Important Questions for CBSE Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry

Important Questions of Chemistry Class 11 Chapter 1 – Some Basic Concepts of Chemistry

Chemistry is the experimental study of atoms, molecules, compounds, chemical equations and their reactions. One who has command over all these concepts excels in Chemistry. 

Some basic concepts of Chemistry include the study of atoms, molecules, compounds, moles etc. which lays a strong foundation for other chapters of Class 11 and Class 12 Chemistry textbooks. The other vital topics included in Chapter 1 Class 11 Chemistry are:

• Importance Of Chemistry
• Nature Of Matter
• Properties Of Matter And Their Measurement
• Uncertainty In Measurement
• Laws Of Chemical Combinations
• Dalton’s Atomic Theory
• Atomic And Molecular Masses
• Mole Concept And Molar Masses
• Percentage Composition
• Stoichiometry And Stoichiometric Calculations

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Solving questions is crucial for any student to revise chapters and understand their weak sections. Extramarks’ expert team of Chemistry teachers have created Important Questions Class 11 Chemistry Chapter 1 and have given step-by-step solutions for all questions covered. The questions are collected from many different sources such as NCERT textbooks, NCERT Exemplar books, past years’ question papers, etc. The questions are chosen in such a way that all the topics covered in the chapter are well revised while solving our Important Questions Class 11 Chemistry Chapter 1. 

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Chemistry Class 11 Chapter 1 Question Answer

Below are a few of the Important Questions and their solutions which are included in our Chemistry Class 11 Chapter 1 Important Questions:

Question 1. Two students performed the same experiment separately, and each one of them recorded two separate readings of mass which are given below. The right reading of the mass is 3.0 g. Based on the provided data below, mark the correct option for the following statements.

Student Readings

(i) Results of both the students are found neither accurate nor precise.

(ii) Results of student A are found both precise and accurate.

(iii) Results of student B are found neither precise nor accurate.

(iv) Student B’s results are both precise and accurate.

Answer 1. Option (ii) is the correct answer.

Question 2. The measured temperature on the Fahrenheit scale is 200 °F. What would the reading

be on the Celsius scale?

(i) 40 °C

(ii) 94 °C

(iii) 93.3 °C

(iv) 30 °C

Answer 2. Option (iii) is the correct answer.

Question 3. What will be the molarity of the solution that contains 5.85 g of NaCl(s) per 500 mL?

(i) 4 mol L-1

(ii) 20 molL-1

(iii) 0.2 molL-1

(iv) 2molL-1

Answer 3. Option (iii) is the correct answer.

Question 4. When 500 mL of a given 5M solution is diluted to 1500 mL, what would be the molarity of the new solution?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

Answer 4. Option (ii) is the correct answer.

Question 5. The number of atoms in an element’s mole equals the Avogadro number. Which among the following element contains the greatest number of atoms?

(i) 4g He

(ii) 0.40g Ca

(iii) 46g Na

(iv) 12g He

Answer 5. Option (iv) is the correct answer.

Question 6. When the glucose concentration (C6H12O6) in the blood is 0.9 g L-1, what would be the molarity of glucose found in the blood?

(i) 0.5 M

(ii) 50 M

(iii) 0.005 M

(iv) 5 M

Answer 6. Option (iii) is the correct answer.

Question 7. What would be the molality for the solution obtained containing 18.25 g of HCl gas in 500 g of water?

(i) 0.5 m

(ii) 1 M

(iii) 0.1 m

(iv) 1 m

Answer 7. Option (iv) is the correct answer.

Question 8. One mole of any given substance contains 6.022 × 1023 atoms/molecules. A number of molecules of H2SO4 that are present in 100 mL of 0.02M H2SO4 solution are ______.

(i) 12.044 × 1020 molecules

(ii) 12.044 × 1023 molecules

(iii) 1 × 1023 molecules

(iv) 6.022 × 1023molecules

Answer 8. Option (i) is the correct answer.

Question 9. What is the mass percent of the carbon found in carbon dioxide?

(i) 0.034%

(ii) 27.27%

(iii) 3.4%

(iv) 28.7%

Answer 9. Option (ii) is the correct answer.

Question 10. The compound’s empirical formula and molecular mass found are CH2O and 180 g, respectively. What would be the molecular formula of the given compound?

