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

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Questions:

Generic 3

53 x 3 = 159

Calculate the number of atoms/molecules present in the following 100 g of sulphur dioxide

Calculate the number of atoms/molecules present in the following 1.8 g of water

Calculate the number of atoms/molecules present in the following : 10 g of Hg

Calculate the oxidation number of underlined atoms \(\\ { Na }_{ 2 }[\underline { F } e{ (CN) }_{ 6 }]\)

Calculate the oxidation number of underlined atoms \({ C }_{ \underline { 6 } }{ H }_{ 12 }{ O }_{ 6 }\)

Calculate the oxidation number of underlined atoms \(\underline { { S }_{ 2 } } { O }_{ 3 }\)

Calculate the oxidation number of underlined atoms \(\underline { A } sO_{ 3 }^{ 3- }\)

Calculate the oxidation number of underlined atoms \(\underline { C } l{ O }_{ 3 }\)^-

Calculate the oxidation number of underlined atoms \(H_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }\)

10.

Calculate the molecular mass of the following Methane

11.

Calculate the molecular mass of the following Crystalline oxalic acid

12.

Calculate the molecular mass of the following KMnO₄

13.

Elucidate the steps involved in arriving at the molecular formula of a compound.

14.

Write the simplest formula for the following. - H₂O₂

15.

Write the simplest formula for the following. - H₂O

16.

Write the simplest formula for the following. - C₆H₁₂O₆

17.

Write the simplest formula for the following. - N₂O₄

18.

How are 0.5 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

19.

What is the simplest formula of the compound which has the following percentage composition? C = 80%; H = 20%

20.

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B \(\longrightarrow \) 3C + 4D, when 5 moles of A react with 6 moles of B, then
(i) Which is the limiting reagent
(ii) Calculate the amount of C formed.

21.

What do you understand by stoichiometric coefficients in a chemical equation?

22.

Why is it necessary to balance a chemical equation?

23.

The percentage of all the elements present o in a compound is 95. What does it indicate?

24.

Define stoichiometry.

25.

Calculate the oxidation number of underlined atoms \({ K }_{ 2 }\underline { Mn } { O }_{ 4 }\)

26.

An iron nail is placed in copper sulphate solution taken in the beaker. Observe it for some time ?. Find the changes that takes place and why ?

27.

Place an iron piece in a moist atmosphere and observe it after two days. Is there any deposition of new substance? Why does it happen? What is this phenomenon called?

28.

A piece of cut apple becomes brown. Why? Can you prevent it by a simple method

29.

The organic compound Vitamin-C, has the following composition by mass: 40.92% C, 4.58% H, and the rest is oxygen. Determine its molecular formula. Molar mass of the substance is 176 g mol^-1

30.

An organic compound used for welding operation contains the following composition by mass: C = 92.3%, H = 7.7%. Find out the molecular formula of the compound. At STP, 10.0 L of this gas is found to weigh 11.6g.

31.

A compound contains 11.99 % N, 13.70 % O, 9.25 % B and 65.06 % F. Find its empirical formula

32.

3.24g of titanium reacts with oxygen to form 5.40g of the metal oxide. Find the empirical formula of the metal oxide?

33.

Calculate the equivalent mass of potassium dichromate in acid medium
[K₂Cr₂O₇ + 4H₂SO₄ )\(\rightarrow\)K₂SO₄ + Cr₂(SO₄)₃ +₄H₂O + 3(O) 3 x 16 = 48 294 g]

34.

Calculate equivalent mass of the Magnesium hydroxide

35.

Calculate equivalent mass of the Calcium hydroxide

36.

Calculate equivalent mass of the Ammonium hydroxide

37.

Calculate equivalent mass of the Aluminium hydroxide

38.

Calculate equivalent mass of the Sodium hydroxide

39.

Balance the following reaction by oxidation number method.

40.

A compound on analysis was found to contain C = 34.6%; H = 3.85% and O = 61.55%. Calculate its empirical formula.

41.

What will be the molecular formula for the compound, whose empirical formula is CH₂Ci and molar mass is 98.96 g ?

42.

Calculate the number of moles in the following 65.6 mg of oxygen

43.

Calculate the number of moles in the following 4.66 mg of silicon

44.

