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

1 MARK QUESTIONS 

1. What is chemistry?

Ans: Chemistry is the scientific study of the composition, characteristics, and interactions of matter.

2. How has chemistry contributed to the nation's development?

Ans: Weather patterns, brain function, computer operation, chemical industries, manufacturing, fertilizers, alkalis, acids, salts, dyes, polymers, medicines, soaps, detergents, metals, alloys, and other fields of chemistry have all contributed to the national economy.

3. Differentiate solids, liquids & gases in terms of volume & shapes.

Ans: The differences between solids, liquids, and gases are:

4. Name the different methods that can be used for the separation of components of a mixture.

Ans: Physical procedures such as handpicking, filtrations, crystallization, distillation, and others can be used to separate the components of a mixture.

5. Classify the following as pure substances and mixtures – Air, glucose, gold, sodium and milk.

Ans: From the substances given in the question Glucose, Gold, and Sodium are the pure substances while air milk is the mixtures.

6. What is the difference between molecules and compounds? Give examples of each.

Ans: The difference is tabulated below:

7. How can we separate the components of a compound?

Ans: The constituents of a compound cannot be separated by physical methods. They can only be separated by chemical methods. 

8. How are physical properties different from chemical properties?

Ans: Color, odor, and other physical properties can be measured or observed without changing the substance's identity or composition, whereas chemical properties require a chemical change to be measured. 

9. What are the two different systems of measurement?

Ans: The different systems of measurement are the English system and the metric system.

10. What is the SI unit of density?

Ans: $\text{Kg }{{\text{m}}^{\text{-3}}}$ or $\text{Kg/}{{\text{m}}^{\text{3}}}$ is the SI unit of density.

11.What are the reference points in a thermometer with the Celsius scale?

Ans: The thermometers with Celsius scale are calibrated form ${{\text{0}}^{\text{o}}}\text{C}$  to $\text{10}{{\text{0}}^{\text{o}}}\text{C}$  where there two temperatures are the freezing and boiling of water.

12. What is the SI unit of volume? What is the other common unit which is not an SI unit of volume?

Ans: The SI unit of volume is ${{\text{m}}^{\text{3}}}$ while litre (L) is the common unit which is not an SI unit.

13. What is the difference between precision and accuracy?

Ans: The difference between precision and accuracy are: 

14. What do you understand by significant figures?

Ans: Significant figures are used to define those numbers which have some uncertainty in the form of digits. Considering the following example, if we have a 5.756 value, then it has 4 significant figures.

15. State law of definite proportions.

Ans: Law of definite proportions also known by the name of the law of constant proportions which states that a given element always contains exactly the same proportion of elements by weight.

16. State Avogadro’s law.

Ans: Avogadro’s law states that at the same temperature and pressure, gases that have equal volume will contain an equal number of molecules.

17. Define one atomic mass unit (amu).

Ans: One atomic mass unit is defined as the mass that is exactly equivalent to 1/12th of the mass of a carbon atom, whereas the mass of a carbon atom is 12.0107 u.

18. What is formula mass?

Ans: When a material has a three-dimensional structure and does not include discrete molecules as component particles, the molecular mass is calculated by summing the atomic masses of all the individual atoms present in that composition.

19. What is the value of one mole?

Ans: A mole of a material or particle is defined as having exactly $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ particles, which can be atoms, molecules, or ions, with $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ being Avogadro's number.

20. At NTP, what will be the volume of molecules of $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}\text{ }{{\text{H}}_{\text{2}}}$?

Ans: Under NTP circumstances, $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ hydrogen molecules will contain precisely 22.4 litres of hydrogen.

21. Calculate the number of molecules present in 0.5 moles of $\text{C}{{\text{O}}_{\text{2}}}$?

Ans: 1 mole of $\text{C}{{\text{O}}_{\text{2}}}$ contains exactly $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ molecules, then 0.5 moles will contain: $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}\text{  }\!\!\times\!\!\text{  0}\text{.5 = 3}\text{.011  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ 

So, 0.5 moles of $\text{C}{{\text{O}}_{\text{2}}}$ contains $\text{3}\text{.011  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ molecules.

