NCERT Solutions for Class 11 Chemistry: Chapter 1 (Some Basic Concepts of Chemistry) are provided in this page for the perusal of Class 11 chemistry students. These solutions can also be downloaded in a PDF format for free by clicking the download button provided above. 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 chemistry 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 examinations.
NCERT Chemistry – Class 11, 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 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 on calculating the molecular weight of compounds.
- Numerical problems on calculating mass per cent 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 Chemistry NCERT Solutions provided in this page for Class 11 (Chapter 1) provide detailed explanations on the steps to be followed while solving the numerical value questions that are frequently asked in examinations. The subtopics covered under the chapter are listed below.
NCERT Chemistry Class 11 Chapter 1 Subtopics (“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 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.
NCERT Solutions for Class 11 Chemistry Chapter 1
Exercise
Q1. Find out the value of molecular weight of the given compounds:
(i)
Ans.
(i)
Molecular weight of methane,
= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)
= [1(12.011 u) +4 (1.008u)]
= 12.011u + 4.032 u
= 16.043 u
(ii)
Molecular weight of water,
= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u +16.00 u
= 18.016u
So approximately
= 18.02 u
(iii)
= Molecular weight of carbon dioxide,
= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)
= [1(12.011 u) + 2(16.00 u)]
= 12.011 u +32.00 u
= 44.011 u
So approximately
= 44.01u
Q2. Sodium Sulphate (
Ans.
Now for
Molar mass of
= [(2 x 23.0) + (32.066) + 4(16.00)]
=142.066 g
Formula to calculate mass percent of an element =
Therefore, Mass percent of the sodium element:
=
= 32.379
=32.4%
Mass percent of the sulphur element:
=
= 22.57
=22.6%
Mass percent of the oxygen element:
=
=45.049
=45.05%
Q3. Find out the empirical formula of an oxide of iron having 69.9% Fe and 30.1% O_{2} by mass.
Ans.
Percent of Fe by mass = 69.9 % [As given above]
Percent of O_{2} by mass = 30.1 % [As given above]
Relative moles of Fe in iron oxide:
=
=
= 1.25
Relative moles of O in iron oxide:
=
=
= 1.88
Simplest molar ratio of Fe to O:
= 1.25: 1.88
= 1: 1.5
Therefore, empirical formula of iron oxide is
Q4. Find out the amount of CO_{2} that can be produced when
(i) 1 mole carbon is burnt in air.
(ii) 1 mole carbon is burnt in 16 g of O_{2}.
(iii) 2 moles carbon are burnt in 16 g O_{2}.
Ans.
(i) 1 mole of carbon is burnt in air.
1 mole of carbon reacts with 1 mole of O_{2} to form one mole of CO_{2}.
Amount of
(ii) 1 mole of carbon is burnt in 16 g of O_{2}.
1 mole of carbon burnt in 32 grams of O_{2} it forms 44 grams of
Therefore, 16 grams of O_{2} will form
= 22 grams of
(iii) 2 moles of carbon are burnt in 16 g of O_{2}.
Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O_{2} will react with 6g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18g of carbon (1.5 mol) will not undergo combustion.
Q5. Find out the mass of
Ans.
0.375 Maqueous solution of
= 1000 mL of solution containing 0.375 moles of
Therefore, no. of moles of
=
= 0.1875 mole
Molar mass of sodium acetate =
Therefore, mass that is required of
=
= 15.38 gram
Q6. A sample of HNO_{3} has a density of
Ans.
Mass percent of HNO_{3} in sample is 69 %
Thus, 100 g of HNO_{3} contains 69 g of HNO_{3} by mass.
Molar mass of HNO_{3}
= { 1 + 14 + 3(16)}
= 1 + 14 + 48
Now, No. of moles in 69 g of
=
= 1.095 mol
Volume of 100g HNO_{3} solution
=
=
= 70.92mL
=
Concentration of HNO_{3}
=
= 15.44mol/L
Therefore, Concentration of HNO_{3} = 15.44 mol/L
Q7. How much Cu (Copper) can be obtained from 100 gram of
Ans.
1 mole of
Molar mass of
= (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00
= 159.5 gram
159.5 gram of
Therefore, 100 gram of
=
=39.81 gram
Q8. The mass percent of iron and oxygen in an oxide of iron is 69.9 and 30.1 calculate the molecular formula of the oxide of iron.
Ans.