Calculate the number of moles in the following 7.85 g of copper

45.

Calculate Equivalent mass of the Phosphorous acid

46.

Calculate Equivalent mass of the Crystalline oxalic acid

47.

Calculate Equivalent mass of the Acetic acid

48.

Calculate Equivalent mass of the Nitric acid

49.

Calculate Equivalent mass of the Hydrochloric acid

50.

1.05g of a metal gives on oxidation 1.5g of its oxide. Calculate its equivalent mass.

51.

Write down the formulae for calculating the equivalent mass of an acid, base and oxidizing agent.

52.

Why are the atomic masses of most of the elements fractional?

53.

(i) If an acid is monobasic, how will you relate their equivalent mass and molecular mass.
(ii) What is the basicity of H₄ P₂ O₇ ?
(iii) Give any two examples for dibasic acids.

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Answers Key:

Generic 3

53 x 3 = 159

Molecular mass of SO₂= 64
64 g of sulphur dioxide contains = 6.023 \(\times\) 10²³
Molecules of SO₂
∴ 100g of SO₂ contains = \(\frac { 100\times 6.023\times { 10 }^{ 23 } }{ 64 } \)
= 9.41
Molecules of SO₂

1 mole of water = 18 g mol^-1
18 g of water contains 6.023 \(\times\)10²³ molecules of water
1.8 g of water contains = \(\frac { 1.8\times 6.023\times { 10 }^{ 23 } }{ 18 } \)
= 0.602 \(\times\)10²³
= 6.02 \(\times\) 10²⁴
Molecules of water

Atomic mass of Hg = 200 gmol-1
200 g of mercury contains 6.023 \(\times\) 10²³ atoms of mercury.
10g of mercury contains = \(\frac { 10\times 6.023\times { 10 }^{ 23 } }{ 200 } \)
= 0.301 \(\times\)10²³
= 3.01\(\times\) 10²⁴
atoms of mercury

2(1) + x + 6(-1) = 0
2 + x - 6 = 0
x - 4 = 0
x = 4
Oxidation number of Fe in Na₂[Fe(CN)₆ ] is +4

6x + 12(1) + 6(-2) = 0
6x + 12 -12 = 0
6x = 0
x = 0
Oxidation number of C in C₆H₁₂O₆ is zero.

2x + 3(-2) = 0
2x - 6 = 0
2x = 6
x = 3
Oxidation number of S in S₂O₃ is +3.

X + 3(-2) = -3
x - 6 = -3
x = -3 + 6
x = +3
Oxidation number of As in \(\underline { A } sO_{ 3 }^{ 3- }\) is +3

x + 3(-2) = -1
x - 6 = -1
Oxidation number of Cl in \(\underline { C } l{ O }_{ 3 }\)^-is +5.

4(1) + 2x + 7(-2) = 0
4 + 2x - 14 = 0
2x - 10 = 0
2x = 10
x = 5
Oxidation number of P in H₄P₂O₇ is +5.

10.

C ⟶ 1 \(\times\)12 =12
H ⟶ 4\(\times\) 1 = 4
\(\underline { \overline { 16 } } \)
∴ Molecular mass of CH₄ = 16

11.

\(\overset { COOH }{ \underset { COOH }{ |\quad \quad \quad } } .2{ H }_{ 2 }O\)
C ⟶ 1 \(\times\) 12 = 24
O ⟶ 4 \(\times\) 16 = 64
H₂ \(\times\) 1 = 2
\(\underline { \overline { 90 } } \)
4\(\times\) 1 = 4
2 \(\times\)16 = 32
\(\underline { \overline { 126 } } \)
∴ Molecular mass of oxalic acid 126

12.

1 atomic mass of K = 1 \(\times\) 39 = 39
Mn = 1\(\times\) 55 = 55
O = 4\(\times\) 16 = 63
\(\overline { \underline { 158 } } \)
∴ Molecular mass of KMnO₄ = 158

13.

Step 1: Moles of the element = \(\frac { percentage\ of\ element }{ at.mass } \)
Step 2: Simplest molar ratio = \(\frac { molar\ ratio }{ minium\ molar\ ratio } \)
Step 3: Empirical Formula
Step 4; \(\frac { Molar\ mass }{ Empirical\ formula\ mass } \) = n
Molecular formula = n x Empirical formula n is integer such as 1, 2, 3...etc

14.