22. 1L of a gas at STP weighs 1.97g. What is molecular mass?

Ans: Molecular mass can be calculated by multiplying the weight by 22.4, so the 22.4 L of gas will weigh: 

$\text{1}\text{.97  }\!\!\times\!\!\text{  22}\text{.4 = 44}\text{.1 g}$ 

Hence, the molecular mass is 44.1 g.

23. What is stoichiometry?

Ans: Stoichiometry is formed by combining two Greek words: stoikhein, which means element, and metron, which means measurement. As a result, we may say that stoichiometry is concerned with calculating the masses of reactants and products in chemical processes.

24. The substance which gets used up in any reaction is called _________ .

Ans: Limiting reagent

25. What is 1 molal solution?

Ans: One molal solution is defined as a solution containing one mole of a solute per kilogram or 1000 g of solvent.

2 MARKS QUESTION

1. How can we say that sugar is solid and water is liquid?

Ans: Sugar's constituent particles are densely packed, and it also has its own volume and form, making it a solid, whereas water's constituent particles are not as densely packed. It has a defined volume but no defined form, therefore it is classified as a liquid.

2. How is matter classified at macroscopic level?

Ans: Macroscopic classification of matter is given as follows:

3. Classify the following substances as elements, compounds and mixtures – water, tea, silver, steel, carbon dioxide and platinum.

Ans: From the substances given in the question water and carbon dioxide are compounds, silver and platinum are elements while tea and steel are mixture.

4. Write seven fundamental quantities and their units.

Ans: The seven fundamental quantities and their SI units are listed as follows:

5. What is the difference between mass & weight? How is mass measured in the laboratory?

Ans: The difference between mass and weight is:

The mass of a material is generally determined in the laboratory using an analytical balance.

6. How is volume measured in the laboratory? Convert 0.5L into mL  and $\text{30 c}{{\text{m}}^{\text{3}}}$ to $\text{d}{{\text{m}}^{\text{3}}}$ .

Ans: In laboratories, volume of a liquid is generally measured by using burette, graduated cylinder, pipette etc. 

As,

$\text{1 L = 1000 mL}$

so, 0.5 L will be equal to:-

$\text{0}\text{.5 L = 0}\text{.5  }\!\!\times\!\!\text{  1000 mL}$ 

$\text{0}\text{.5 L = 500 mL}$ 

Now, $\text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 d}{{\text{m}}^{\text{3}}}$ 

 So, $\text{30 c}{{\text{m}}^{\text{3}}}$ will be equal to:-

$\text{30 c}{{\text{m}}^{\text{3}}}\text{ = }\dfrac{\text{1}}{\text{1000}}\text{  }\!\!\times\!\!\text{  30d}{{\text{m}}^{\text{3}}}$ 

$\text{30 c}{{\text{m}}^{\text{3}}}\text{ = 0}\text{.03 d}{{\text{m}}^{\text{3}}}$ 

7. Convert $\text{3}{{\text{5}}^{\text{o}}}\text{C}$ to $^{\text{o}}\text{F}$  and K.

Ans: To convert $\text{3}{{\text{5}}^{\text{o}}}\text{C}$to $^{\text{o}}\text{F}$  

We will use the following formula,

$^{\text{o}}\text{F=}\dfrac{\text{9}}{\text{5}}{{\text{(}}^{\text{o}}}\text{C)+32}$

Putting the value of 35 in $^{\text{o}}\text{C}$, we get,  

$^{\text{o}}\text{F = }\dfrac{\text{9}}{\text{5}}\text{ (35) + 32}$ 

$\text{63 + 32 = 9}{{\text{5}}^{\text{o}}}\text{F}$ 

Now, to convert $\text{3}{{\text{5}}^{\text{o}}}\text{C}$ to K,

We will us the following relationship,

$\text{K =}{{\text{ }}^{\text{o}}}\text{C + 273}\text{.15}$ 

Putting the values, we get:

