Basic Concepts of Chemistry and Chemical Calculations

Download Tamil Nadu 11th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Multiple Choice Question

5 x 1 = 5

The equivalent mass of potassium permanganate in alkaline medium is:
MnO₄^- + 2H₂O + 3e^-\(\rightarrow\) MnO₂ + 4OH^-

(a)

31.6

(b)

52.7

(c)

(d)

None of these

Which one of the following is used as a standard for atomic mass?

(a)

₆C¹²

(b)

₇C¹²

(c)

₆C¹³

(d)

₆C¹⁴

Which of the following is/are true with respect to carbon -12 ?

(a)

relative atomic mass is 12 u

(b)

the oxidation number of carbon is +4 in all its compounds.

(c)

1 mole of carbon-12 contain 6.022 x 10²² carbon atoms.

(d)

All of these

Which of the following compound(s) has /have a percentage of carbon same as that in ethylene (C₂H₄)?

(a)

propene

(b)

ethyne

(c)

benzene

(d)

ethane

The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25⁰ c and 1 atm pressure) is 1.1g. The molar mass of the gas is _______.

(a)

66.25 g mol^-1

(b)

44 g mol^-1

(c)

24.5 g mol^-1

(d)

662.5 g mol^-1

Generic 2

1 x 2 = 2

What do you understand by the term oxidation number ?

Generic 3

4 x 3 = 12

How many moles of ethane is required to produce 44 g of CO_2(g) after combustion.

What is the difference between molecular mass and molar mass ? Calculate the molecular mass and molar mass for carbon monoxide.

The density of carbon dioxide is equal to 1.965 kgm^-3 at 273 K and 1 atm pressure. calculate the molar mass of CO₂.

10.

Distinguish between oxidation and reduction.

Generic 5

3 x 5 = 15

11.

Balance the following equations by oxidation number method
i) \({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+KI+{ H }_{ 2 }SO_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }({ SO }_{ 4 })+{ I }_{ 2 }+{ H }_{ 2 }O\)
ii) \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)
iii) \(Cu+{ HNO }_{ 3 }\longrightarrow Cu\left( { No }_{ 3 } \right) _{ 2 }+{ No }_{ 2 }+{ H }_{ 2 }O\)
iv) \({ KMn }O_{ 4 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ MnSO }_{ 4 }+{ CO }_{ 2 }+{ H }_{ 2 }O\)

12.

The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of Al = 27 u atomic mass of O = 16 u )
2Al + Fe₂O₃ \(\longrightarrow \) Al₂O₃ + 2Fe; If in this process, 324 g of aluminum is allowed to react with 1.12 kg of ferric oxide
i) Calculate the mass of Al₂O₃ formed
ii) How much of the excess reagent is left at the end of the reaction ?

13.

Mass of one atom of an element is 6.645 x 10^-23g. How many moles of element are there in 0.320 kg.

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Multiple Choice Question

5 x 1 = 5

(b)
52.7

(a)
₆C¹²

(a)
relative atomic mass is 12 u

(a)
propene

(b)
44 g mol^-1

Generic 2

1 x 2 = 2

It is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules.

Generic 3

4 x 3 = 12

The balanced equation for the combustion of ethane
C₂H₆ + \(\frac { 7 }{ 2 } \)O₂ \(\longrightarrow \) 2CO₂ +3H₂O
2C₂H₆ + 7O₂ \(\longrightarrow \) 4CO₂ + 6H₂O
To produce 4 moles of CO₂, 2 moles of ethane is required
To produce 1 moles (44 g) of CO₂ required
Number of moles of ethane
\(=\frac{2 \mathrm{~mol} \text { ethane }}{4 \not \mathrm{molCO}_{2}} \times 1 \not \mathrm{molCO}_{2}\)
= \(\frac { 1 }{ 2 } mole\quad of\quad ethane\)
= 0.5 mole of ethane

1) The unit of molecular mass is atomic mass unit [amu]. The unit of molar mass is gram per mole.
2) Molecular mass is the mass of one molecule while molar mass is the mass of one mole of molecules (6.022 x 10²³)
(i) Molecular mass of CO₂ = 1(C) + 2(0) = 12 + 32 = 44 amu
or 7.304 x 10^-23 g
(ii) Molar mass of CO₂ = 44 g mol^-1.

