CBSE Class 11 Chemistry Some Basic Concepts Of Chemistry Worksheet

Some Basic Concepts Of Chemistry MCQ Questions with Answers Class 11 Chemistry

Question- Many countries use Fahrenheit scale for expressing temperature of atmosphere. If temperature in any such country is measured 41°F then what is its value in celcius scale and would you expect hot or cold atmosphere in that country?
(a) 15°C, cold
(b) 25°C, normal
(c) 5°C, cold
(d) 41°C, hot
Answer-(c)

Some Basic Concepts Of Chemistry MCQ Questions with Answers Class 11 Chemistry

Question- Molecular mass is defined as the

(a) mass of one atom compared with the mass of one molecule

(b) mass of one atom compared with the mass of one atom of hydrogen

(c) mass of one molecule of any substance compared with the mass of one atom of C-12

(d) None of the above

Answer-(c)

Question-The law of definite proportions was given by –

(a) John Dalton

(b) Humphry Davy

(c) Proust

(d) Michael Faraday

Answer-(c)

Question-Given P = 0.0030m, Q = 2.40m, R = 3000m, Significant figures in P, Q and R are respectively

(a) 2, 2, 1

(b) 2, 3, 4

(c) 4, 2, 1

(d) 4, 2, 3

Answer-(b)

Question-20 g of CaCO₃ on heating gave 8.8 g of CO₂ and 11.2 g of CaO. This is in accordance with

(a) The law of conservation of mass.

(b) The law of constant composition.

(c) The law of reciprocal proportion.

(d) None of these

Answer-(a)

Question-In which of the following number all zeros are significant?

(a) 0.0005

(b) 0.0500

(c) 50.000

(d) 0.0050

Answer-(c)

Question-The correctly reported answer of addition of 29.4406, 3.2 and 2.25 will have significant figures

(a) 3

(b) 4

(c) 2

(d) 5

Answer-(a)

Question- Arrange the numbers in increasing no. of significant figures.
0.002600, 2.6000, 2.6, 0.260

(a) 2.6 < 0.260 < 0.002600 < 2.6000

(b) 2.6000 < 2.6 < 0.002600 < 0.260

(c) 0.260 < 2.6 < 0.002600 < 2.6000

(d) 0.002600 < 0.260 < 2.6 < 2.6000

Answer-(a)

Question-The number of significant figures for the three numbers 161 cm, 0.161 cm, 0.0161 cm are

(a) 3,4 and 5 respectively

(b) 3,4 and 4 respectively

(c) 3,3 and 4 respectively

(d) 3,3 and 3 respectively

Answer-(d)

Question- Dimension of pressure are same as that of

(a) Energy

(b) Force

(c) Force per unit volume

(d) Energy per unit volume

Answer-(d)

Question-Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates

(a) law of reciprocal proportions

(b) law of constant proportions

(c) law of multiple proportions

(d) law of equivalent proportions

Answer-(c)

Question- One mole of a gas occupies a volume of 22.4 L. This is derived from

(a) Berzelius’ hypothesis

(b) Gay-Lussac’s law

(c) Avogadro’s law

(d) Dalton’s law

Answer-(c)

Question-Which of the following is the best example of law of conservation of mass?

(a) 12 g of carbon combines with 32 g of oxygen to form 44 g of CO₂

(b) When 12 g of carbon is heated in a vacuum there is no change in mass

(c) A sample of air increases in volume when heated at constant pressure but its mass remains unaltered

(d) The weight of a piece of platinum is the same before and after heating in air

Answer-(a)

Question-In an experiment 4.2 g of NaHCO₃ is added to a solution of acetic acid weighing 10.0 g, it is observed that 2.2 g of CO₂ is released into the atmosphere. The residue left behind is found to weigh 12.0 g The above observations illustrate

(a) law of definite proportions.