(i) C9H18O9

(ii) CH2O

(iii) C6H12O6

(iv) C2H4O2

Answer 10. Option (iii) is the correct answer.

Question 11. When the density of the solution is 3.12 g mL-1, the mass of 1.5 mL solution in the significant figures is _______.

(i) 4.7g

(ii) 4680 × 10-3 g

(iii) 4.680g

(iv) 46.80g

Answer 11. Option (i) is the correct answer

Question 12. Which of the following statements given about the compound is incorrect?

(i) A molecule of the given compound has atoms of the different elements.

(ii) A compound can’t be separated into its constituent elements through physical separation methods.

(iii) A compound retains the required physical properties of its constituent elements.

(iv) The ratio of atoms for the different elements in the compound is fixed.

Answer 12. Option (iii) is the correct answer.

Question 13. Which among the following statements is right about the reaction given below:

4Fe(s) + 3O2(g) → 2Fe2O3(g)

(i) The total mass of given iron and oxygen in reactants = total mass of iron and oxygen in the product; thus, it follows the law of conservation of mass.

(ii) The total mass of the given reactants = the total mass of product; thus, the law of multiple proportions followed.

(iii) Amount of Fe2O3 could increase by taking any of the reactants given (iron or oxygen) in excess.

(iv) Amount of Fe2O3 produced will decrease when the amount of any of the given reactants (iron as well as oxygen) taken is in excess.

Answer 13. Option (i) is the correct answer.

Question 14. Which among the following reactions is incorrect as per the law of conservation of the mass.

(i) 2Mg(s) + O2(g) →2MgO(s)

(ii) C3H8(g) + O2(g) → CO2(g) + H2O(g)

(iii) P4(s) + 5O2(g) → P4O10(s)

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O (g)

Answer 14. Option (ii) is the correct answer.

Question 15. The following statements indicate the law of the multiple proportions.

(i) Sample of given carbon dioxide taken from any source will always have carbon and oxygen in the ratio of 1:2.

(ii) Carbon forms the two oxides, namely CO2 and CO. Here, masses of oxygen that combine with the fixed mass of carbon are in the simple ratio of 2:1.

(iii) When magnesium burns in the given oxygen, the amount of magnesium that took for the reaction is the same as the amount of magnesium found in the magnesium oxide formed.

(iv) At the given constant temperature and pressure, 200 mL of hydrogen would combine with 100 mL of oxygen to produce 200 mL of water vapour.

Answer 15. Option (ii) is the correct answer.

Question 16. One mole of oxygen gas formed at STP is equal to _______.

(i) 6.022 × 1023 molecules of the oxygen

(ii) 6.022 × 1023 atoms of the oxygen

(iii) 16 g of the oxygen

(iv) 32 g of the oxygen

Answer 16. Option (i) and (iv) are the correct answers.

Question 17. Sulphuric acid reacts with the given sodium hydroxide as follows as given below:

H2SO4 + 2NaOH → Na2SO4+ 2H2O 

if 1L of 0.1M sulphuric acid solution is allowed to react with the given 1L of the 0.1M sodium hydroxide solution, the amount of the sodium sulphate formed will be, and its molarity in the solution got

(i) 0.1 mol L-1

(ii) 7.10 g

(iii) 0.025 mol L-1

(iv) 3.55 g

Answer 17. Option (ii) and (iii) are the right answers.

Question 18. Which among the following pairs has the same number of atoms?

(i) 16 g of O2(g) as well as 4 g of H2(g)

(ii) 16 g of O2 as well as 44 g of CO2

(iii) 28 g of N2 as well as 32 g of O2

(iv) 12 g of C(s) as well as 23 g of Na(s)

Answer 18. Option (iii) and (iv) are the right answers.

Question 19. Which among the following solutions has same concentration?

(i) 20 g of the NaOH in 200 mL of the solution

(ii) 0.5 mol of KCl in 200 mL of the solution

(iii) 40 g of NaOH in 100 mL of the solution

(iv) 20 g of KOH in 200 mL of the solution

Answer 19. Option (i) and (ii) are the right answers.

Question 20. 16 g of given oxygen has the same number of molecules as in

(i) 16 g of CO

(ii) 28 g of N2

(iii) 14 g of N2

(iv) 1.0 g of H2

Answer 20. Option (iii) and (iv) are the right answers.

