Class 11 NCERT solution of chapter 1- Some concepts of chemistry

Here class 11 chemistry NCERT solutions of chapter 1- Some basic concepts of chemistry are explained for helping class 11 students in their preparation of the exams. All the questions of chapter 1 of class 11 chemistry are explained by an expert of the subject by a step by step method so that every student could understand the solutions of each NCERT question of the chemistry textbook of class 11. These NCERT solutions will help you in clearing your concept of the basic laws of the chemistry.

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NCERT solutions of class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

Class 11 NCERT solutions of Physics and Chemistry

Chapter 1 – Some concepts of the chemistry

Chapter 1-Physical World

Q1. Calculate the molar mass of the following.

Ans.(i) CH₄

 

The molecular mass of carbon is = 12.011 u, the molecular mass of hydrogen = 1.oo8 u

The molecular mass of CH₄= 1 × 12.011 u + 4× 1.008 u = 12.011 u + 4.032 u = 16.043 u

(ii) H₂O

The molecular mass of hydrogen is = 1.008 u, the molecular mass of oxygen = 16.oo u

The molecular mass of H₂O= 2 × 1.008 u + 1× 16.00 u =2. 016 u + 16.00 u = 18.016 u

(iii) CO₂

The molecular mass of carbon is = 12.011 u, the molecular mass of oxygen = 16.oo u

The molecular mass of CO₂= 1× 12.011 u + 2× 16.00 u = 12.011 u + 32.00 u = 44.011 u

Q2. Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Ans. Mass of Na in (Na₂SO₄) = 2× 23.0 = 46 gm/mole of sodium sulphate

Molar mass of (Na₂SO₄) = 2 × 23.0 u + 1 × 32.066 u + 4 × 16.00 u = 46 u + 32.066 u + 64 u = 142.066 u

 

32.379 ≈ 32.38 %

Mass of sulphur in (Na₂SO₄) = 1 ×32.066 = 32.066 u

 

22.57 ≈ 22.6 %

Mass of  oxygen  in (Na₂SO₄) = 4× 23.0 = 92 gm/mole of sodium sulphate

45.049 ≈ 45.05 %

Q3.Determine the empirical formula of an oxide of iron, which has 69.9%  iron and 30.1 % dioxygen by mass.

Ans. The given % of Fe = 69.9%

The atomic  mass of Fe = 55.85 u

Relative no. of moles of Fe

The given % of  O₂ = 30.1%

The atomic mass of Relative no. of moles of Fe16 u

Relative no. of moles of O₂ = 30.1%

The molar ratio between Fe and O

1.25: 1.88

Simple whole number molar ratio

2 : 3

Hence the empirical formula of an oxide of iron is Fe₂O₃

Q4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Ans. (i) 1 mole of carbon is burnt in air.

The combustion equation of carbon is following

C(s) + O₂ (g) ⇒ CO₂(g)

12 gm carbon (1 mole) reacts with 32 gm oxygen (1 mole) produces 44 gm of Carbon dioxide(1 mole)

Therefore 44 gm carbon dioxide is produced.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

1 mole of carbon when reacts with 32 gm oxygen produces  carbon di oxide= 44 gm

Therefore 1 mole of carbon when reacts with 16 gm oxygen produces carbon dioxide

Hence carbon dioxide produces = 22 gm

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

1 mole of carbon when reacts with 32 gm of dioxygen produces carbon dioxide = 44 gm

2 moles of carbon = 2 × 12 = 24 gm carbon

12 gm carbon corresponds to 32 gm related to the production of 44 gm carbon dioxide(1 mole)

16 gm of oxygen corresponds to the production of carbon dioxide = 0.5 mole

Since oxygen is limiting reactant here

So, 16 gm oxygen will react with limiting the amount of carbon = 12 × 0.5 = 6 gm carbon

Rest of carbon = 24 -6 = 18 gm will not react

Therefore carbon dioxide produces = 0.5 × 44 = 22 gm

Q5. Calculate the mass of sodium acetate (CH₃COONa)  required to make 500 mL of 0.375 molar aqueous solution. The molar mass of sodium acetate is 82.0245 g mol^–1.

Ans. 0.375 M aqueous solution means that 1000 ml of the solution contains sodium acetate = 0.375 mole

No. of moles of sodium acetate (CH₃COONa) in 500 ml solution

The molar mass of sodium acetate = 82.0245 g mol^–1

m = 82.0245 × 0.1875 = 15.38 gm

Therefore required mass of sodium acetate is 15.38 gm

 

Q6.Calculate the concentration of nitric acid in moles per litre in a sample that has a density of 1.41 g mL^-1 and the mass per cent of nitric acid in it is being 69%.

