NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry

NCERT Solutions Class 11 Chemistry Chapter 1 – Free PDF Download

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are provided on this page for Class 11 Chemistry students. These solutions can also be downloaded in a PDF format for free by clicking the download button provided below. The free Class 11 Chemistry NCERT Solutions provided by BYJU’S have been prepared by seasoned academics and chemistry experts, keeping the needs of Class 11 students in mind. The NCERT Solutions provided on this page contain step-by-step explanations in order to help students answer similar questions that might be asked in their board examinations.

The expert tutors have created NCERT Solutions for Class 11 in simple and understandable language in accordance with the latest CBSE Syllabus 2023-24 and its guidelines. The basic concepts are covered in the NCERT Class 11 Solutions with the aim of providing a quality learning experience for Class 11 students. You can download the Class 11 Chemistry NCERT Solutions Chapter 1 PDF from BYJU’S to complete a part of the CBSE syllabus before the board exam.

Download NCERT Solutions Class 11 Chemistry Chapter 1 PDF

ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 01
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 02
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 03
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 04
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 05
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 06
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 07
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 08
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 09
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 10
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 11
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 12
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 13
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 14
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 15
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 16
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 17
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 18
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 19
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 20
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 21
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 22
ncert solutions for class 11 chemistry jan18 chapter 1 some basic concepts of chemistry 23

NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry

“Some Basic Concepts of Chemistry” is the first chapter in the Class 11 Chemistry syllabus as prescribed by NCERT. The chapter touches upon topics such as the importance of Chemistry, atomic mass, and molecular mass. Some basic laws and theories in Chemistry, such as Dalton’s atomic theory, Avogadro’s law and the law of conservation of mass, are also discussed in this chapter.

The types of questions provided in the NCERT Class 11 Chemistry textbook for Chapter 1 include:

  • Numerical problems in calculating the molecular weight of compounds.
  • Numerical problems in calculating mass percent and concentration.
  • Problems on empirical and molecular formulae.
  • Problems on molarity and molality.
  • Other problems related to the mole concept (such as percentage composition and expressing concentration in parts per million).

The NCERT Solutions of Chemistry provided on this page for Class 11 Chapter 1 contains detailed explanations of the steps to be followed while solving the numerical value questions that are frequently asked in the examinations. The subtopics covered in the chapter are listed below.

Subtopics of NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

  1. Importance of Chemistry
  2. Nature of Matter
  3. Properties of Matter and Their Measurement  
  4. The International System of Units (SI)
  5. Mass and Weight
  6. Uncertainty in Measurement
  7. Scientific Notation
  8. Significant Figures
  9. Dimensional Analysis
  10. Laws of Chemical Combinations    
  11. Law of Conservation Of Mass    
  12. Law of Definite Proportions
  13. Law of Multiple Proportions  
  14. Gay Lussac’s Law of Gaseous Volumes  
  15. Avogadro’s Law
  16. Dalton’s Atomic Theory
  17. Atomic and Molecular Masses   
  18. Atomic Mass
  19. Average Atomic Mass
  20. Molecular Mass
  21. Formula Mass
  22. Mole Concept and Molar Masses
  23. Percentage Composition    
  24. Empirical Formula for Molecular Formula
  25. Stoichiometry and Stoichiometric Calculations                
  26. Limiting Reagent
  27. Reactions in Solutions

Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 1

Exercise

Q1. Calculate the molar mass of the following:

(i)

\(\begin{array}{l}CH_{4}\end{array} \)
      (ii)
\(\begin{array}{l}H_{2}O\end{array} \)
      (iii)
\(\begin{array}{l}CO_{2}\end{array} \)

Ans.

(i) 

\(\begin{array}{l}CH_{4}\end{array} \)
:

Molecular mass of

\(\begin{array}{l}CH_{4}\end{array} \)
= Atomic mass of C + 4 x Atomic mass of H

= 12 + 4 x 1

= 16 u

(ii)

\(\begin{array}{l}H_{2}O\end{array} \)
:

Molar mass of water

\(\begin{array}{l}H_{2}O\end{array} \)

Atomic mass of H = 1

Atomic mass of O = 16

H2O = 2×H+1×O

Molar mass of water = 2×1+16 = 18g/mol

(iii)

\(\begin{array}{l}CO_{2}\end{array} \)
:

Molecular mass of

\(\begin{array}{l}CO_{2}\end{array} \)
= Atomic mass of C + 2 x Atomic mass of O

= 12 + 2 × 16

= 44 u

Q2. Calculate the mass per cent of different elements present in sodium sulphate (

\(\begin{array}{l}Na_{2}SO_{4}\end{array} \)
) .

Ans.

Now for

\(\begin{array}{l}Na_{2}SO_{4}\end{array} \)
.