Molar mass = density x Molar volume
= 1.965 x 2.24 x 10^-2
= 4.4016 x 10^-2 kg/mol
= 4.4016 x 10^-2 x 10³ g/mol
= 44.016 g/mol

10.

	Oxidation	Reduction
1.	Addition of oxygen	Addition of Hydrogen
2.	Removal of Hydrogen	Removal of oxygen
3.	Addition of an electronegative element.	Addition of an electro positive element
4.	Removal of an electro positive element	Removal of an electro negative element
5.	Loss of electron	Gain of electron
6.	Increase in oxidation state / number	Decrease in oxidation state/ number.

Generic 5

3 x 5 = 15

11.

(i) \({ K }_{ 2 }\overset { +6 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \uparrow } }{ Cr_{ 2 } } } { O }_{ 7 }+K\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \downarrow }{ I } } } +{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ \overset { +3 }{ Cr } }_{ 2 }({ SO }_{ 4 })_{ 3 }+\overset { 0 }{ I } _{ 2 }+{ H }_{ 2 }O\)
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \) K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
K₂Cr₂O₇ + 6KI + H₂SO₄ \(\longrightarrow \) K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
K₂Cr₂O₇ + 6KI + 7H₂SO₄ \(\longrightarrow \) 4k₂SO₄ + Cr₂(SO₄)₃ + ₃I₂ + 7H₂
ii) \({ K }Mno_{ 4 }+{ Na }_{ 2 }{ So }_{ 3 }\longrightarrow { MnO }_{ 2 }+{ Na }_{ 2 }{ So }_{ 4 }+KOH\)
\({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \) 2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ + 2KOH
iii) \(Cu+{ HNO }_{ 3 }\longrightarrow Cu\left( { No }_{ 3 } \right) _{ 2 }+{ No }_{ 2 }+{ H }_{ 2 }O\)
\(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O
iv) \({ KMn }O_{ 4 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ MnSO }_{ 4 }+{ CO }_{ 2 }+{ H }_{ 2 }O\)
\({ K }\overset { +7 }{ \underset { \underset { 2\times { 3e }^{ - } }{ \downarrow } }{ M } } n{ O }_{ 4 }+{ H }_{ 2 }\overset { -1 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ C_{ 2 } } } } { O }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+\overset { +2 }{ M } n{ SO }_{ 4 }+\overset { +4 }{ C } { O }_{ 2 }+{ H }_{ 2 }O\)
2KMnO₄ + 5 H₂C2O₄ + H₂S0₄ \(\longrightarrow \) Mn0₂ + Na₂S0₄ + KOH
2KMnO₄+ 5 H₂C₂O₄ + H₂S0₄ \(\longrightarrow \) K₂SO₄ + 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5 H₂C₂O₄ + 3H₂S0_{4 \(\longrightarrow \)} K₂S0₄ + 2MnS0₄ + 10C0₂ + 8 H₂0

12.

2Al + Fe₂O₃ \(\longrightarrow \) Al₂O₃ + 2Fe

	Reactants		Products
	Al	Fe₂O₃	Al₂O₃	Fe
Amount of reactant allowed to react	324 g	1.12 kg	-	-
Number of moles allowed to react	\(\frac { 324 }{ 27 } =12mol\)	\(\frac { 1.12\times { 10 }^{ 3 } }{ 160 } =7mol\)	-	-
Stoichiometric Co-efficient	2	1	1	2
Number of moles consumed during reaction	12 mol	6 mol	-	-
Number of moles of reactant unreacted and number of moles of product formed	-	1 mol	6 mol	12 mol

Molar mass of Al₂O₃format = 6 mol x 102 g mol^-1 = 612 g
[ Al₂O₃: (2 x 27) + 3(16) = 54 + 48 = 102] = 612 g
Excess reagent = Fe₂O₃
Amount of excess reagent left at the end of the reaction = 1 mol x 160 g mol^-1
= 160g [ Fe₂O₃: (2 x 56) + (3 x 16) = 112 + 48 = 160] = 160 g

13.

mass of one atom = 6.645 x 10^-23g
\(\therefore\) mass of 1 mole of an atom = 6.645 x 10^-23 g x 6.022 x 10²³= 40 g
\(\therefore\) number of moles of element in 0.320 kg = \(\frac { 1\quad mole }{ 40g } \times 0.320kg\)
= \(\frac { 1mol\times 320g }{ 40g } \)
= 8 mol