Ans. The given mass per cent of nitric acid is = 69 %

There is 69 gm of nitric acid in 100 gm solution of nitric acid

Molar mass of  HNO₃) = 1 + 14 + 3 × 16 = 63 gm/mole

Mass = 69 gm

No of moles in 69 gm of nitric acid are = 1.095 mol

The density of nitric acid is = 1.41 gm/l

The volume of 100 gm nitric acid

= 70.92 ml

 

Now let’s find the concentration of nitric acid in moles /litre also known as the molarity of the solution

Conc. of nitric acid in moles /litre

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄ )? (Atomic mass of Cu= 63.5 amu)

Ans.

1 mole of CuSo₄ contains 1 mole (1 g atom) of Cu

Molar Mass of CuSo₄ = 63.5 + 32 +4 × 16 = 159.5 g mol^-1

Thus, Cu that can be obtained from 159.5 g  of CuSo_{4 =} 63.5 g

Therefore Cu that can be obtained from 100 g of CuSo₄

Q8.Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol^-1(Atomic mass: Fe = 55.85, O = 16.00 amu) Calculation of Empirical Formula. See Q3.

Ans. Empirical formula mass of Fe₂O₃ = 2 × 55.85 + 3 × 16.00 = 159.7 mol^-1

Hence, molecular formula is same as empirical formula, viz, Fe₂0₃ 

 

Q9.Calculate the atomic mass (average) of chlorine using the following data:

 

	%  Natural Abundance	Molar Mass
	75.77	34.9689
	24.23	36.9659

 

Ans.

Fractional abundance of  ³⁵Cl = 0.7577,Molar mass = 34.9689

Average atomic mass of Cl

Therefore, average atomic mass =(0.7577) (34.9689 amu) + (0.2423) (36.9659 amu)

= 26.4959 + 8.9568 = 35.4527

Therefore the average atomic mass of chlorine =35.4527 u

Q10. In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms (ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane

Ans.(i)1 mole of C₂H₆ contains 2 moles of carbon atoms

3 moles of C₂H₆ will C-atoms =6 moles

(ii) 1 mole of C₂H₆ contains H-atoms = 18 moles

3 moles of C₂H₆  will contain H-atoms = 18 moles

(iii) In 1 mole of ethane the number of molecules of ethane = 6.023 × 10²³

In 3 moles of ethane the number of molecules of ethane = 3 × 6.023 × 10²³=18 .069 × 10²³

Q11.What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L ^-1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Ans.  No of moles of sugar

Mass of sugar  = 20 gm

Molar mass of sugar(C₁₂H₂₂O₁₁) =  12 × 12 + 22 × 1 + 11 ×  16 = 144 + 22 + 176 = 342 gm/mole

No. of moles of sugar

Molarity of sugar

Volume of solution = 2L

Hence concentration of sugar (C₁₂H₂₂O₁₁) = 0.0293 mole/litre

 

Q12. If the density of methanol is 0.793 kg L ^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Ans. The density of methanol = 0.793 kg L^-1) = 793 gm L^-1)

Molar mass of methanol (CH₃OH) = 12 + 3 + 16 + 1 = 32 gm

According to titration equation

Where   and    are the initial molarity and initial volume of the solution and and are the final molarity and volume of the solution.

  = 24.78 mol/L,   = ?,    = 0.25 M = 0.25 mol/L,   = 2L

24.78 ×   = 0.25 × 2.5

 

0.02522 L = 25.22 ml

Hence required volume to be added is = 25.22 ml

Q13.Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below:1 Pa = 1 Nm^-2_.If mass of air at sea level is 1034 g cm^-2,calculate the pressure in pascal.

Ans. Pressure is defined as follows

   

Where F is the force, A is the area, P is the pressure

Mass of air in per unit area is given = 1034 g cm^-2)

So, A = 1 cm² = 10^-4 m²

m = 1034 gm = 1.034 kg, g = 9.8 m/s²

F = mg  = 1.034 × 9.8 kg ms^-1= 10.1332 kg ms^-1

1 N = 1 kg m s^-2

1 Pa = 1 Nm^-2

1 Pa = 1 kg m^-1s^-2

Therefore the pressure of air in pascal = 1.01332 x 10^-5 Pa

Q14.What is the SI unit of mass? How is it defined?

Ans. SI unit of mass is kilogram.

Q15.  Match the following prefixes with their multiples:

Prefixes	Multiples
Micro	106
Deca	109
Mega	10-6
Giga	10-15
Femto	10

Ans.

Prefixes	Multiples
Micro	10-6
Deca	10
Mega	106
Giga	109
Femto	10-15

 

Q16. What do you mean by significant figures?

Ans. The Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in the experimented value.

As an example suppose the result of an experiment is 23.5, total significant figures are 3

103.50, in this outcome the significant figures are 5

0.0056, in this outcome the significant figures are 2 because 0’s before 5 are not significant.

5600. has only 2 significant figure, trailing 0’s in a number without decimal are not significant.

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NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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