K = 35 + 273.15

K = 308.15

8. What does the following prefixes stand for:

1.  Pico

Ans: $\text{Pico = 1}{{\text{0}}^{\text{-12}}}$ 

2.  Nano

Ans: $\text{Nano = 1}{{\text{0}}^{\text{-9}}}$ 

3.  Centi

Ans: $\text{Centi = 1}{{\text{0}}^{\text{-2}}}$ 

4.  Deci

Ans: $\text{Deci = 1}{{\text{0}}^{\text{-1}}}$ 

9. Explain the law of multiple proportions with an example.

Ans: The law of multiple proportions states that if two elements can combine to form more than one compound, the masses of one element which combine with a fixed mass of another element are in a ratio of small whole numbers.

For example:

Hydrogen and oxygen can combine to form water (whose chemical formula is ${{\text{H}}_{\text{2}}}\text{O}$) as well as hydrogen peroxide (whose chemical formula is ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$). 

Here, the masses of oxygen (16g and 32 g) combines with a fixed mass of hydrogen (2g) element bear a simple ratio which is 16:32 = 1:2

10. Write Postulates of Dalton’s atomic theory.

Ans: Postulates of Dalton’s atomic theory are as follows–

1. Matter consists of indivisible atoms.

2. All atoms of an element have a similar atomic mass. But atoms of different elements have different atomic masses.

3. Compounds are formed when atoms of different elements combine in a fixed ratio. 

4. Chemical reaction involves the reorganization of atoms. These are neither created nor destroyed.

11. Calculate the molecular mass of-

${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}\text{, }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{, }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{, }{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$ 

Ans: The molecular mass is the sum of the atomic masses of the individual elements present in a molecule. The molecular masses of the given compounds are calculated as follows with the help of the molar masses of the elements.

The molar mass of C= 12 

The molar mass of H= 1 

The molar mass of O= 16 

The molar mass of S= 32 

The molar mass of P= 31 

\[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}\text{ = (2 }\!\!\times\!\!\text{ 12) + (6 }\!\!\times\!\!\text{ 1) = 30 g/ mol}\] 

\[{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ = (12 }\!\!\times\!\!\text{ 12) + (22 }\!\!\times\!\!\text{ 1) + (11 }\!\!\times\!\!\text{ 16) = 342 g/ mol}\] 

\[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ = (2 }\!\!\times\!\!\text{ 1) + (32) + (16 }\!\!\times\!\!\text{ 4) = 98 g/ mol}\] 

\[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\text{ = (3 }\!\!\times\!\!\text{ 1) + (31) + (16 }\!\!\times\!\!\text{ 4) = 98 g/ mol}\] 

12. Give one example each of molecule in which empirical formula and molecular formula are

1. Same

Ans: Molecule having same molecular formula and the empirical formula is Carbon dioxide, that is $\text{C}{{\text{O}}_{\text{2}}}$.

2. Different

Ans: When molecular formula and empirical formula are different, the example of such molecule is, 

Hydrogen peroxide: the molecular formula is ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and empirical formula is HO.

13. Calculate the number of moles in the following masses:

1. 7.85g of Fe

Ans: Given 7.85g of Fe

56g of Fe contains $\text{6}\text{.022  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{23}}}$ atoms = 1 mole

56g of Fe = 1 mole

So, $\text{7}\text{.85 g of Fe = }\dfrac{\text{1}}{\text{56}}\text{  }\!\!\times\!\!\text{  7}\text{.85 = 0}\text{.14 moles}$

2. 7.9 mg of Ca

Ans: As, 40g of Ca = $\text{40  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{3}}}\text{ mg}$ of Ca

40g of Ca contain 1 mole of Ca

Or we can write $\text{4  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}\text{ mg Ca = 1 mole}$ 

