115\times { 10 }^{ 23 }\)
= 0.5 moles
Molar volumes of 1 mole of SO₂ = 2.24 \(\times\)10^-2m³
Molar volumes of 0.5 moles of SO₂ = 2.24 \(\times\) 10^-2 \(\times\)0.5
= 1.12 \(\times\) 10^-2 m³

37.

Molar mass of formic acid = 46 g
Molar volume of 46 g (1 mole) of formic acid = 2.24 \(\times\) 10^-2 m3
Molar volume of 460 g of (10 moles) of formic acid = \(\frac { 2.24\times { 10 }^{ -2 }\times 460 }{ 46 } \)
= 2.24 \(\times\) 10^-2m³

38.

Molar mass of methane = 16 g
Molar volume of 16g (1 mole) of methane = 2.24 \(\times\) 10-² m³
volume of 5 moles(80 g) of methane = \(\frac { 2.24\times { 10 }^{ -2 }\times 80 }{ 16 } \)
= 11.2 \(\times\) 10^-2 m³

39.

Molar mass of CO₂ = 44 g
Molar volume of 44 g (1 mole) of CO₂
= 2.24 \(\times\) 10^-2 m³
The volume of 88g (2 moles) = \(\frac { 2.24\times { 10 }^{ -2 }\times 88 }{ 44 } \)
= 4.48 \(\times\) 10^-2 m³

40.

Bicarbonate ion HCO₃^-
Molar mass of HCO3-^- = 61
Equivalent mass of ion = \(\frac{Molar \ Mass}{Charge \ of \ ion}\)
Equivalent mass of HCO₃^-= \(\frac{61}{1}=61\)

41.

No. of moles of Na₂CO₃ = 0.2 mole
Molar mass of Na₂CO₃ = 106g/mol
Mass = No of moles x molar mass of Na₂CO₃
= 0.2 x 106
= 21.2g

42.

Mass of glucose 720g
Molecular weight of glucose (C₆H₁₂O₆) = 180
No. of moles = \(\frac {Mass}{Molar \ Mass}\)
\(=\frac{720}{180}=4 Moles\)

43.

Density of CO₂ = 1.977 Kgm^-3
PV = nRT
No of moles = \(\frac{Mass}{Molar \ Mass}\)
PV = \(\frac{Mass}{Molar \ Mass} \times R \times T\)
Density = \(\frac {Mass}{V}\)
Molar Mass of CO₂ = \(\frac{D \times R \times T}{P}\)
Standard Temperature = 273 K
Standard Pressure = 760mm of Hg = 1 atm
=\(\frac {1.977 \times 0.0821 \times 273}{1}\)
= 44

44.

Atomic mass of potassium = 39
No. of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles = \(\frac { 19.5 }{ 39 } \) = 0.5 moles

45.

Molar mass of MgO = 40
No .of moles = \(\frac { Mass }{ Molar\ mass } \)
No .of moles = \(\frac { 90 }{ 40 } \) = 2.25 moles

46.

Molar mass of ethanol = 46
No.of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles (n) = \(\frac { 46 }{ 46 } \) = 1 mole

47.

Molar mass of sodium hydroxide = 40
No.of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles (n) = \(\frac { 120 }{ 40 } \)
= 3 moles

48.

Molar mass of calcium chloride = 111
No. of moles = \(\frac { Mass }{ Molar\ mass } \)
No. of moles = \(\frac { 50 }{ 111 } \)
= 0.450 moles

49.

Molecular mass of FeSO47H2O
Atomic mass of Fe= 55.845
Atomic mass of S = 32.065
Atomic mass of O = 15.994 x 11 = 63.304
Atomic mass of H = 1.00794 x 14 = 5.076
Molecular of FeSO₄7H₂O = 55.945 + 32.065 + (4 x 15.994) + 7 x (1.0079 x 2 + 15.9994)
= 278.014g/mol

50.

One mole C₂H₆ contains 2 moles of carbon atoms
∴ In 3 moles of C₂H₆
Number of moles of carbon atom = 2 x 3 = 6 moles

51.

Mass of Na₂CO₃= 424 x 106g
No of moles (n) = \(\frac{Mass \ of \ the \ substance}{Molar \ mass \ of \ the \ substance}\)
\(=\frac{425\times10^{6}}{106}\)
= \(4.009 moles \times 10^{6}\)
Mass o fCH₃0_H = \(320\times10^{6}\)
No of moles = \(\frac{Mass \ of \ the \ substance}{Molar \ mass \ of \ the \ substance}\)
\(=\frac{320\times10^{6}}{32}\)
= 10\(\times 10^{6}\)
Methyl alcohol is produced more

52.

Molecular mass of acetic acid = 60
60g of acetic acid contains = 6.023 \(\times\) 10²³
Molecules of acetic acid
∴ 1000g of acetic acid contains
= \(\frac { 100\times 6.023\times { 10 } }{ 60 } \)
= 100 \(\times\) 10²³
Molecules of acetic acid

Generic 5

2 x 5 = 10

53.

Content	Reactant		Product
	x	y	l	m
Stoichiometric coefficient	2	3	4	1
No. of moles allowed to react	8	15	-	-
No. of moles of reactant reacted and product formed	8	12	16	4
No. of moles of un-reacted reactants and the product formed	-	3	16	4

Limiting reagent : x
Product formed : 16 moles of l & 4 moles of m
Amount of excess reactant : 3 moles of y

54.

Element	Percentage Composition	\(\frac{Relative No.of atoms Percentage}{Atomic mass}\)	\(\frac{Atomic Mass Percentage}{Relative No.of atoms }\)	Simple Ratio
X	32%	2	16	\(\frac{2}{0.5}=\) 4
Y	24%	1	24	\(\frac{1}{0.5}=\) 2
Z	44%	0.5	88	\(\frac{0.5}{0.5}=\) 1
Empirical formula (X₄y₂Z)

Empirical formula mass = (16 x 4) + (24 x 2) + 88 = 64 + 48 + 88 = 200
n = \(\frac{Molar \quad mass}{calculated \ emphirical \ formula\ mass}\)
n = \(\frac{400}{200}\)
= 2
Molecular formula (X₄y₂Z)₂ = X₈Y₄Z₂