Here,
Mass percent of Fe = 69.9%
Mass percent of O = 30.1%
No. of moles of Fe present in oxide
=
= 1.25
No. of moles of O present in oxide
=
=1.88
Ratio of Fe to O in oxide,
= 1.25: 1.88
=
=
=
Therefore, the empirical formula of oxide is
Empirical formula mass of
= [2(55.85) + 3(16.00)] gr
= 159.69 g
Therefore n =
= 0.999
= 1(approx)
The molecular formula of a compound can be obtained by multiplying n and the empirical formula.
Thus, the empirical of the given oxide is
Q9. Find out the atomic mass (average) of chlorine using the following data:
Percentage Natural Abundance | Molar Mass | |
75.77 | 34.9689 | |
24.23 | 36.9659 |
Ans.
Average atomic mass of Cl.
=[(Fractional abundance of
=[{(
= 26.4959 + 8.9568
= 35.4527 u
Therefore, the average atomic mass of Cl = 35.4527 u
Q10. In 3 moles of ethane (
(a) No. of moles of C- atoms
(b) No. of moles of H- atoms.
(c) No. of molecules of C_{2}H_{6}.
Ans.
(a) 1 mole
= 2 * 3
= 6
(b) 1 mole
= 3 * 6
= 18
(c) 1 mole
= 3 * 6.023 *
= 18.069 *
Q11. What is the concentration of sugar
Ans.
Molarity (M) is as given by,
=
=
=
=
=
= 0.02925 mol
Therefore, Molar concentration = 0.02925 mol
Q12. The density of CH_{3}OH (methanol) is 0.793 kg
Ans.)
Molar mass of
= (1 * 12) + (4 * 1) + (1 * 16)
= 32 g
= 0.032 kg
Molarity of the solution
=
= 24.78 mol
(From the definition of density)
Q13. Pressure is defined as force per unit area of the surface. Pascal, the SI unit of pressure is as given below:
1 Pa = 1 N
Assume that mass of air at the sea level is 1034 gcm^{-2}. Find out the pressure in Pascal.
Ans.
As per definition, pressure is force per unit area of the surface.
P =
=
= 1.01332 ×
Now,
1 N = 1 kg m
Then,
1 Pa = 1
= 1
Pa = 1
Q14. Write SI unit for mass. Also define mass.
Ans.
Si Unit: Kilogram (kg)
Mass:
“The mass equal to the mass of the international prototype of kilogram is known as mass.”
Q15. Match the prefixes with their multiples in the table given below:
Prefixes | Multiples | |
(a) | femto | 10 |
(b) | giga | |
(c) | mega | |
(d) | deca | |
(e) | micro |
Ans.
Prefixes | Multiples | |
(a) | femto | |
(b) | giga | |
(c) | mega | |
(d) | deca | 10 |
(e) | micro |
Q16. What are 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 the shows the uncertainty of the result is known as significant figures.”
Q17. A sample of drinking water was found to be highly contaminated with
(a) Express in terms of percent by mass.
(b) Calculate the molality of chloroform in the given water sample.
Ans.
(a) 1 ppm = 1 part out of 1 million parts.
Mass percent of 15 ppm chloroform in H_{2}O
=
=
(b) 100 gram of the sample is having 1.5 ×
1000 gram of the sample is having 1.5 ×
=
Molar mass (
= 12 + 1 + 3 (35.5)
= 119.5 gram
Therefore, molality of
= 1.25 ×
Q18. Express the given number in scientific notation:
(a) 0.0047
(b) 235,000
(c)8009
(d)700.0
(e) 5.0013
Ans.
(a) 0.0047= 4.7 ×
(b) 235,000 = 2.35 ×
(c) 8009= 8.009 ×
(d) 700.0 = 7.000 ×
(e) 5.0013 = 5.0013
Q19. Find the number of significant figures in the numbers given belowt.
(a) 0.0027
(b) 209
(c)6005
(d)136,000
(e) 900.0
(f)2.0035
Ans.
(i) 0.0027: 2 significant numbers.
(ii) 209: 3 significant numbers.
(iii)6005: 4 significant numbers.
(iv)136,000:3 significant numbers.
(v) 900.0: 4 significant numbers.
(vi)2.0035: 5 significant numbers.
Q20. Round up the given numbers upto 3 significant numbers.
(a) 35.217
(b) 11.4108
(c)0.05577
(d)2806
Ans.
