Basic Concepts Of Chemistry And Chemical Calculations Model Question Paper In 11th Chemistry | QB365

11th Standard Syllabus & Materials

11th Standard

Tamilnadu 11th Standard Tamil மொழி கலை -செய்யுள் - ஒவ்வொரு புல்லையும் Important Questions And Answers Study Material - QB365

NEW

11th Standard

Tamilnadu 11th Standard Tamil கேடில் விழுச்செல்வம் - உரைநடை - தமிழகக் கல்வி வரலாறு Important Questions And Answers Study Material - QB365

NEW

11th Standard

Tamilnadu 11th Standard Tamil பீடு பெற நில் - இலக்கணம் - பகுபத உறுப்புகள் Important Questions And Answers Study Material - QB365

NEW

11th Standard

Tamilnadu 11th Standard Tamil பீடு பெற நில் - செய்யுள் - குறுந்தொகை Important Questions And Answers Study Material - QB365 Set B

NEW

11th Standard

Tamilnadu 11th Standard Tamil பீடு பெற நில் - செய்யுள் - குறுந்தொகை Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

Tamilnadu 11th Standard Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set B

Basic Concepts Of Chemistry And Chemical Calculations Model Question Paper In 11th Chemistry

Published on: 26/07/2018

In this question paper, the questions are prepared from the concept of Chemistry and Chemical Calculations.
The questions are prepared from the book back and previous year questions. Please click here, to get the answer of this question paper.

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 + Answers key

Take MCQ Chemistry Test

QB365 Student Learing Managemnet System

Std: 11th Standard - Chemistry

chapter: - Model Question paper

Visit QB365-SLMS - https://www.qb365.in/slms/ Complete Questions and Answers

Questions:

Multiple Choice Question

8 x 1 = 8

The number of moles of H, in 2.24 litre of hydrogen gas at STP is ____________

(a)

(b)

0.1

(c)

0.01

(d)

0.001

How many molecules of hydrogen is required to produce 4 moles of ammonia?

(a)

15 moles

(b)

20 moles

(c)

6 moles

(d)

4 moles

How many equivalents of sodium sulphate is formed when sulphuric acid is completely neutralized by a base NaOH __________.

(a)

0.2

(b)

(c)

0.1

(d)

The number of molecules in 16g of methane is _________

(a)

3.023 x 10²³

(b)

6.023 x 10²³

(c)

16/6.023 x 10²³

(d)

6.023/3 x 10²³

Two elements X and Y ( atomic mass of X = 75; Y = 16) combine to give a compound having 76% of X. The formula of the compound is?

(a)

(b)

X₂Y

(c)

X₂Y₂

(d)

X₂Y₃

How many moles of magnesium phosphate Mg₃(PO₄)₂ Will Contain 0.25 moles of oxygen atoms?

(a)

0.02

(b)

3.125 x 10^-2

(c)

1.25 x 10^-2

(d)

2.5 x 10^-2

Which among the following statement(s) describe an element?
i) It is a pure substance which could be split into two or more simpler substance.
ii) It is a pure substance which cannot be split into simpler substance
iii) It's composition is not uniform
iv) All the above

(a)

only (iv)

(b)

only (ii)

(c)

(ii) and (iii)

(d)

(i) and (iii)

What is the mass of precipitate formed when 50 ml of 8.5 % solution of AgNO₃ is mixed with 100 ml of 1.865 % potassium chloride solution ?

(a)

3.59 g

(b)

(c)

14 g

(d)

28 g

Generic 3

6 x 3 = 18

Define the following:
(i) Element
(ii) Compound.

10.

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

11.

Calculate the oxidation number of underlined atoms of the following :
NO₃^-

12.

How much mass (in gram units) is represented by the following ?
0.2 mol of NH₃

13.

Calculate the mass of the following : 1 atom of silver

14.

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.

Generic 5

4 x 5 = 20

15.

\({ 2NH }_{ 3 }\left( g \right) +{ CO }_{ 2 }\left( g \right) \rightarrow \underset{Urea}{H_2N}-\overset { \underset { || }{ O } }{ C } -{ NH }_{ 2 }\left( aq \right) +{ H }_{ 2 }O(I)\)
In a process, 646 g of ammonia is allowed to react with 1.144 kg of CO₂ to form urea.
(i) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent ?
(ii) Calculate the quantity of urea formed and unreacted quantity of the excess reagent. The balanced equation is
\(\overset { { 2NH }_{ 3 }+{ CO }_{ 2 } }{ \underset { { H }_{ 2 }NCON{ H }_{ 2 }+{ H }_{ 2 }O }{ \downarrow } } \)

16.