Molar mass of

\(\begin{array}{l}Na_{2}SO_{4}\end{array} \)

= [(2 x 23.0) + (32.066) + 4(16.00)]

=142.066 g

Formula to calculate mass percent of an element =

\(\begin{array}{l}\frac{Mass\;of\;that\;element\;in\;the\;compound}{Molar\;mass\;of\;the\;compound}\times 100\end{array} \)

 

Therefore, mass percent of the sodium element:

=

\(\begin{array}{l}\frac{46.0g}{142.066g}\times 100\end{array} \)

= 32.379

= 32.4%

Mass percent of the sulphur element:

=

\(\begin{array}{l}\frac{32.066g}{142.066g}\times 100\end{array} \)

= 22.57

= 22.6%

Mass percent of the oxygen element:

=

\(\begin{array}{l}\frac{64.0g}{142.066g}\times 100\end{array} \)

= 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.

Given there is an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass:

Relative moles of iron in iron oxide:

=

\(\begin{array}{l}\frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}\end{array} \)

=

\(\begin{array}{l}\frac{69.9}{55.85}\end{array} \)

= 1.25

 

Relative moles of oxygen in iron oxide:

=

\(\begin{array}{l}\frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}\end{array} \)

=

\(\begin{array}{l}\frac{30.1}{16.00}\end{array} \)

= 1.88

 

The simplest molar ratio of iron to oxygen:

 1.25: 1.88 ⇒ 1: 1.5 ⇒ 2: 3

Therefore, the empirical formula of the iron oxide is

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)
.

 

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.

\(\begin{array}{l}C+O_{2}\rightarrow CO_{2}\end{array} \)

1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2.

Amount of

\(\begin{array}{l}CO_{2}\end{array} \)
produced = 44 g

 

(ii) 1 mole of carbon is burnt in 16 g of O2.

1 mole of carbon burnt in 32 grams of O2 it forms 44 grams of

\(\begin{array}{l}CO_{2}\end{array} \)
.

Therefore, 16 grams of O2 will form

\(\begin{array}{l}\frac{44\times 16}{32}\end{array} \)

= 22 grams of

\(\begin{array}{l}CO_{2}\end{array} \)

 

(iii) 2 moles of carbon are burnt in 16 g of O2.

Here again, dioxygen is the limiting reactant. 16g of dioxygen can combine only with 0.5mol of carbon. CO2 produced again is equal to 22g.

Q5. Calculate the mass of sodium acetate

\(\begin{array}{l}(CH_{3}COONa)\end{array} \)
required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.

 

Ans.

0.375 M aqueous solution of

\(\begin{array}{l}CH_{3}COONa\end{array} \)

= 1000 mL of solution containing 0.375 moles of

\(\begin{array}{l}CH_{3}COONa\end{array} \)

Therefore, no. of moles of

\(\begin{array}{l}CH_{3}COONa\end{array} \)
in 500 mL

=

\(\begin{array}{l}\frac{0.375}{1000}\times 500\end{array} \)

= 0.1875 mole

Molar mass of sodium acetate =

\(\begin{array}{l}82.0245\;g\;mol^{-1}\end{array} \)

Therefore, the mass of

\(\begin{array}{l}CH_{3}COONa\end{array} \)

=

\(\begin{array}{l}(82.0245\;g\;mol^{-1})(0.1875\;mole)\end{array} \)

= 15.38 grams

 

Q6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%.

Ans.

Mass percent of HNO3 in sample is 69 %

Thus, 100 g of HNO3 contains 69 g of HNO3 by mass.

Molar mass of HNO3

= { 1 + 14 + 3(16)}

\(\begin{array}{l}g.mol^{-1}\end{array} \)

= 1 + 14 + 48

\(\begin{array}{l}= 63g\;mol^{-1}\end{array} \)

 

Now, no. of moles in 69 g of

\(\begin{array}{l}HNO_{3}\end{array} \)
:

=

\(\begin{array}{l}\frac{69\:g}{63\:g\:mol^{-1}}\end{array} \)

= 1.095 mol

 

Volume of 100g HNO3 solution

=

\(\begin{array}{l}\frac{Mass\;of\;solution}{density\;of\;solution}\end{array} \)

=

\(\begin{array}{l}\frac{100g}{1.41g\;mL^{-1}}\end{array} \)

= 70.92mL

=

\(\begin{array}{l}70.92\times 10^{-3}\;L\end{array} \)

 

Concentration of HNO3

=

\(\begin{array}{l}\frac{1.095\:mole}{70.92\times 10^{-3}L}\end{array} \)

= 15.44mol/L

Therefore,

Concentration of HNO3 = 15.44 mol/L

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

Ans.

1 mole of

\(\begin{array}{l}CuSO_{4}\end{array} \)
contains 1 mole of Cu.