15.

H₂O

16.

CH₂O

17.

NO₂

18.

(i) 0.5 mol refers to the number of moles of Na₂CO₃ which is weight / molecular weight.
(ii) 0.5 M refers to the molarity of Na₂CO₃ solution which is number of moles of solute dissolved per litre of solution.

19.

Element	%	Relative No.of Moles	Simple whole moles	simplest whole number ratio
C	80	\(\frac { 80 }{ 12 } \) = 6.66	\(\frac { 6.66 }{ 6.66 } \) = 1	1
H	20	\(\frac { 20 }{ 1.008 } \)= 19.84	\(\frac { 19.84 }{ 6.66 } \)= 2.9	3

The empirical formula is CH₃.

20.

2A + 4B \(\longrightarrow \) 3C +4D
According to the above equation, 2 mols of 'A' require 4 mols of'B' for the reaction.
Hence, for 5 mols of 'A', the moles of 'B' required
= 5 mol of A x \(\frac { 4\ mol\ of\ B }{ 2\ mol\ of\ A } \)
= 10 mol B
But we have only 6 mols of ' B'. hence, ' B ' is the limiting reagent. So amount of ' C ' formed is determined by amount of ' B'.
Since 4 mols of 'B' give 3 mols of 'C'. Hence 6 mols of 'B' will give
6 mol of B x \(\frac { 3\ mol\ of\ C }{ 4\ mol\ of\ B } \)

21.

The coefficients of reactants and products involved in a chemical equation represented by the balanced form are known as stoichiometric coefficients
N_2(g) + 3H₂_(g) \(\longrightarrow \) 2NH_3(g)
The stoichiometric coefficients are 1, 3 and 2 respectively

22.

A chemical equation has to be balanced to satisfy the law of conservation of mass. According to this law mass is conserved during the change from I I the reactant to product.

23.

This indicates that the compound contains oxygen. Its percentage is given as 100 - 95 = 5

24.

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles.

25.

\({ K }_{ 2 }\underline { Mn } { O }_{ 4 }\)
Oxidation number of Mn be x
2 (1) + x + 4 (-2) = 0
2 + x - 8 = 0
x - 6 = 0
x = 6
Oxidation number of Mn in K₂ MoO₄ is +6.

26.

When iron nail is dipped in copper sulphate solution, the colour of copper sulphate turns from blue to light green and reddish brown deposits is formed on iron nail. This is because iron is more reactive than copper, so it displaces Cu from CuS0₄ solution. The displacement reaction can be written as
CuSO₄ + Fe \(\rightarrow\) FeS0₄ + Cu

27.

When iron is exposed to moist air, the iron reacts with oxygen in the presence of moisture to form a reddish - brown chemical compound, iron - oxide. This phenomenon is called rusting. A new substance Iron (III) oxide is formed.
4 Fe(OH)₂+ O₂ + xH₂O\(\rightarrow\)2 Fe₂O₃ (x + 4)H₂O

28.

Apple turns brown when cut since the surface is exposed to air and undergoes oxidation. It can be prevented by dipping sliced apples in lemon juice. Lemon juice is an antioxidant which takes in all the available oxygen and prevents it from reaching the apple's tissues

29.

Element	Percentage	Atomic mass	Relative No. of moles	Simple ratio	Simplest whole Number Ratio
C	40.92	12	\(\frac{40.92}{14}=3.41\)	\(\frac{3.41}{3.406}=1.001\)	3
H	4.58	1	\(\frac{4.58}{1}=4.58\)	\(\frac{4.58}{3.406}=1\)	4
O	100-[40.92+4.58]	16	\(\frac{54.5}{10}=3.406\)	\(\frac{3.406}{3.406}=1\)	3

30.