Question 21. Which among the following terms is unitless?

(i) Molality

(ii) Molarity

(iii) Mole fraction

(iv) Mass percent

Answer 21. Option (iii) and (iv) are the right answers.

Question 22. One of the following statements of Dalton’s atomic theory given below:

“Compounds are formed for the following when atoms of different elements combine in a fixed ratio.”

Which among the following laws is not related to this statement?

(i) Law of conservation of mass

(ii) Law of definite proportions

(iii) Law of multiple proportions

Answer 22. Option (i) and (iv) are the correct answers.

Question 23. What would be the mass of one atom of given C-12 in grams?

Answer 23. 1 mole of given carbon atom = 12g= 6.022 × 1023 atoms.

Question 24. How many significant figures should be there to answer the following given calculations?

2.5 1.25 3.5/2.01

Answer 24. Two significant figures should be there in this.

As the least number of significant figures from the given figure is 2 (in 2.5 and 3.5).

Question 25. What is the symbol for the SI unit of the given mole? How can the mole be defined?

Answer 25. The symbol for the SI unit for the given mole is mol. The mole is defined as the amount of substance that contains as many entities as there are atoms in the given 12g carbon.

Question 26. What is the difference between the term molality and molarity?

Answer 26. Molarity is the provided number of moles of solute dissolved in 1 litre of the solution. Molality is the provided number of moles of solute present in 1kg of the solvent.

Question 27. Calculate the mass percent of the calcium, phosphorus and oxygen in the given calcium phosphate Ca3(PO4)

Answer 27. Molecular mass of the given Ca3(PO4) = 3*40+2*31+8*16 =310

Mass percent of the given Ca = 3*40/310*100 = 38.71%

Mass percent of the given P = 2*31/310*100 = 20%

Mass percent of the given O = 8*16/310 = 41.29%

Question 28. 45.4 L of dinitrogen, when reacted with 22.7 L of the given dioxygen and 45.4 L of the nitrous oxide, was formed. The reaction is given below as:

2N2(g) + O2(g) → 2N2O(g)

Which law obey for the following experiment? Write the statement of the given law.

Answer 28. The above experiment proves the Gay-Lussac’s law which states that the given gases combine or are produced in the chemical reaction in a simple whole-number ratio by volume, provided that all the given gases are at the same temperature and pressure.

Question 29. If two given elements can combine to form more than one compound, the masses of one element, which combines with the fixed mass of the other element, are in the whole-number ratio.

(a) Is this statement true?

(b) If yes, according to which given law?

(c) Give one example related to this law.

Answer 29. (a) Yes, the statement is true.

(b) As per the law of multiple proportions

(c), hydrogen and oxygen react to form given water and hydrogen peroxide

H2 + 1/2O2 → H2O

H2 + O2 → H2O2

Masses of oxygen which combine with the fixed mass of hydrogen are in the ratio of 16:32 

Question 30. Calculate the average atomic mass of the given hydrogen using the following data :

Answer 30. 

Average atomic mass is equal to

= 99.985*1+0.015*2/100

= 099.985*1+0.015*2/100

=1.00015u

Question 31. Hydrogen gas is prepared in the laboratory by reacting the given diluted HCl with granulated zinc. The following reaction takes place.

Zn + 2HCl → ZnCl2 + H2

Calculate the volume of the given hydrogen gas liberated at STP if 32.65 g of zinc reacts with HCl. 1 mol of the given gas occupies 22.7 L volume at STP; atomic mass of the Zn = 65.3 u.

Answer 31. 1 mol of the given gas occupies = 22.7L Volume at STP atomic mass of Zn = 65.3u

From the above-given equation,

65.3g of Zn if reacts with HCl produces = 22.7L H2 at STP

Thus, 32.65g of Zn, when reacted with the given HCl, will produce 

= 22.7 * 32.65/65.3 

= 11.35L of H2 at STP

Question 32. The density of 3 molal solutions of given NaOH is 1.110 g mL–1. Calculate the required molarity of the solution.