Molar mass of

\(\begin{array}{l}CuSO_{4}\end{array} \)

= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 grams

159.5 grams of

\(\begin{array}{l}CuSO_{4}\end{array} \)
contains 63.5 grams of Cu.

Therefore, 100 grams of

\(\begin{array}{l}CuSO_{4}\end{array} \)
will contain
\(\begin{array}{l}\frac{63.5\times 100g}{159.5}\end{array} \)
of Cu.

=

\(\begin{array}{l}\frac{63.5\times 100}{159.5}\end{array} \)

=39.81 grams

 

Q8. Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively. 

Ans.

Here,

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

 

No. of moles of Fe present in oxide

=

\(\begin{array}{l}\frac{69.90}{55.85}\end{array} \)

= 1.25

 

No. of moles of O present in oxide

=

\(\begin{array}{l}\frac{30.1}{16.0}\end{array} \)

=1.88

 

Ratio of Fe to O in oxide,

= 1.25: 1.88

=

\(\begin{array}{l}\frac{1.25}{1.25}:\frac{1.88}{1.25}\end{array} \)

=

\(\begin{array}{l}1:1.5\end{array} \)

=

\(\begin{array}{l}2:3\end{array} \)

Therefore, the empirical formula of oxide is

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

 

Empirical formula mass of

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

= [2(55.85) + 3(16.00)] g

= 159. 7g

The molar mass of 

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)
= 159.69g

Therefore n =

\(\begin{array}{l}\frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}\end{array} \)

= 0.999

= 1(approx)

 

The molecular formula of a compound can be obtained by multiplying n with the empirical formula.

Thus, the empirical of the given oxide is

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)
and n is 1.

Therefore, the molecular formula of the oxide is 

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

 

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

Percentage Natural Abundance Molar Mass
\(\begin{array}{l}_{}^{35}\textrm{Cl}\end{array} \)
75.77 34.9689
\(\begin{array}{l}_{}^{37}\textrm{Cl}\end{array} \)
24.23 36.9659

 

Ans.

Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689  

Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659 

Average Atomic mass = (0.7577 x 34.9689)amu + (0.2423 x 36.9659) 

= 26.4959 + 8.9568 = 35.4527

 

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

(iii) Number of molecules of ethane

Ans.

(i) 1 mole of

\(\begin{array}{l}C_{2}H_{6}\end{array} \)
contains two moles of C- atoms.

\(\begin{array}{l}∴\end{array} \)
No. of moles of C- atoms in 3 moles of
\(\begin{array}{l}C_{2}H_{6}\end{array} \)
.

= 2 x 3

= 6

(ii) 1 mole of

\(\begin{array}{l}C_{2}H_{6}\end{array} \)
contains six moles of H- atoms.

\(\begin{array}{l}∴\end{array} \)
No. of moles of H- atoms in 3 moles of
\(\begin{array}{l}C_{2}H_{6}\end{array} \)
.

= 3 x 6

= 18

(iii) 1 mole of

\(\begin{array}{l}C_{2}H_{6}\end{array} \)
contains 1 mole of ethane- atoms.

\(\begin{array}{l}∴\end{array} \)
No. of molecules in 3 moles of
\(\begin{array}{l}C_{2}H_{6}\end{array} \)
.

= 3 x 6.023 x

\(\begin{array}{l}10^{23}\end{array} \)

= 18.069 x

\(\begin{array}{l}10^{23}\end{array} \)

 

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 2L?

 

Ans.

Molarity (M) is as given by,

=

\(\begin{array}{l}\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}\end{array} \)

=

\(\begin{array}{l}\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}\end{array} \)

=

\(\begin{array}{l}\frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}\end{array} \)

=

\(\begin{array}{l}\frac{\frac{20\;g}{342\;g}}{2\;L}\end{array} \)

=

\(\begin{array}{l}\frac{0.0585\;mol}{2\;L}\end{array} \)

= 0.02925 mol

\(\begin{array}{l}L^{-1}\end{array} \)

Therefore, Molar concentration = 0.02925 mol

\(\begin{array}{l}L^{-1}\end{array} \)

 

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.)

Molar mass of methanol (CH3OH)

= 32 gmol-1 = 0.032 kgmol-1

molarity of the given solution

\(\begin{array}{l}=\frac{W_{2}in kg}{M_{w_{2}}\times V_{(sol)}L}=\frac{d_{sol}(kgL^{-1})}{Mw_{2}(kg)}\\=\frac{0.793kgL^{-1}}{0.032kgmol^{-1}}= 24.78 M\\\underset{(Given solution)}{Applying M_{1}\times V_{1}}= \underset{(solution to be prepared)}{M_{2}V_{2}}\end{array} \)

24.78 x V1 = 0.25 x 2.5 L

or V1 = 0.02522L = 25.22mL

 

Q13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal

Ans.