Therefore, $\text{7}\text{.9 mg of Ca = }\dfrac{\text{7}\text{.9}}{\text{4  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{4}}}}$ 

$\text{= 1}\text{.97  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}\text{ moles}$ 

14. How much potassium chlorate should be heated to produce 2.24 L of oxygen at NTP?  

Ans: The reaction for heating of potassium chlorate is:

\[\text{2 KCl}{{\text{O}}_{\text{3}}}\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\text{2 KCl + 3}{{\text{O}}_{\text{2}}}\] 

From the reaction, it is evident that 2 moles of potassium chlorate liberate 3 moles of oxygen.

Therefore, we have:

67.2 L of oxygen is produced from 245g of $\text{KCl}{{\text{O}}_{\text{3}}}$ 

Then, 2.24L of oxygen will be produced from = $\dfrac{\text{245}}{\text{67}\text{.2}}\text{  }\!\!\times\!\!\text{  2}\text{.24}$ 

$\text{=8}\text{.17 g of KCl}{{\text{O}}_{\text{3}}}$

15. Write an expression for molarity and molality of a solution.

Ans: Molarity is the number of moles of solute per litre of a solution, that is,

$\text{Molarity = }\dfrac{\text{number of moles of solutes}}{\text{Volume of solution in Litres}}$ 

While molality is the number of moles od solute per kilogram of a solvent, that is,

$\text{Molality = }\dfrac{\text{number of moles of solutes}}{\text{Mass of solvent in kg}}$ 

16. Calculate the weight of lime CaO obtained by heating 200kg of 95% pure limestone $\text{CaC}{{\text{O}}_{\text{3}}}$ .

Ans: 100 kg impure sample has pure $\text{CaC}{{\text{O}}_{\text{3}}}$ = 95% = 95 kg

Therefore, 200kg impure sample has pure $\text{CaC}{{\text{O}}_{\text{3}}}$ = $\dfrac{\text{95  }\!\!\times\!\!\text{  200}}{\text{100}}\text{ = 190 kg}$ 

From the below reaction:

\[\text{CaC}{{\text{O}}_{\text{3}}}\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\text{ CaO + C}{{\text{O}}_{\text{2}}}\] 

We can observe that 100kg $\text{CaC}{{\text{O}}_{\text{3}}}$ will give CaO = 56 kg

Therefore, 190 kg $\text{CaC}{{\text{O}}_{\text{3}}}$ will give CaO = $\dfrac{\text{56  }\!\!\times\!\!\text{  190}}{\text{100}}\text{ = 106}\text{.4 kg}$ 

17. 4 litres of water added to 2L of 6 molar HCl solution. What is the molarity of the resulting solution? 

Ans: Let the initial volume ${{\text{V}}_{\text{1}}}\text{ = 2 L}$ 

The final volume, ${{\text{V}}_{\text{2}}}\text{ = 4 L + 2 L = 6 L}$ 

Given, Initial Molarity, ${{\text{M}}_{\text{1}}}\text{ = 6 M}$ 

Let, Final molarity = ${{\text{M}}_{\text{2}}}$ 

Using the following relationship, ${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=}{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$ 

$\text{6 M  }\!\!\times\!\!\text{  2 L = }{{\text{M}}_{\text{2}}}\text{  }\!\!\times\!\!\text{  6 L}$ 

We have, ${{\text{M}}_{\text{2}}}\text{=}\dfrac{\text{6 M  }\!\!\times\!\!\text{  2 L}}{\text{6 L}}\text{ = 2 M}$ 

18. What volume of 10M HCl and 3M HCl should be mixed to obtain 1L of 6M HCl solution?

Ans: Let the required volume og 10M HCl be V liters.

The, the required volume of 3M HCl be (1 – V) liters.

Using the resultant Molarity formula,

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{ + }{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}\text{ = }{{\text{M}}_{\text{3}}}{{\text{V}}_{\text{3}}}$ .