Answer 32. 3 molal solution of given NaOH = 3 moles of given NaOH dissolved in 1000g water

3 mole of given NaOH = 3*40g = 120g

Density of the solution = 1.110gmL-1

thus, Volume = mass/density = 1120g/1.110gmL-1 =1.009L

Molarity of the given solution = 3/1.009 = 2.97M

Question 33. The volume of the solution changes with a change in the temperature. Then, would the molality of the solution be affected by the temperature? Give the reason for your answer.

Answer 33. Mass does not change because the temperature changes. Thus, the molality of the solution does not change.

Molality = moles of the solute/ weight of the solvent (in g) *1000

Question 34. If 4 g of the NaOH dissolves in 36 g of H2O, calculate the mole fraction of each

component in the solution. Here, determine the molarity of the solution (specific

gravity of solution is 1g mL–1).

Answer 34. Mole fraction of the H2O = number of moles of the H2O/ Total no: of moles (H2O+NaOH)

No: of moles of the H2O = 36/18=2 moles

No: of moles of the NaOH = 4/40=0.1mol

Total number of the moles = 2+0.1= 2.1

Mole fraction of the H2O = 2/2.1 = 0.952

Mole fraction of the NaOH = 0.1/2.1 = 0.048

Mass of the solution = Mass of H2O + Mass of NaOH = 36+4=40G

Volume of the solution = 40/1 = 40mL

Molarity = 0.1/0.04 = 2.5M

Question 35. The reactant completely consumed in the reaction is called the limiting reagent.

In the reaction 2A + 4B → 3C + 4D, if 5 moles of A react with 6 moles of B,

then

(i) which is the limiting reagent?

(ii) calculate the amount of C formed?

Answer 35. (i) B would be the limiting reagent as it gives the lesser amount of product.

(ii) Let B completely consumed 

4 mol of B gives 3 mol of C

6 mol of B will give 3/4 *6 mol C =4.5 mol C

Question 36. Match The Following Type

Answer 36. A → b

B → c

C → a

D → e

E → d

Question 37. Match the following

Answer 37. (i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

Question 38. Assertion (A): Ethene’s empirical mass is half its molecular mass.

Reason (R): The empirical formula presents the simplest whole-number

the ratio of various atoms present in the compound.

(i) Both A as well as R are true, and R is the correct explanation of A.

(ii) A is true, as well as R is false.

(iii) A is false, as well, as R is true.

(iv) Both A as well as R are false.

Answer 38. Option (i) is correct.

Question 39. Assertion (A): One atomic mass unit has defined as one-twelfth of the mass of

one carbon-12 atom.

Reason (R): Carbon-12 isotope is the most abundant isotope of carbon

and has been chosen as the standard.

(i) Both A and R are true, and R is the correct explanation of A.

(ii) Both A and R are true, but R is not the correct explanation of A.

(iii) A is true, but R is false.

(iv) Both A and R are false.

Answer 39. Option (ii) is correct. Carbon-12 is considered a standard for defining atomic and molecular mass.

Question 40. Assertion (A): Significant figures for the 0.200 is 3 here; for 200 it is 1.

Reason (R): Zero at the end or the right of the number is significant, provided

they are not on the right side for the decimal point.

(i) Both A and R are true, as well as R is the correct explanation of A.

(ii) Both A and R are true, as well as R is not a correct explanation of A.

(iii) A is true, as well as R is false.

(iv) Both A as well as R are false.

Answer 40. Option (iii) is correct. Significant figures for the 0.200 = 3 and for 200 =1

Zero at the end of the number without a decimal point may or may not be significant depending upon the accuracy of the measurement.

Question 41. Assertion (A): Combusting 16 g of methane gives 18 g of water.

Reason (R): In the combustion of methane, water is one of the products.

(i) Both A and R are true, but R is not the correct explanation of A.

(ii) A is true, but R is false.

(iii) A is false, but R is true.

(iv) Both A and R are false.

Answer 41. Option (iii) is correct.

16g of CH4 on complete combustion will give 36g of water.

Question 42. A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The

gas transfers to another vessel at a constant temperature, where

the pressure becomes half of the original pressure. Calculate

(i) the volume of the new vessel.

(ii) a number of molecules of dioxygen.

Answer 42. (i) Moles of the oxygen = 1.6/32 = 0.05mol

At STP, 1 mol of the O2 = 22.4L

now, volume of O2 = 22.4 × 0.05 = 1.12L

V1 = 1.12L

V2 =?