Balance the following equations by oxidation number method.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂

17.

Balance the following equations by ion electron method.
i) \({ KMn }O_{ 4 }+{ SnCl }_{ 2 }+HCI\longrightarrow MnCI_{ 2 }+{ SnCI }_{ 4 }+{ H }_{ 2 }O+KCI\)
ii)
iii)
iv)

18.

What is the empirical formula of the following?
i) Fructose (C₆ H₁₂O₆) Found in honey
ii) Caffeine (C₈H₁₀N₄ O₂) a substance found in tea and Coffee

Assertion and reason

1 x 2 = 2

19.

Assertion: Equal volumes of all the gases do not contain equal number of atoms
Reason: Atom is the smallest particle which takes part in chemical reactions.
Codes:
(a) Both assertion and reason are correct and reason is the correct explanation for assertion
(b) Both assertion and reason are correct but reason is not the correct explanation for assertion
(c) Assertion is true but reason is false.
(d) Both assertion and reason is false.

(a)

Both assertion and reason are correct and reason is the correct explantion for assertion

(b)

Both assertion and reason are correct but reason is not the correct explantion for assertion

(c)

Assertion is true but reason are false

(d)

Both assertion and reason are false

Generic 2

2 x 2 = 4

20.

What is meant by Plasma state? Give an example.

21.

Calculate the equivalent masses of the following - HNO₃

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

Answers Key:

Multiple Choice Question

8 x 1 = 8

(b)
0.1

(b)
2

(b)
6.023 x 10²³

(d)
X₂Y₃

(b)
3.125 x 10^-2

(b)
only (ii)

(a)
3.59 g

Generic 3

6 x 3 = 18

(i) An element consists of only one type of atom ..An atom is the smallest electrically neutral - particle, being made up of fundamental particles, namely electrons, protons and neutrons.
(ii) Compounds are made up of molecules which contain two or more atoms of different elements. Properties of compounds are different.from those of their constituent elements. The constituents of a compound is present in a fixed ratio by weight.

10.

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

11.

NO₃^-
x + 3(-2) = -1
x - 6 = -1
x = -1 + 6
x = +5
Oxidation number of N in NO₃^-is +5

12.

Molar mass of NH₃= (1 x 14 + 3 x 1) = 17g mol^-1
Mass of 0.2 mol of NH₃ = 0.2 mol x 17g mol^-1
= 3.4 g

13.

Molecular mass of silver (Ag) = 107.87 u
Molar mass of Ag = 107.87 g mol^-1
\(\therefore\) Mass of 1 atom of Ag = \(\frac { Molar\ mass }{ Avogadro's\ number } \)
= \(\frac { 107.87g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 17.91 x 10^-23g
Mass of 1 atom of Ag = 17.91 x 10^-23 g.

14.

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

Generic 5

4 x 5 = 20

15.

(i) The entire quantity of ammonia is consumed in the reaction. So ammonia is the Iimiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.
(ii) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide.
Amount of carbon dioxide leftover
= number of moles of CO₂left over x molar mass of CO₂
= 7 moles x 44 g mol^-1
= 308 g

	Reactants		Products
	NH₃	CO₂	Urea	H₂O
Stoichiometric coefficients	2	1	1	1
Number of moles of reactants allowed to react \(N=\frac { Mass }{ Molar\quad mass } \)	\(\frac { 646 }{ 17 } =38\) moles	\(\frac { 1144 }{ 44 } =26\)moles	-	-
Actual number of moles consumed during reaction Ratio (2:1)	38 moles	19 moles	-	-
No.of moles of product thus formed	-	-	19 moles	19 moles
No.of moles of reactant left at the end of the reaction	-	7 moles	-	-

16.

Step - 1 : To find atoms undergoing change in O.N.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 2 : To find the total increase and decrease in O.N.
K₂Cr₂O7 + CrI₃ (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI ⟶ I₂ (increase of 1 unit / atom = total increase 1 x 6 = 6)
Step - 3 : To balance the total increase and decrease in O.N, multiply HI by 6.,
K₂Cr₂O₇ + 6 HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 4 : To balance all atoms other than '0' and 'H'
K₂Cr₂O₇ + 6 HI ⟶ 2KI + 2Crl₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in O.N). Hence to balance the iodine atoms add 8HI to LHS.
i.e., K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI ⟶ 2KI + 2Crl₃ + 7H₂O + 3I₂

17.

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii)

iii)

iv)