Pressure is the force (i.e., weight) acting  per unit area

But weight = mg

∴ Pressure = Weight per unit area

\(\begin{array}{l}=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\\=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\times \frac{1kg}{1000g}\times \frac{100cm\times 100cm}{1m\times 1m}\times \frac{1N}{kgms^{-2}}\times \frac{1Pa}{1Nm^{-2}}\\= 1.01332\times 10^{^{5}}Pa\end{array} \)

 

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

Ans.

The SI unit of mass is kilogram (kg). A kilogram is equal to the mass of a platinum-iridium cylinder kept at the International Bureau of Weights and Measures at Sèvres, France.

 

Q15. Match the following prefixes with their multiples:

 

  Prefixes Multiples
(a) femto 10
(b) giga
\(\begin{array}{l}10^{-15}\end{array} \)
(c) mega
\(\begin{array}{l}10^{-6}\end{array} \)
(d) deca
\(\begin{array}{l}10^{9}\end{array} \)
(e) micro
\(\begin{array}{l}10^{6}\end{array} \)

 

Ans.

Prefixes Multiples
(a) femto
\(\begin{array}{l}10^{-15}\end{array} \)
(b) giga
\(\begin{array}{l}10^{9}\end{array} \)
(c) mega
\(\begin{array}{l}10^{6}\end{array} \)
(d) deca 10
(e) micro
\(\begin{array}{l}10^{-6}\end{array} \)

 

 

Q16. What do you mean by significant figures?

Ans.

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

e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3.

Therefore, “the total number of digits in a number with the last digit that shows the uncertainty of the result is known as significant figures.”

 

Q17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.

Ans.

(i) 1 ppm = 1 part out of 1 million parts.

Mass percent of 15 ppm chloroform in H2O

=

\(\begin{array}{l}\frac{15}{10^{6}} \times 100\end{array} \)

=

\(\begin{array}{l}\approx\end{array} \)
1.5 ×
\(\begin{array}{l}10^{-3}\end{array} \)
%

 

(ii) 

\(\begin{array}{l}Molarity = \frac{15/119.5}{10^{6}\times 10^{-3}}= 1.25 \times 10^{-4}\end{array} \)

 

Q18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

 

Ans.

(i) 0.0048= 4.8 ×

\(\begin{array}{l}10^{-3}\end{array} \)

(ii) 234,000 = 2.34 ×

\(\begin{array}{l}10^{5}\end{array} \)

(iii) 8008= 8.008 ×

\(\begin{array}{l}10^{3}\end{array} \)

(iv) 500.0 = 5.000 ×

\(\begin{array}{l}10^{2}\end{array} \)

(v) 6.0012 = 6.0012 ×

\(\begin{array}{l}10^{0}\end{array} \)

 

Q19. How many significant figures are present in the following?

(a) 0.0025

(b) 208

(c) 5005

(d) 126,000

(e) 500.0

(f) 2.0034

Ans.

(a) 0.0025: 2 significant numbers.

(b) 208: 3 significant numbers.

(c) 5005: 4 significant numbers.

(d) 126,000:3 significant numbers.

(e) 500.0: 4 significant numbers.

(f) 2.0034: 5 significant numbers.

 

Q20. Round up the following upto three significant figures:

(a) 34.216

(b) 10.4107

(c)0.04597

(d)2808

 

Ans.

(a) The number after round up is: 34.2

(b) The number after round up is: 10.4

(c)The number after round up is: 0.0460

(d)The number after round up is: 2810

 

Q21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

  Mass of dioxygen Mass of dinitrogen
(i) 16 g 14 g
(ii) 32 g 14 g
(iii) 32 g 28 g
(iv) 80 g 28 g

 

(a) Which law of chemical combination is obeyed by the above experimental data?
Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm3

 

Ans.

(a)

Here if we fix the mass of dinitrogen at 14g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16g, 32g, 32g, and 80g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5.

Hence, the given experimental data obeys the Law of Multiple Proportions.

(b)

i. 

\(\begin{array}{l}1 km = 1 km \times \frac{1000m}{1km}\times \frac{100cm}{1m}\times \frac{10mm}{1cm}= 10^{6}mm\\1km = 1 km \times \frac{1000m}{1km}\times \frac{1pm}{10^{-12}m}= 10^{^{15}}pm\end{array} \)

ii. 

\(\begin{array}{l}1 mg = 1 mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}= 10^{-6}kg\\1mg = 1mg\times \frac{1g}{1000mg}\times \frac{1ng}{10^{-9}g}=10^{6}ng\end{array} \)

iii. 