(b) law of conservation of mass

(c) law of multiple proportions

(d) None of these

Answer-(b)

Question-Irrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of

(a) conservation of mass

(b) multiple proportions

(c) constant composition

(d) constant volume

Answer-(c)

Question-The number of significant figures in 10.3106 g is

(a) 2

(b) 3

(c) 1

(d) 6

Answer-(d)

Question-The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same.The illustrate the law of

(a) constant proportions

(b) conservation of mass

(c) multiple proportions

(d) reciprocal proportions

Answer-(a)

Question-In compound A, 1.00g of nitrogen unites with 0.57g of oxygen. In compound B, 2.00g of nitrogen combines with 2.24g of oxygen. In compound C, 3.00g of nitrogen combines with 5.11g of oxygen. These results obey the following law

(a) law of constant proportion

(b) law of multiple proportion

(c) law of reciprocal proportion

(d) Dalton’s law of partial pressure

Answer-(b)

Question- If the density of a solution is 3.12 g mL^–1, the mass of 1.5 mL solution in significant figures is______.

(a) 4.7 g

(b) 4680 × 10^–3 g

(c) 4.680 g

(d) 46.80 g

Answer-(a)

Question-Equal volumes of two gases A and B are kept in a container at the same temperature and pressure. Avogadro’s law is invalid if

(a) the gases are reactive

(b) the gases are non-reactive

(c) gas A has more number of molecules than gas B.

(d) None of these

Answer-(d)

Question- Which of the following statements indicates that law of multiple proportion is being followed.

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

(b) Carbon forms two oxides namely CO2 and CO, where masses fo oxygen which combine with fixed mass of carbon are in the simple ration 2 : 1.

(c) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.

(d) At constant temperature and pressure 200 mL of hydrogen will combine combine with 100 mL oxygen to produce 200 mL of water vapour.

Answer-(b)

Some Basic Concepts Of Chemistry MCQ Questions with Answers Class 11 Chemistry

Question- The modern atomic weight scale is based on

(a) O¹⁶

(b) C¹²

(c) H¹

(d) C¹³

Answer-(b)

Question- Calculate the volume at STP occupied by 240 gm of SO₂.

(a) 64

(b) 84

(c) 59

(d) 73

Answer-(b)

Question-  O₂, N₂ are present in the ratio of 1 : 4 by weight. The ratio of number of molecules is :

(a) 7 : 32

(b) 1 : 4

(c) 2 : 1

(d) 4 : 1

Answer-(a)

Question-  The number of gram molecules of oxygen in 6.02 × 10²⁴ CO molecules is

(a) 10 gm molecules

(b) 5 gm molecules

(c) 1 gm molecules

(d) 0.5 gm molelcules

Answer-(b)

Question-If the mass of the one atom is found to be 2.324784×10–23g, then this atom can be ?

(a) Oxygen

(b) Carbon

(c) Fluorine

(d) Nitrogen

Answer-(d)

Question- The number of moles of oxygen in one litre of air containing 21% oxygen by volume, under standard conditions are

(a) 0.0093 mole

(b) 0.21 mole

(c) 2.10 mole

(d) 0.186 mole

Answer-(a)

Question- Empirical formula of hydrocarbon containing 80% carbon and 20% hydrogen is :

(a) CH₃

(b) CH₄

(c) CH

(d) CH₂

Answer-(a)

Question- One litre oxygen gas at S.T.P will weigh :

(a) 1.43 g

(b) 2.24 g

(c) 11.2 g

(d) 22.4 g

Answer-(a)

Question-  20.0 kg of H₂(g) and 32 kg of O₂(g) are reacted to produce H₂O(l). The amount of H2O (l) formed after completion of reaction is

(a) 62 kg

(b) 38 kg

(c) 42 kg

(d) 72 kg

Answer-(d)

Question- Number of moles of NaOH present in 2 litre of 0.5 M NaOH is :

(a) 1.5

(b) 2.0

(c) 1.0

(d) 2.5

Answer-(c)

Question- At S.T.P. the density of CCl₄ vapours in g/L will be nearest to:

(a) 6.87

(b) 3.42

(c) 10.26

(d) 4.57

Answer-(a)

Question- How many moles of Al₂(SO₄)₃ would be in 50 g of the substance ?

(a) 0.083 mole

(b) 0.952 mole

(c) 0.481 mole

(d) 0.140 mole

Answer-(d)

Question- A hydrocarbon is composed of 75% carbon. The empirical formula of the compound is

(a) CH₂

(b) CH₃

(c) C₂H₅

(d) CH₄

Answer-(d)

Question-  The number of molecules in 8.96 litre of a gas at 0ºC and 1 atm. pressure is approximately

(a) 6.023 × 10²³

(b) 12.04 × 10²³

(c) 18.06 × 10²³

(d) 24.08 × 10²²

Answer-(d)

Question- Which has maximum number of molecules?