Putting the values, we get:

$\text{10  }\!\!\times\!\!\text{  V + 3  }\!\!\times\!\!\text{  (1-V) = 6  }\!\!\times\!\!\text{  1}$ 

10V + 3 – 3V = 6

7 V = 3

$\text{V = }\dfrac{\text{3}}{\text{7}}\text{ = 0}\text{.428 L = 428 ml}$ 

Then the volume of 10M HCl required = 428 ml and volume of 3M HCl required will be:

1000ml – 428ml = 572ml

3 MARKS QUESTION

1. How many significant figures are present in

1.  $\text{4}\text{.01  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{2}}}$ 

Ans: There are rules that must be followed for counting the number of significant figures in a given number. 

If a number terminates in zeros, but these zeros do not reach the right side of the decimal point, these zeros might be significant or not important.

So, the given number is $\text{4}\text{.01  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{2}}}$, therefore, there are three significant figures in this.

2.  8.256

Ans: All the non-zero digits and the zeros are important between the non-zero digits.

So, the given number is 8.256, therefore, there are four significant figures in this.

3. 100

Ans: If a number terminates in zeros, but these zeros do not reach the right side of the decimal point, these zeros might be significant or not important.

So, the given number is 100, therefore, there is only one significant figure in this.

2. Vitamin C is essential for the prevention of scurvy. Combustion of

0.2000g of vitamin C gives 0.2998g  of $\text{C}{{\text{O}}_{\text{2}}}$ and 0.819g of ${{\text{H}}_{\text{2}}}\text{O}$ . What is the empirical formula of Vitamin C?

Ans: The empirical formula is the simplest form of any molecular formula, and it can be also the same as the molecular formula.

First, we have to find the percentage of carbon, hydrogen, and oxygen in the given amount of compounds. These are given below:

Percentage of carbon = $\dfrac{\text{2}}{\text{18}}\text{  }\!\!\times\!\!\text{  0}\text{.2998  }\!\!\times\!\!\text{  }\dfrac{\text{100}}{\text{0}\text{.2}}\text{= 40}\text{.88}$ 

Percentage of Hydrogen = $\dfrac{\text{2}}{\text{18}}\text{  }\!\!\times\!\!\text{  0}\text{.819  }\!\!\times\!\!\text{  }\dfrac{\text{100}}{\text{0}\text{.2}}\text{ = 4}\text{.55}$ 

Percentage of Oxygen = $100\text{ - (40}\text{.88 + 4}\text{.55) = 54}\text{.57}$ 

So, we have the percentage of all the elements, now we can find the empirical formula as in the table given below:

So, the empirical formula of Vitamin C = ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{3}}}$.

Important Questions For Class 11 Chemistry Chapter 1 - Free PDF Download

Some Basic Concepts of Chemistry is a chapter that deals with the theoretical and practical aspects and the role of chemistry in each student's life. This chapter presents the definition, balanced equations, empirical formula and molecular formula for a compound, and numerical, which may be difficult for students to understand. Lack of practice and a solid course plan and curriculum may act as a barrier preventing you from securing good scores.

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Chemistry Class 11 Chapter 1 Important Questions

Some Basic Concepts of Chemistry Class 11 Important Questions will help students gain comprehensive knowledge and understanding of the concepts and theories under this chapter. Continuous practice of the essential questions for class 12 chemistry chapter 1 will help you implement the plan and tactics to enhance preparation. A comprehensive brief of the chapter for students to learn are as follows:

Chemistry

Chemistry is often referred to as the Central Science because it plays a significant role in connecting the physical sciences, such as Chemistry with Applied Science and Life Sciences such as Engineering and Medicine. Chemistry is defined as the study of interaction, composition, and properties of matter.