\(\begin{array}{l}1mL = 1mL\times \frac{1L}{1000mL}=10^{-3}L\\1mL = 1cm^{3} \\ =1cm^{3}\times \frac{1dm\times 1dm\times 1dm}{10cm\times 10cm\times 10cm}= 10^{-3}dm^{3}\end{array} \)

 

Q22. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns.

Ans.

Time taken = 2 ns

= 2 ×

\(\begin{array}{l}10^{ -9 }\end{array} \)
s

Now,

Speed of light = 3 ×

\(\begin{array}{l}10^{ 8 }\end{array} \)
\(\begin{array}{l}ms^{ -1 }\end{array} \)

We know that,

Distance = Speed x Time

So,

Distance travelled in 2 ns = speed of light x time taken

= (3 ×

\(\begin{array}{l}10^{ 8 }\end{array} \)
)(2 ×
\(\begin{array}{l}10^{ -9 }\end{array} \)
)

= 6 ×

\(\begin{array}{l}10^{ -1 }\end{array} \)
m

= 0.6 m

 

Q23. In a reaction
A + B2 →  AB2
Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

 

Ans.

Limiting reagent:

It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limits the amount of product formed.

 

(i) 300 atoms of A + 200 molecules of B

1 atom of A reacts with 1 molecule of B. Similarly, 200 atoms of A reacts with 200 molecules of B, so 100 atoms of A are unused. Hence, B is the limiting reagent.

 

(ii) 2 mol A + 3 mol B

1 mole of A reacts with 1 mole of B. Similarly, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused. Hence, A is the limiting reagent.

 

(iii) 100 atoms of A + 100 molecules of Y

1 atom of A reacts with 1 molecule of Y. Similarly, 100 atoms of A reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting reagent.

 

(iv) 5 mol A + 2.5 mol B

1 mole of A reacts with 1 mole of B. Similarly 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of A is unused. Hence, B is the limiting reagent.

 

(v) 2.5 mol A + 5 mol B

1 mole of A reacts with 1 mole of B. Similarly, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of B is unused. Hence, A is the limiting reagent.

 

Q24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + H2(g)→ 2NH3 (g)

(i) Calculate the mass of

\(\begin{array}{l}NH_{ 3 }\end{array} \)
produced if
\(\begin{array}{l}2 \; \times \;10^{ 3 }\end{array} \)
g N2 reacts with
\(\begin{array}{l}1 \; \times \;10^{ 3 }\end{array} \)
g of H2?

 (ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass.

 

Ans.

(i) 1 mol of N2 i.e., 28 g reacts with 3 moles of H2 i.e., 6 g of H2

∴ 2000 g of N2 will react with H2

\(\begin{array}{l}\frac{6}{28}\times 200g = 428.6g\end{array} \)

Thus, N2 is the limiting reagent while H2 is the excess reagent

2 mol of N2 i.e., 28 g of N2 produces NH3 = 2 mol

= 34 g

Therefore, 2000 g will produces NH3

\(\begin{array}{l}\frac{34}{28}\times 2000 g\end{array} \)

= 2428.57 g

(ii) H2 will remain unreacted

(iii) Mass left unreacted = 1000g – 428.6g = 571.4g

 

Q25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Ans.

Molar mass of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)
:

= (2 × 23) + 12 + (3 × 16)

= 106 g

\(\begin{array}{l}mol^{ -1 }\end{array} \)

1 mole of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)
means 106 g of
\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)

Therefore, 0.5 mol of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)

=

\(\begin{array}{l}\frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol \end{array} \)
\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)

= 53 g of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)

0.5 M of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)
= 0.5 mol/L
\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)

Hence, 0.5 mol of

\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)
is in 1 L of water or 53 g of
\(\begin{array}{l}Na_{ 2 }CO_{ 3 }\end{array} \)
is in 1 L of water.

 

Q26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Ans.

Reaction:

\(\begin{array}{l}2H_{ 2 }\;(g)  \; + \; O_{ 2 }\; (g)  \; \rightarrow \; 2H_{ 2 }O\; (g) \end{array} \)

 

2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of water vapour.

Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of water vapour.

 

Q27. Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

 

Ans.

(i) 28.7 pm

1 pm =

\(\begin{array}{l}10^{ -12 } \; m\end{array} \)

28.7 pm = 28.7 ×

\(\begin{array}{l}10^{ -12 } \; m\end{array} \)

= 2.87 ×

\(\begin{array}{l}10^{ -11 } \; m\end{array} \)

 

(ii) 15.15 pm

1 pm =

\(\begin{array}{l}10^{ -12 } \; m\end{array} \)

15.15 pm = 15.15 ×

\(\begin{array}{l}10^{ -12 } \; m\end{array} \)

= 1.515 ×

\(\begin{array}{l}10^{ -11 } \; m\end{array} \)

 

(iii) 25365 mg

1 mg =

\(\begin{array}{l}10^{ -3 } \; g\end{array} \)

1 mg = 10-6 kg

25365 mg = 25365 x 10-6 kg

25365 mg = 2.5365 ×

\(\begin{array}{l}10^{ -2 } \; kg\end{array} \)

 

Q28. Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of

\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

 

Ans.