(a) 7 gm N₂

(b) 2 gm H₂

(c) 16 gm NO₂

(d) 16 gm O₂

Answer-(b)

Question- The empirical formula of a compound is CH₂. One mole of this compound has a mass of 42 grams. Its molecular formula is :

(a) C₃H₆

(b) C₃H₈

(c) CH₂

(d) C₂H₂

Answer-(a)

Question- What is the mass of 1 molecule of CO.

(a) 2.325 × 10^–23

(b) 4.65 × 10^–23

(c) 3.732 × 10^–23

(d) 2.895 × 10^–23

Answer-(b)

Question- In a hydrocarbon, mass ratio of hydrogen and carbon is 1:3, the empirical formula of hydrocarbon is

(a) CH₄

(b) CH₂

(c) C₂H

(d) CH₃

Answer-(a)

Question- An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%.The empirical formula of the compound would be :

(a) CH₃O

(b) CH₂O

(c) CHO

(d) CH₄O

Answer-(a)

Question-  10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

(a) 3 mol

(b) 4 mol

(c) 1 mol

(d) 2 mol

Answer-(b)

Question- What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene ?

(a) 2.8 kg

(b) 6.4 kg

(c) 9.6 kg

(d) 96 kg

Answer-(c)

Some Basic Concepts Of Chemistry MCQ Questions with Answers Class 11 Chemistry

Question- What is the molarity of 0.2N Na₂CO₃ solution?

(a) 0.1 M
(b) 0 M
(c) 0.4 M
(d) 0.2 M

Answer-(a)

Question- Volume of water needed to mix with 10 mL 10N HNO₃ to get 0.1 N HNO₃ is :

(a) 1000 mL
(b) 990 mL
(c) 1010 mL
(d) 10 mL

Answer-(b)

Question-  One kilogram of a sea water sample contains 6 mg of dissolved O₂. The concentration of O₂ in the sample in ppm is

(a) 0.6
(b) 6.0
(c) 60.0
(d) 16.0

Answer-(b)

Question- Which of the following statements are correct ?
(i) Both solids and liquids have definite volume.
(ii) Both liquids and gases do not have definite shape.
(iii) Both solids and gases take the shape of the container.

(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (i), (ii) and (iii)

Answer-(c)

Question- With increase of temperature, which of these changes?

(a) Molality
(b) Weight fraction of solute
(c) Molarity
(d) Mole fraction

Answer-(c)

Question-  Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?

(a) 2.70 M
(b) 1.344 M
(c) 1.50 M
(d) 1.20 M

Answer-(b)

Question-  If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

(a) decrease twice
(b) increase two fold
(c) remain unchanged
(d) be a function of the molecular mass of the substance

Answer-(a)

Question-  How much time (in hours) would it take to distribute one Avogadro number of wheat grains if 1020 grains are distributed each second ?

(a) 0.1673
(b) 1.673
(c) 16.73
(d) 167.3

Answer-(b)

Question- The molar solution of H₂SO₄ is equal to :

(a) N/2 solution
(b) N solution
(c) 2N solution
(d) 3N solution

Answer-(a)

Question- Arrange the following in the order of increasing mass (atomic mass: O = 16, Cu = 63, N = 14)
I. one atom of oxygen
II. one atom of nitrogen
III. 1 × 10–10 mole of oxygen
IV. 1 × 10–10 mole of copper

(a) II < I < III < IV
(b) I < II < III < IV
(c) III < II < IV < I
(d) IV < II < III < I

Answer-(a)

Question- The increasing order of molarity with 25 gm each of NaOH, LiOH, Al(OH)₃, KOH, B(OH)₃ in same volume of water?

(a) Al(OH)₃ < B(OH)₃ < KOH < NaOH < LiOH
(b) LiOH < NaOH < KOH < B(OH)₃ < Al(OH)₃
(c) LiOH < NaOH < B(OH)₃ < KOH < Al(OH)3
(d) NaOH < LiOH < B(OH)₃ < Al(OH)₃ < KOH

Answer-(a)

Question- What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene ?