(a) The number after round up is: 35.2
(b) The number after round up is: 11.4
(c)The number after round up is: 0.0560
(d)The number after round up is: 2810
Q21. When dioxygen and dinitrogen react together, they form various compounds. The information is given below:
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 |
(1) In the data given above, which chemical combination law is obeyed? Also give the statement of the law.
(2) Convert the following:
(a) 1 km =_____ mm = _____ pm
(b) 1 mg = _____ kg = _____ ng
(c) 1 mL = _____ L =_____
Ans.
(1) If we fix the mass of N_{2} at 28 g, the masses of N_{2} that will combine with the fixed mass of N_{2} are 32 gram, 64 gram, 32 gram and 80 gram.
The mass of O_{2} bear whole no. ratio of 1: 2: 2: 5. Therefore, the given information obeys the law of multiple proportions.
The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”
(2) Convert:
(a) 1 km = ____ mm = ____ pm
- 1 km = 1 km *
$\frac{ 1000 \; m }{ 1 \; km }$ ×$\frac{ 100 \; cm }{ 1 \; m }$ *$\frac{ 10 \; mm }{ 1 \; cm }$
- 1 km = 1 km *
$\frac{ 1000 \; m }{ 1 \; km }$ *$\frac{1 \; pm}{10^{ -12 } \; m}$
Therefore, 1 km =
(b) 1 mg = ____ kg = ____ ng
- 1 mg = 1 mg *
$\frac{ 1 \; g }{ 1000 \; mg }$ *$\frac{ 1 \; kg }{ 1000 \; g }$
1 mg =
- 1 mg = 1 mg *
$\frac{ 1 \; g }{ 1000 \; mg }$ *$\frac{ 1 \; ng }{ 10^{ -9 } \; g }$
1 mg =
Therefore, 1 mg =
(c) 1 mL = ____ L = ____
- 1 mL = 1 mL *
$\frac{1 \; L}{1000 \; mL}$
1 mL =
- 1 mL = 1
$cm^{ 3 }$ = 1 *$\frac{1 \; dm \; \times \; 1 \; dm \; \times \;1 \; dm }{10 \; cm \; \times \; 10 \; cm \; \times \; 10 \; cm } cm^{ 3 }$
1 mL =
Therefore, 1 mL =
Q22. What is the distance covered by the light in 2 ns if the speed of light is 3 ×
Ans.
Time taken = 2 ns
= 2 ×
Now,
Speed of light = 3 ×
So,
Distance travelled in 2 ns = speed of light * time taken
= (3 ×
= 6 ×
= 0.6 m
Q23. In the reaction given below:
Find the limiting reagent if it is present in the reactions given below:
(a) 2 mol X + 3 mol Y
(b) 100 atoms of X + 100 molecules of Y
(c) 300 atoms of X + 200 molecules of Y
(d) 2.5 mol X + 5 mol Y
(e) 5 mol X + 2.5 mol Y
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 limiting the amt. of products formed.
(a) 2 mol X + 3 mol Y
1 mole of X reacts with 1 mole of Y. Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. Hence, X is limiting agent.
(b) 100 atoms of X + 100 molecules of Y
1 atom of X reacts with 1 molecule of Y. Similarly, 100 atoms of X reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting agent.
(c) 300 atoms of X + 200 molecules of Y
1 atom of X reacts with 1 molecule of Y. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. Hence, Y is limiting agent.
(d) 2.5 mol X + 5 mol Y
1 mole of X reacts with 1 mole of Y. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. Hence, X is limiting agent.
(e) 5 mol X + 2.5 mol Y
1 mole of X reacts with 1 mole of Y. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. Hence, Y is limiting agent.
Q24. H_{2} and N_{2} react with each other to produce NH_{3} according to the given chemical equation
(a) What is the mass of
(b) Will the reactants N_{2} or H_{2 }remain unreacted?
(c) If any, then which one and give it’s mass.
Ans.
(a) Balance the given equation:
Thus, 1 mole (28 g) of N_{2} reacts with 3 mole (6 g) of H_{2} to give 2 mole (34 g) of
Given:
Amt of H_{2} =
Therefore,
28 g of
Therefore, mass of
=
(b)
(c) Mass of H_{2} unreacted
=
= 571.4 g
Q25. 0.50 mol
Ans.
Molar mass of
= (2 × 23) + 12 + (3 × 16)
= 106 g
1 mole of
Therefore, 0.5 mol of
=
= 53 g of
0.5 M of
Hence, 0.5 mol of
Q26. If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of vapour would be obtained?
