NCERT Solutions-Some Basic Concepts of Chemistry Class 11 Notes

February 11, 2025December 10, 2024 by Neeraj Anand, Param Anand

NCERT Solutions for Class 11 Chemistry Chapter Some Basic Concepts of Chemistry are provided for Class 11 Chemistry students. It 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

Importance of Chemistry
Nature of Matter
Properties of Matter and Their Measurement
The International System of Units (SI)
Mass and Weight
Uncertainty in Measurement
Scientific Notation
Significant Figures
Dimensional Analysis
Laws of Chemical Combinations
Law of Conservation Of Mass
Law of Definite Proportions
Law of Multiple Proportions
Gay Lussac’s Law of Gaseous Volumes
Avogadro’s Law
Dalton’s Atomic Theory
Atomic and Molecular Masses
Atomic Mass
Average Atomic Mass
Molecular Mass
Formula Mass
Mole Concept and Molar Masses
Percentage Composition
Empirical Formula for Molecular Formula
Stoichiometry and Stoichiometric Calculations
Limiting Reagent
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.

(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)

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

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 ×

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

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 ×

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

= 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