(i) 1 g of Au (s)

=

\(\begin{array}{l}\frac{ 1 }{ 197 }\end{array} \)
mol of Au (s)

=

\(\begin{array}{l}\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }\end{array} \)
atoms of Au (s)

= 3.06

\(\begin{array}{l}\times \; 10^{ 21 }\end{array} \)
atoms of Au (s)

 

(ii) 1 g of Na (s)

=

\(\begin{array}{l}\frac{ 1 }{ 23 }\end{array} \)
mol of Na (s)

=

\(\begin{array}{l}\frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 }\end{array} \)
atoms of Na (s)

= 0.262

\(\begin{array}{l}\times \; 10^{ 23 }\end{array} \)
atoms of Na (s)

= 26.2

\(\begin{array}{l}\times \; 10^{ 21 }\end{array} \)
atoms of Na (s)

 

(iii) 1 g of Li (s)

=

\(\begin{array}{l}\frac{ 1 }{ 7 }\end{array} \)
mol of Li (s)

=

\(\begin{array}{l}\frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 }\end{array} \)
atoms of Li (s)

= 0.86

\(\begin{array}{l}\times \; 10^{ 23 }\end{array} \)
atoms of Li (s)

= 86.0

\(\begin{array}{l}\times \; 10^{ 21 }\end{array} \)
atoms of Li (s)

 

(iv)1 g of

\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

=

\(\begin{array}{l}\frac{ 1 }{ 71 }\end{array} \)
mol of
\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

(Molar mass of

\(\begin{array}{l}Cl_{ 2 }\end{array} \)
molecule = 35.5 × 2 = 71 g
\(\begin{array}{l}mol^{ -1 }\end{array} \)
)

=

\(\begin{array}{l}\frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 }\end{array} \)
atoms of
\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

= 0.0848

\(\begin{array}{l}\times \; 10^{ 23 }\end{array} \)
atoms of
\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

= 8.48

\(\begin{array}{l}\times \; 10^{ 21 }\end{array} \)
atoms of
\(\begin{array}{l}Cl_{ 2 }\end{array} \)
(g)

 

Therefore, 1 g of Li (s) will have the largest no. of atoms.

 

Q29. Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Ans.

Mole fraction of

\(\begin{array}{l}C_{ 2 }H_{ 5 }OH\end{array} \)

=

\(\begin{array}{l}\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}\end{array} \)

0.040 =

\(\begin{array}{l}\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}\end{array} \)
——(1)

 

No. of moles present in 1 L water:

\(\begin{array}{l}n_{ H_{ 2 }O} \; = \; \frac{ 1000 \; g}{18 \; g \; mol^{ -1 }}\end{array} \)
\(\begin{array}{l}n_{ H_{ 2 }O}\end{array} \)
= 55.55 mol

 

Substituting the value of

\(\begin{array}{l}n_{ H_{ 2 }O}\end{array} \)
in eq (1),

\(\begin{array}{l}\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55}\end{array} \)
= 0.040

\(\begin{array}{l}n_{C_{ 2 }H_{ 5 }OH}\end{array} \)
= 0.040
\(\begin{array}{l}n_{C_{ 2 }H_{ 5 }OH}\end{array} \)
+ (0.040)(55.55)

0.96

\(\begin{array}{l}n_{C_{ 2 }H_{ 5 }OH}\end{array} \)
= 2.222 mol

\(\begin{array}{l}n_{C_{ 2 }H_{ 5 }OH}\end{array} \)
=
\(\begin{array}{l}\frac{ 2.222 }{ 0.96 } \; mol\end{array} \)
\(\begin{array}{l}n_{C_{ 2 }H_{ 5 }OH}\end{array} \)
= 2.314 mol

 

Therefore, molarity of solution

=

\(\begin{array}{l}\frac{ 2.314 \; mol }{ 1 \; L }\end{array} \)

= 2.314 M

 

Q30. What will be the mass of one 12C atom in g?

Ans.

1 mole of carbon atoms

=

\(\begin{array}{l}6.023 \; \times \; 10^{ 23 }\end{array} \)
atoms of carbon

= 12 g of carbon

Therefore, mass of 1 atom of

\(\begin{array}{l}_{}^{ 12 }\textrm{ C }\end{array} \)

=

\(\begin{array}{l}\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}\end{array} \)

=

\(\begin{array}{l}1.993 \; \times \; 10^{ -23 } g\end{array} \)

 

Q31. How many significant figures should be present in the answer of the following calculations?