P1 = 1atm

P2 = ½ = 0.5atm

As per, Boyle’s law, P1V1 = P2V2

thus, Substituting the value

V2 = 1 × 1.12/0.5 = 2.24L

(ii) Number of molecules in the 1.6g or 0.005mol = 6.022 × 1023 × 0.05 = 3.011 ×  1022

Question 43. Calcium carbonate reacts with the aqueous HCl to give CaCl2 and CO2 according

to the reaction given below:

CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

What would be form masses of CaCl2 if 250 mL of 0.76 M HCl reacts with

1000 g of the CaCO3? Name the limiting reagent. Calculate the number of the moles

of CaCl2 formed in the reaction.

Answer 43. No: of moles of the HCl taken = MV/1000 = 0.76*250/1000 = 0.19

No: of moles of the CaCO3 = Mass/Molar mass = 1000/100 = 10

1. if CaCO3 is completely consumed

1 mol of CaCO3 = 1 mol CaCl2

10 mol CaCO3 = 10mol CaCl2

2. if HCl will consume completely.

2 mol HCl = 1 mol CaCl2

0.19mol of HCl = ½  × 0.19mol CaCl2 = 0.095 mol CaCl2

HCl would be the limiting reagent, and the number of moles of CaCl2 formed will be 0.095mol

Question 44. A box contains a few identical red coloured balls that have to label as A, each weighing

2 grams. Another box contains an identical blue coloured ball and labelled as B,

each weighing 5 grams. Consider the combinations of AB, AB2, A2B and A2B3

and show that the law of the multiple proportions is applicable.

Answer 44. 

Masses of B, which combine with a fixed mass of A, are

10g, 20g, 5g, 15g

2: 4: 1 : 3

This is the simple whole-number ratio.

Question 45. 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 of A + 3 mol of B

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

(iv) 5 mol of A + 2.5 mol of B

(v) 2.5 mol of A + 5 mol of B

 Answer 45. Limiting reagent:

It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causing the reaction to stop and limiting 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 react 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 react 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 react with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting reagent.

 (iv) 5 mol of A + 2.5 mol of B

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

 (v) 2.5 mol of A + 5 mol of B

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

Question 46. Define the law of multiple proportions. Explain it with the two examples. How does this law point to the existence of atoms?

Answer 46. Dalton first studied the law of multiple proportions in 1803, and it can be found and follows.

If two elements combine to form two or more chemical compounds, the masses of one of the elements combine with the fixed mass of the other in a simple ratio.

For example, hydrogen reacts with oxygen to form two compounds: water and hydrogen peroxide.

In this case, the masses of the oxygen (i.e. 16g and 32g) that combine with the fixed mass of hydrogen (2g) have a simple ratio, i.e. 16:32 or 1:2.

As we all know, if compounds mix in different proportions, they form different compounds. For example, if hydrogen mixes with a different proportion of oxygen, it forms water or hydrogen peroxide.

It demonstrates that there are constituents that combine in a specific manner. This constituent could be atoms. As a result, the law of multiple proportions demonstrates the existence of atoms that combine to form the molecules.

Question 47. A box contains a few identical red coloured balls, labelled as A, each weighing equal to 2 grams. The other box contains identical blue coloured balls, labelled for the following as B, each weighing 5 grams. Consider the following combinations AB, AB2, A2B and A2B3, and show that the multiple proportions law applies.

Answer 47. 

When the two elements combine to form the given two or more compounds, the different masses of one element, which combine with a fixed mass of the other, bear a simple ratio to one another, as per the law of multiple proportions.

The mass of B, if combined with the fixed mass of A (say 1g), is 2.5g, 5g, 1.25g, and 3.75g. They have a 2:4:1 ratio, which is a simple whole-number ratio. Hence. The multiple proportions law is applicable.

Question 48. Assertion (A): an atomic mass unit defined as one-twelfth of the mass of a given one carbon-12 atom.

Reason (R): Carbon-12 isotope has the most abundant isotope of the given carbon and choice as standard.

(i) Both A as well as R are true, and R is the correct explanation of A.

(ii) Both A as well as R are true, but R is not the correct explanation for A.