Element	Percentage	Atomic mass	Relative No. of moles	Simple ratio mole	Simplest whole Number Ratio
C	92.3	12	\(\frac{92.3}{12}=7.7\)	\(\frac{7.7}{7.7}=1\)	1
H	7.7	1	\(\frac{7.7}{1}=7.7\)	\(\frac{7.7}{7.7}=1\)	1

Empirical formula is CH
Molecular formula = n x emprical formula
Emperical formula mass (1\(\times\)12) + (1\(\times\)1) 12 + 1 = 13
\(n=\frac{Molecular \ mass}{Empirical\ formula \ mass}\)
Molar mass = \(\frac{wt. of \ the \ substance \ x \ Molar \ volume}{vol.of\ the \ substance}\)
Molar volume at STP = 2.24\(\times\)10^-2 m³ = 22.4 I = 22400 ml
Molar mass of the gas at STP =\(\frac{11.6 \times 22.4}{10}=25.9=26\)
\(n=\frac{26}{13}=2\)
Molecular formula = n\(\times\) (emp. formula) = 2\(\times\)(CH) = C₂ H₂

31.

Element	Percentage	Atomic mass	Relative No. of moles	Simple ratio	Simplest whole Number Ratio
N	11.99	14	\(\frac{11.99}{14}=0.856\)	\(\frac{0.856}{0.856}=1\)	1
O	13.70	16	\(\frac{13.70}{16}=0.856\)	\(\frac{0.856}{0.856}=1\)	1
B	9.25	10	\(\frac{9.25}{10}=0.925\)	\(\frac{0.925}{0.856}=1\)	1
F	65.06	19	\(\frac{65.06}{19}=3.424\)	\(\frac{3.424}{0.856}=4\)	4

Empirical formula of the compound in NOBF₄

32.

Weight of Titanium = 3.24 g
Weight of metal oxide = 5.40 g
Weight of Oxygen = (5.40 - 3.24).
= 2.16 g

Element	Percentage	Atomic mass	Relative No. of moles	Simple ratio mole	Simplest whole Number Ratio
Ti	3.24	48	\(\frac{3.24}{48}=0.0675\)	\(\frac{0:067}{0:067}=1\)	1
O	2.16	16	\(\frac{2.16}{16}=0.135\)	\(\frac{0.135}{0.067}=2\)	2

∴ The empirical formula is Ti O₂

33.

48 parts by mass of oxygen are made available from 294 parts by mass of K₂Cr₂O₇
∴ 8 parts by mass of oxygen will be furnished by =\(\frac{294\times8}{48}=49\)
Equivalent mass of K₂Cr₂O₇ = 49 g equiv^-1

34.

Magnesium hydroxide Mg(OH)²
equivalent mass of Magnesium hydroxide = \(\frac{58}{2}=29\)

35.

Calcium hydroxide
equivalent mass of Ca(OH)₂ =\(\frac{74}{2}=37\)

36.

Ammonium hydroxide
equivalent mass of NH₄OH = \(\frac{35}{1}=35\)

37.

Aluminium hydroxide
equivalent mass of AI(OH)₃ = \(\frac{78}{3}=26\)

38.

NaOH
equivalent mass of NaOH =\(\frac{40}{1}=40\)

39.

MnO₄^-1 + H₂S + H⁺ \(\longrightarrow\)Mn²⁺ + S (Acidic Medium)
(i) Write oxidation number of elements
\(\underset {(+7)(-2)}{ { MnO }_{4 }^{ -1 } } +\underset {(+1) (-2) }{ {H}_{2} } S\longrightarrow\underset{+2}{{Mn}^{2+}}+\underset{0}{S} \)
(ii) Balance the number of atoms of the elements in which oxidation number changes
\(\underset {(+7)}{ { MnO }_{4 }^{ -1 } } +\underset {(-2) }{ {H}_{2} } S\longrightarrow\underset{(+2)}{{Mn}^{2+}}+\underset{0}{S} \)
(iii) Decide the oxidation and reduction reaction on the basis of difference of oxidation number. Increase in oxidation number by 2(Oxidation)

Decrease in oxidation number by 5(Reduction)
(iv) On multiplying oxidation reaction by 5 and reduction reaction by 2 to balance the change in oxidation number.
\(2{Mn }_{ 4 }^{ -1 }+H_2S\longrightarrow{2Mn}^{2+}+5S\)
Balance the electric charge and atoms which do not change in oxidation number (spectators).
\(2{Mn }_{ 4 }^{ -1 }+5H_2S+6{H}^{+}\longrightarrow{2Mn}^{2+}+5S+8{H}_{2}O\)
2(-1) 5(0) + 6(+1) 2(+2) + 5(0) + 8(0)
-2 + 6 = +4
+4 = +4
In the above reaction the reactants and products are balanced in terms of electric charge and mass equivalence.