Ans.
Reaction:
2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour.
Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of vapour.
Q27. Convert the given quantities into basic units:
(i) 29.7 pm
(ii) 16.15 pm
(iii) 25366 mg
Ans.
(i) 29.7 pm
1 pm =
29.7 pm = 29.7 ×
= 2.97 ×
(ii) 16.15 pm
1 pm =
16.15 pm = 16.15 ×
= 1.615 ×
(iii) 25366 mg
1 mg =
25366 mg = 2.5366 ×
25366 mg = 2.5366 ×
Q28. Which of the given below have the largest no. of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of
Ans.
(i) 1 g Au (s)
=
=
= 3.06
(ii) 1 g Na (s)
=
=
= 0.262
= 26.2
(iii) 1 g Li (s)
=
=
= 0.86
= 86.0
(iv)1 g of
=
(Molar mass of
=
= 0.0848
= 8.48
Therefore, 1 g of Li (s) will have the largest no. of atoms.
Q29. What is the molarity of the solution of ethanol in water in which the mole fraction of ethanol is 0.040?
(Assume the density of water to be 1)
Ans.
Mole fraction of
=
0.040 =
No. of moles present in 1 L water:
Substituting the value of
0.96
Therefore, molarity of solution
=
= 2.314 M
Q30. Calculate the mass of 1
Ans.
1 mole of carbon atoms
=
= 12 g of carbon
Therefore, mass of 1
=
=
Q31. How many significant numbers should be present in answer of the given calculations?
(i)
(ii) 5 × 5.365
(iii) 0.012 + 0.7864 + 0.0215
Ans.
(i)
Least precise no. of calculation = 0.112
Therefore, no. of significant numbers in the answer
= No. of significant numbers in the least precise no.
= 3
(ii) 5 × 5.365
Least precise no. of calculation = 5.365
Therefore, no. of significant numbers in the answer
= No. of significant numbers in 5.365
= 4
(iii) 0.012 + 0.7864 + 0.0215
As the least no. of decimal place in each term is 4, the no. of significant numbers in the answer is also 4.
Q32. Calculate molar mass of Argon isotopes according to the data given in the table.
Isotope | Molar mass | Abundance |
35.96755 |
0.337 % | |
37.96272 |
0.063 % | |
39.9624 |
99.600 % |
Ans.
Molar mass of Argon:
= [
= [0.121 + 0.024 + 39.802]
= 39.947
Q33. What is the number of atoms in the following compounds?
(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 =
Therefore, 52 mol of Ar = 52 ×
=
(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 =
52 u of He =
= 13 atoms of He
(iii) 52 g of He
4 g of He =
52 g of He =
=
Q34. A welding fuel gas contains hydrogen and carbon. If we burn a small sample, we get 3.38 g of carbon dioxide and 0.69 g of water. A volume of 10 L (at STP) of this welding gas weighs 11.6 g.
Find:
(i) Empirical formula
(ii) Molar mass of the gas, and
(iii) Molecular formula
Ans.
(i) Empirical formula
1 mole of
Therefore, 3.38 g of
=
= 0.9217 g
18 g of water contains 2 g of hydrogen
Therefore, 0.690 g of water will contain hydrogen
=
= 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
=
= 92.32 %
% of H in the compound
=
= 7.68 %
Moles of C in the compound,
=
= 7.69
Moles of H in the compound,
=
= 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, and
Weight of 10 L of gas at STP = 11.6 g
Therefore, weight of 22.4 L of gas at STP
=
= 25.984 g
(iii) Molecular formula
Empirical formula mass:
CH = 12 + 1
= 13 g
n =
=
= 2
Therefore, molecular formula is
Q35. Calcium carbonate reacts with aqueous HCl and gives
Calculate the mass of
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 contins 27.375 g of HCl
Therefore, amt of HCl present in 25 mL of solution
=
= 0.6844 g
Given chemical reaction,
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of
Therefore, amt of
=
= 0.9639 g
Q36. Chlorine is prepared by adding manganese dioxide with hydrochloric acid acc. to the reaction.
How many grams of HCl react with 5 g of manganese dioxide?
Ans.
1 mol of
4 mol of HCl = 4 × 36.5 = 146 g
1 mol of
5 g of
=
= 8.4 g HCl
Therefore, 8.4 g of HCl will react with 5 g of
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