(i)

\(\begin{array}{l}\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }\end{array} \)

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

 

Ans.

(i)

\(\begin{array}{l}\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }\end{array} \)

Least precise number = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 0.112

= 3

 

(ii) 5 × 5.364

Least precise number = 5.364

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.364

= 4

 

(iii) 0.0125 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4. Hence, the no. of significant numbers in the answer is also 4.

 

Q32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope Molar mass Abundance
\(\begin{array}{l}^{36}Ar\end{array} \)
35.96755
\(\begin{array}{l}g \; mol^{ -1 }\end{array} \)
0.337 %
\(\begin{array}{l}^{38}Ar\end{array} \)
37.96272
\(\begin{array}{l}g \; mol^{ -1 }\end{array} \)
0.063 %
\(\begin{array}{l}^{40}Ar\end{array} \)
39.9624
\(\begin{array}{l}g \; mol^{ -1 }\end{array} \)
99.600 %

 

Ans.

Molar mass of Argon:

= [

\(\begin{array}{l}( 35.96755 \; \times \; \frac{ 0.337 }{ 100 })\end{array} \)
+
\(\begin{array}{l}( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })\end{array} \)
+
\(\begin{array}{l}( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })\end{array} \)
]

= [0.121 + 0.024 + 39.802]

\(\begin{array}{l}g \; mol^{ -1 }\end{array} \)

= 39.947

\(\begin{array}{l}g \; mol^{ -1 }\end{array} \)

 

Q33. Calculate the number of atoms in each of the following

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He

 

Ans.

(i) 52 moles of Ar

1 mole of Ar =

\(\begin{array}{l}6.023 \; \times \; 10^{ 23 }\end{array} \)
atoms of Ar

Therefore, 52 moles of Ar = 52 ×

\(\begin{array}{l}6.023 \; \times \; 10^{ 23 }\end{array} \)
atoms of Ar

=

\(\begin{array}{l}3.131 \; \times \; 10^{ 25 }\end{array} \)
atoms of Ar

 

(ii) 52 u of He

1 atom of He = 4 u of He

OR

4 u of He = 1 atom of He

1 u of He =

\(\begin{array}{l}\frac{ 1 }{ 4 }\end{array} \)
atom of He

52 u of He =

\(\begin{array}{l}\frac{ 52 }{ 4 }\end{array} \)
atom of He

= 13 atoms of He

 

(iii) 52 g of He

4 g of He =

\(\begin{array}{l}6.023 \; \times \; 10^{ 23 }\end{array} \)
atoms of He

52 g of He =

\(\begin{array}{l}\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }\end{array} \)
atoms of He

=

\(\begin{array}{l}7.829 \; \times \; 10^{ 24 }\end{array} \)
atoms of He

 

Q34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.  Find:

(i) Empirical formula

(ii) Molar mass of the gas, and

(iii) Molecular formula

 

Ans.

(i) Empirical formula

1 mole of

\(\begin{array}{l}CO_{ 2 }\end{array} \)
contains 12 g of carbon

Therefore, 3.38 g of

\(\begin{array}{l}CO_{ 2 }\end{array} \)
will contain carbon

=

\(\begin{array}{l}\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g\end{array} \)

= 0.9218 g

 

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

=

\(\begin{array}{l}\frac{ 2 \; g }{ 18 \; g } \; \times 0.690\end{array} \)

= 0.0767 g

 

As hydrogen and carbon are the only elements of the compound. Now, the total mass is:

= 0.9217 g + 0.0767 g

= 0.9984 g

 

Therefore, % of C in the compound

=

\(\begin{array}{l}\frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 100\end{array} \)

= 92.32 %

% of H in the compound

=

\(\begin{array}{l}\frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 100\end{array} \)

= 7.68 %

 

Moles of C in the compound,

=

\(\begin{array}{l}\frac{ 92.32 }{ 12.00 }\end{array} \)

= 7.69

 

Moles of H in the compound,

=

\(\begin{array}{l}\frac{ 7.68 }{ 1 }\end{array} \)

= 7.68

 

Therefore, the ratio of carbon to hydrogen is,

7.69: 7.68

1: 1

 

Therefore, the empirical formula is CH.

 

(ii) Molar mass of the gas

Weight of 10 L of gas at STP = 11.6 g

Therefore, weight of 22.4 L of gas at STP

=

\(\begin{array}{l}\frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L\end{array} \)

= 25.984 g

\(\begin{array}{l}\approx\end{array} \)
26 g

 

(iii) Molecular formula

Empirical formula mass:

CH = 12 + 1

= 13 g

n =

\(\begin{array}{l}\frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}\end{array} \)

=

\(\begin{array}{l}\frac{ 26 \; g }{ 13 \; g}\end{array} \)

= 2

Therefore, molecular formula = 2 x CH =

\(\begin{array}{l}C_{ 2 }H_{ 2 }\end{array} \)
.