40.

Element	%	\(\frac { Percentage\quad mass }{ At.mass } \)	Molar Ratio	Simplest Whole Number Ratio
C	34.6	34.6/12 = 2.88	2.88/2.88 = 1	3
H	3.85	3.85/1 = 3.85	3.85/2.88 = 1.335	4
O	61.55	61.55/16 = 3.85	3.85/2.88 = 1.335	4

The empirical formula of the compound = C₃H₄O₄

41.

Empirical formula = CH₂CI
Empirical formula mass = 12.01 + 2 x 1.008 + 35.453
= 49.48 g
n = \(\frac { molecular\quad mass }{ empirical\quad formula\quad mass } \)
\(=\frac { 98.96 }{ 49.48 } \)
= 2(n)
Molecular formula = n x Empirical formula
= 2 x CH₂CI = C₂H₄Cl₂

42.

Moles of oxygen = \(\frac { Mass\ of\ oxygen }{ atomic\ mass } \)
= \(\frac { 65.6\times { 10 }^{ -6} }{ 16} \)
= 4.1 x 10^-6 mol

43.

Moles of silicon = \(\frac { Mass\ of\ silicon }{ atomic\ mass } \)
= \(\frac { 4.66\times { 10 }^{ -3 } }{ 28.1 } \)
= 1.658 x 10^-4 mol

44.

Moles of copper = \(\frac { Mass\ of\ copper }{ atomic\ mass } \)
= \(\frac { 7.85 }{ 63.546 } \)
= 0.123 Mol

45.

Phosphorous acid (H₃ P0₃)
equivalent mass of phosphorous acid
= \(\frac{Molar \ mass}{basicity} =\frac{82}{2}=41\)
∴ equivalent mass of H₃P0₃ = 41

46.

Crystalline oxalic acid
equivalent mass of acetic acid = \(\frac{Molar \ mass}{basicity}\)
equivalent mass = \(\frac{126}{2}=63\)

47.

Acetic acid (CH₃COOH)
equivalent mass of acetic acid = \(\frac{Molar \ mass}{basicity}\)
= \(\frac{60}{1}\)
= 60

48.

Nitric acid
equivalent mass of HNO₃ = \(\frac{Molar \ mass}{basicity}\)
\(=\frac{63}{1}\)
= 63

49.

Hydrochloric acid
equivalent mass of an acid
\(=\frac{Molar \ mass \ of \ the\ acid}{Basicity \ of\ the \ acid}\)
Equivalent mass of HCl = \(\frac{36.5}{1}\)
= 36.5

50.

Mass of oxygen = 1.5 - 1.05
= 0.45 g
0.45g of oxygen combines with 1.05 g of metal.
ஃ 8g of oxygen combines with \(\frac{8\times1.05}{0.45}\) g of metaI
= 18.66 g of metal
ஃequivalent mass of metal = 18.66g equ^-1

51.

(i) Equivalent Mass of Acids:
E = \(\frac { Molar\ mass\ of\ the\ acid }{ basicity\ of\ the\ acid } \)
(ii) Equivalent Mass of Bases
E = \(\frac { Molar\ mass\ of\ the\ base }{ Acidity\ of\ the\ base } \)
(iii) Equivalent Mass of Oxidising agent:
E = \(\frac { Molar\ mass\ of\ the\ oxidising\ agent }{ no.\ of\ moles\ of\ electrons\ gained\ by\ one\ mole\ of\ the\ oxidsing\ agent } \)

52.

It is because most of the elements occur in nature as a mixture of isotopes and their atomic masses are the average atomic masses of the isotopes depending in their abundance

53.

(i) If an acid is mono basic, then its equivalent mass = Molecular mass
(ii) Basicity of H₄P₂O₇ is 4 (Tetrabasic acid )
(iii) Examples of dibasic acid are H₂SO₄, H₃PO₃