Matter

The matter is defined as anything that possesses mass, occupies space, and the presence that can be felt by the five senses. Matter exists in three forms, namely, as a solid, liquid, and gas. Solids are substances that possess a definite structure and a definite shape like sugar, iron, etc. Liquids are substances that have a definite volume but lack a definite form and take the shape of the vessel in which they are put - for example, mercury, milk, water, etc. Gases are substances that can neither possess definite shape or definite volume like hydrogen, oxygen, nitrogen, etc.

Macroscopic Classification of Matter

The matter is classified as:

1. Mixtures: Mixtures consist of two or more substances in any ratio that are present in it which are called its components. A mixture can be heterogeneous or homogeneous.

2. Pure Substances: It is a material that contains only one substance. A pure substance is classified into elements and compounds. 

Properties of Matter

A substance holds unique characteristics or attributed properties. The properties of matter can be classified into two primary categories- physical properties and Chemical properties. Physical properties can be observed or measured without changing the substance's composition and without changing the identity of the substances such as melting point, boiling point, colour, odour, etc. Chemical properties are the chemical changes that produce a characteristic reaction such as combustibility, basicity or acidity, etc.

Measurements

Physical Quantities: Physical Quantities are those that come across during our scientific studies. A physical quantity consists of two parts- The number and The Unit. A unit is referred to as the standard of reference chosen to measure any physical quantity such as Unit of length, Unit of density, Unit of time, Unit of electric current, etc.

Law of Chemical Combination

• Law of Conservation of Life.

• Law of Multiple Proportion.

• Gay Lussac's Law of gaseous Volumes.

• Law of Definite or Constant Proportions.

• Avogadro's Law.

• Law of Reciprocal Proportions.

Chemical Reactions

Chemical reactions are the rearrangement of atoms. Atoms from different molecules rearranged to form other molecules.

Stoichiometry

The study of chemical reactions and the calculations related to the reactions is known as Stoichiometry. The coefficients that are used to balance the chemical reactions are known as the Stoichiometric Coefficients.

Dilution Law

When a solution is diluted, more solutions are added, and the solute's moles remain unchanged. If the volume of a solution with a Molarity of M is changed from V1 to V2, it can be stated that M1V1 = M2V2 can solve the concentration or volume of the concentrated or dilute solution.

Introduction to Equivalent Concept

Equivalent Concept helps understand the processes in chemistry and the chemical reactions derived into simple format through the Equivalent Concept.

Valency Factor (Z): The number of 'H' ions supplied by one molecule or a mole of an acid or the number of -OH ions provided by one mole of a base of one molecule.

Equivalent Mass: The mass of an acid or base that furnishes one mol H+ or OH- is Equivalent Mass.

Equivalents: The equivalents are equal to the weight of acids or bases taken by the Equivalent weight.

Law of Chemical Equivalence

The Law of Chemical Equivalence states that the number of gram equivalents of reactant and the number of gram equivalents of product in a reaction is equal. For example in a reaction, aA + bB = cC + dD irrespective of the stoichiometric coefficients, one eq. of A with one eq. of B gives one eq. of C and one eq. of D.

Important Questions of Chemistry Class 11 Chapter 1

To get a better overview of the Important Questions of Some Basic Concepts of Chemistry Class 11, we have enlisted a few important and potential questions that are most likely to appear in the examination:

1. Draw a comparison between solids, liquids, and gases regarding their shape and volume.

2. State how chemical properties differ in their characteristics from the physical properties.

3. Define S.I and state the S.I unit of density.

4. What do you mean by significant figures?

5. Solve the numerical by finding the volume needed for making 2.5 litres of its 0.25 M solution, given that the density of methanol is 0.793 kg L–1.

6. State the difference between weight and mass. Explain how mass is measured in a laboratory.

7. State the Law of Multiple proportions along with a suitable example. 

8. Determine the mass of one 12C atom in grams?

9. Solve the numerical by calculating the distance covered by light in 2.00 ns if the light speed is 3.0 × 108 m s–1.

10. Explain how many water vapour volumes would be produced if ten volumes of dihydrogen gas react with five volumes of dioxygen gas?

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