 

Q35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) →  CaCl2(aq) + CO2 (g) + H2O(l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Ans.

0.75 M of HCl

≡ 0.75 mol of HCl are present in 1 L of water

≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

=

\(\begin{array}{l}\frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL\end{array} \)

= 0.6844 g

 

Given chemical reaction,

\(\begin{array}{l}CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq)  \; \rightarrow \; CaCl_{ 2 }\;(aq)  \; + \; CO_{ 2 }\; (g)  \; + \; H_{ 2 }O\; (l) \end{array} \)

 

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of

\(\begin{array}{l}CaCO_{ 3 }\end{array} \)
(100 g)

Therefore, amt of

\(\begin{array}{l}CaCO_{ 3 }\end{array} \)
that will react with 0.6844 g

=

\(\begin{array}{l}\frac{ 100 }{ 73 } \; \times \; 0.6844 \; g\end{array} \)

= 0.9375 g

 

Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:

4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Ans.

1 mole of

\(\begin{array}{l}MnO_{2}\end{array} \)
= 55 + 2 × 16 = 87 g

4 mole of HCl = 4 × 36.5 = 146 g

1 mole of

\(\begin{array}{l}MnO_{2}\end{array} \)
reacts with 4 mol of HCl

Hence,

5 g of

\(\begin{array}{l}MnO_{ 2 }\end{array} \)
will react with:

=

\(\begin{array}{l}\frac{146 \; g}{87 \; g} \; \times \; 5 \; g\end{array} \)
HCl

= 8.4 g HCl

Therefore, 8.4 g of HCl will react with 5 g of

\(\begin{array}{l}MnO_{2}\end{array} \)
.

Continue learning about various chemistry topics and Chemistry projects with video lectures by visiting our website or by downloading our App from Google play store. 

Also Access 
NCERT Exemplar for class 11 chemistry Chapter 1
CBSE Notes for class 11 chemistry Chapter 1

About BYJU’S Solutions

The NCERT Solutions that are provided here have been crafted with one sole purpose – to help students prepare for their board examinations and clear them with good results. The NCERT Solutions contain each and every exercise question asked in the NCERT textbook and, therefore, cover many important questions that could be asked in board examinations.

Furthermore, these solutions have been provided by subject matter experts who have made sure to provide detailed explanations for every solution. This makes the solutions provided by BYJU’S very student-friendly and concept-focused. These NCERT Solutions for Class 11 Chemistry can help students develop a strong base for all the important concepts included in the syllabi of both Class 11 as well as competitive exams.

Apart from these solutions, BYJU’S have some of the best subject experts who can guide the students to learn chemistry in a simplified and conceptual manner. In order to help students be successful in their educational journey, BYJU’S tracks the progress of the students by providing regular feedback after periodic assessments. Also, in cases where students face difficulty while going through the NCERT Solutions for Class 11 Chemistry, the BYJU’S support team is always available to clear their doubts.

Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 1

Q1

Why should we download NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry?

The presentation of each solution in Chapter 1 Some Basic Concepts of Chemistry of NCERT Solutions for Class 11 Chemistry, is described in a unique way by the BYJU’S experts in Chemistry. The solutions are explained in understandable language, which improves grasping abilities among students. To score good marks, practising NCERT Solutions for Class 11 Chemistry can help to a great extent. This chapter can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve problems.
Q2

Is BYJU’S website providing answers to NCERT Solutions for Class 11 Chemistry Chapter 1 in a detailed way?

Yes, BYJU’S website provides answers to NCERT Solutions for Class 11 Chemistry Chapter 1 in a step-by-step manner. This helps the students to learn all the concepts in detail, and they can also clear their doubts as well. Regular practising helps them score high in Chemistry board exams.
Q3

Give an overview of questions present in NCERT Solutions for Class 11 Chemistry Chapter 1.

NCERT Solutions for Class 11 Chemistry Chapter 1 has 3 exercises. The concepts of this chapter are listed below.
1. Numerical problems in calculating the molecular weight of compounds.
2. Numerical problems in calculating mass percent and concentration.
3. Problems on empirical and molecular formulae.
4. Problems on molarity and molality.
5. Other problems related to the mole concept (such as percentage composition and expressing concentration in parts per million).
Related Links
NCERT Solutions for Class 11 Physics (All Chapters) NCERT Solutions for Class 11 Chemistry (All Chapters)
NCERT Solutions for Class 11 Maths (All Chapters) NCERT Solutions for Class 11 Biology (All Chapters)

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*

  1. very useful
    easily explained
    (refer byjus only for solutions )

  2. Thank you for the answers with steps

  3. This app is very helpful for me any doubt clear in seconds thanks