(a) 2.8 kg
(b) 6.4 kg
(c) 9.6 kg
(d) 96 kg

Answer- (c)

Question-  Choose correct option based on following statements. Here T stands for true statement and F for false statement.
(i) Homogeneous mixture has uniform composition throughout.
(ii) All components of a heterogeneous mixture are observable to naked eyes.
(iii) All solutions are homogeneous in nature.
(iv) Air is an example of heterogeneous mixture.

(a) TTFF
(b) TFTF
(c) FFTT
(d) TFFF

Answer-(b)

Notes Some Basic Concepts of Chemistry Class 11 Chemistry

• Change of state
o A change of state occurs because heat energy breaks the force of attraction between particles. Kinetic energy of the particle increases.

• Melting point
o The temperature at which a solid melts into a liquid at normal atmospheric pressure.
o At melting point, the temperature does not change until all solid converts into liquid.

• Latent heat
o The heat required to break the force of attraction between the particles at transition temperature. This heat becomes confined within the material and is called the latent heat.
o Amount of heat required to change 1 kg of material to change its state at normal atmospheric pressure at transition temperature is called the latent heat for that transition.

• Sublimation
o Solid ⇌ gas [directly]
o Example: Ammonium chloride

• Effect of change of pressure
o If pressure is applied,
o Melting point → decreases
o Boiling point → increases

• Dry Ice – Solid CO₂ [directly converts to gas]

• Everything around us is composed of matter.

• There are five states of matter- solid, liquid, gaseous, plasma and Bose-Einstein condesate Solid phase
o Permanent change in shape is difficult
o Negligible compressibility
o Definite shape, size, and boundary
o No particle motion

• Liquid phase
o No fixed shape and boundary
o Have a fixed volume
o Low compressibility
o Lesser particle motion

• Gaseous state
o No fixed shape, volume, and boundary
o Highly compressible
o Gases exert pressure
o High particle motion

• Plasma State
o Super-energetic and super-excited particles
o No definite shape and volume
o Most common state of matter in universe
o Influenced by electric and magnetic field

• Bose-Einstein Condensate
o Super-unenergetic and super-cooled particles
o Formed on cooling an extremely low density gas to an extremely low pressure
o Super-fluid and super-conductive

Mass: S.I. Unit is kg.

Volume: S.I. Unit is m3.

Density: S.I. Unit is kg m–3.

Temperature: Three scales are
• °C (degree Celsius)
• °F (degree Fahrenheit)
• K (kelvin)
S.I. Unit is K
• Relation between °F and °C
°F = 9/5 (°C) + 32
• Relation between K and °C
K = °C + 273.15

Laws of chemical combination:

Five laws
• Law of conservation of mass → Matter can neither be created nor destroyed.
• Law of definite proportions → A compound always contains exactly the same proportions of elements by weight.
• Law of multiple proportions → If two elements can combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in small whole number ratios.
• Gay Lussac’s law of gaseous volumes → When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure.
• Avogadro law → At the same temperature and pressure, equal volumes of all gases contain equal number of molecules. 

Dalton’s atomic theory:
• Matter consists of indivisible atoms.
• Atoms of a given element have identical properties including identical mass while those of different elements have different masses.
• Atoms of different elements combine in a fixed ratio to form a compound.
• Atoms are neither created nor destroyed in a chemical reaction.

Atomic Mass
One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. 

Molecular Mass
Molecular mass is the sum of atomic masses of the elements present in a molecule. 

Formula Mass
Formula mass is the mass of an ionic compound.
1 mole of any substance can be defined as:
• Amount of a substance that contains as many particles (atoms, molecules or ions) as there are atoms in 12 g of the ¹²C isotope
• Avogadro number or Avogadro constant (N_A); equal to 6.022 × 10²³ particles Percentage Composition
Mass percent of an element = (Mass of the element in the compound × 100%) / Molar mass of the compound

Empirical formula and molecular formula:

Limiting reagent (Limiting reactant): Limiting reagent is the reactant present in the lesser amount, which gets consumed after sometime. After that, no reaction takes place further, whatever is the amount of the other reactant present.

Expression for concentration of a solution:

Mass per cent (W/W %) = (Mass of solute/Mass of solution) × 100%

Mole fraction of A = No. of moles of A / No. of moles of solution = n_A / (n_A+n_B)

[Here, A and B are the components of the solution]

Molarity (M) = No. of moles of solute / Volume of solution in litres

Molality (m) = No. of moles of solute / Mass of solvent in kg