(iii) A is true, as well as R is false.

(iv) Both A and R are false.

Answer 48. The correct option is (i) Both A and R are true, and R is the correct explanation of A.

As C-12 is used as the standard atom, one atomic mass unit is defined as one-twelfth of the mass of the given one carbon – 12 atom. That is because it has an equal number of protons and neutrons (6) and forms the majority of matter.

Carbon-12 is the most abundant isotope of all carbon’s isotopes.

Question 49. Assertion (A): Significant figures for the number 0.200 are 3, whereas, for the number 200, it is 1.

Reason (R): Zero at the end or right of the number is significant, provided they don’t give on the correct side of the decimal point.

(1) Both A as well as R are true, and R is the correct explanation for A.

(ii) Both A as well as R are true, but R is not the correct explanation for A.

(iii) A is true, as well as R is false.

(iv) Both A as well as R are false.

Answer 49. The correct option is (iii) A is true, as well as R is false.

Zero at the end, as well as to the right of the number, is significant when this is on the correct side of the decimal point. For example, 0.200 has 3 significant figures.

Question 50. Assertion (A): Combustion of 16 g for methane gives 18 g of water.

Reason (R): Water is one of the products in the combustion of methane.

(i) Both A as well as R are true, but R is not the correct explanation for A.

(ii) A is true, as well as R is false.

(iii) A is false, as well, as R is true.

(iv) Both A as well as R are false.

Answer 50. The correct option is (iii) A is false, but R is true.

CH4 + O2 → CO2 + 2H2O

Water produces during the combustion for methane, but 16 g of methane on complete combustion gives 36 g of water.

Question 51. Calculate the molecular mass of the following :

(i) H2O

(ii) CO2

(iii) CH4

Answer 51. (i)CH4 :

The molecular weight of methane, CH4

= (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) H2O :

The molecular weight of water, H2O

= (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) CO2 :

= Molecular weight of carbon dioxide, CO2

= (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

Question 52. The following data obtained if dinitrogen and dioxygen react together to form different compounds:

 Mass of dinitrogen          Mass of dioxygen

(i)14 g                                      16 g

(ii)14 g                                      32 g

(iii)28 g                                      32 g

(iv)28 g                                      80 g

(a) Which chemical combination law obeys the above experimental data? Given its statement.

(b) Fill in the blanks in the following conversions given below:

(i) 1 km = …………………. mm = …………………. pm

(ii) 1 mg = …………………. kg = …………………. ng

(iii) 1 mL = …………………. L = …………………. dm3

Answer 52. (a) Fixing the mass of the dinitrogen as 28 g, masses of the dioxygen combined would be 32, 64, 32 and 80 g in given four oxides. These masses of the dioxygen bear a simple whole number ratio as 2:4:2:5. Therefore, the data given will obey the law of the multiple proportions.

The statement is as follows, two elements always combine in the fixed mass of another, bearing a simple ratio to another to form the two or more chemical compounds.

(b) (i) 1 km =  1km× 1000m/ 1km ×100cm /1m/ 10mm /1cm = 106 mm

1 km =  1km× 1000m / 1km × 1pm/ 10-12 m = 1015 pm

(ii) 1 mg = 1mg ×1g/ 1000mg × 1kg / 1000g = 10-6 kg

1 mg = 1mg ×1g/ 1000mg × 1ng/ 10-9g = 10-6 ng

(iii) 1 mL = 1mL×1L/ 1000mL = 10-3 L

1 mL = 1cm3 = 1cm3× (1dm × 1dm × 1dm/ 10cm × 10cm × 10cm) = 103dm3

Question 53. When the speed of light is 3.0 × 108ms-1, calculate the distance for the following covered by the light in 2.00 ns.

Answer 53. Distance covered will be

= Speed  × Time = 3.0 × 108 ms-1 × 2.00 ns

= 3.0 × 108ms-1 × 2.00 ns ×10-9 s  /1ns = 6.00×10-1 m = 0.600m

Question 54. When ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour will produce?

Answer 54. Dihydrogen gas reacts with the given dioxygen gas as, 

2H2(g) + O2(g) → 2H2O(g)

Therefore, the two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. So, the ten volumes of dihydrogen would react with the five volumes of dioxygen to produce the required ten volumes of water vapour.

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