NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Edited By Sumit Saini | Updated on Sep 10, 2022 03:58 PM IST

NCERT Exemplar Class 11 Chemistry solutions chapter 1 Some Basic Concepts of Chemistry provides detailed explanations on the importance of some of the basic concepts and laws of Chemistry. These laws include Avogadro’s Law, Law of Conservation of Mass, and many more. NCERT Exemplar Class 11 Chemistry chapter 1 solutions have been drafted by the experts in such a way that anyone who goes through it gets a deep insight into the topics related to the basic terms used in Chemistry. The experts have provided solutions for numerous questions in a detailed step-by-step process to ensure understanding of the students on the various topics covered.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 1)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Short Answer Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Assertion and Reason Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type
Main subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 1
Importance of Chemistry
What Will Students Learn?
NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise
Important Topics To Cover For Exams From NCERT Exemplar Class 11 Chemistry Solutions Chapter 1

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 1)

Question:1

Two students performed the same experiment separately, and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements.

Student	Reading
	(i)	(ii)
A	3.01	2.99
B	3.05	2.95

(i) Results of both the students are neither accurate nor precise.
(ii) Results of student A are both precise and accurate.
(iii) Results of student B are neither precise nor accurate.
(iv) Results of student B are both precise and accurate.

Answer:

The answer is the option (ii) Results of student A are both precise and accurate.

Explanation: First, we calculate, the average of both the readings of student A and B. Student A's average is 3.00, and student B's average is also 3.00. The correct reading of the mass is 3.00g. Therefore, both the students have gotten an average value close to the correct value and are accurate. But student A's values differ only by 0.02 and are precise while B's readings are not close to each other and differ by 0.1, and thus are not precise.

Therefore, A has the readings which are both precise and accurate.

Question:2

A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale?
(i) 40 °C
(ii) 94 °C
(iii) 93.3 °C
(iv) 30 °C
Answer:

The answer is the option iii) 93.3 °C
Explanation:
$^{o}F=\frac{9}{5} (^{o}C)+32$
$^{o}C=\frac{5}{9} (^{o}F-32)$
$^{o}C=\frac{5}{9} (200-32)=93.3$

Question:3

What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?
(i) 4 mol L^-1
(ii) 20 mol L^-1
(iii) 0.2 mol L^-1
(iv) 2 mol L^-1
Answer:

The answer is the option (iii) 0.2 mol L^-1
Explanation: First, calculating the number of moles of NaCl $=\frac{5.85}{58.5}=0.1\; mol/L$
$500\; ml=\frac{500}{1000}=0.5L$
Molarity (M) = number of moles of solute/ Volume of solution in litres
Molarity $=\frac{0.1}{0.5}$
Molarity = 0.2 M

Question:4

If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(i) 1.5 M
(ii) 1.66 M
(iii) 0.017 M
(iv) 1.59 M
Answer:

The answer is the option (ii) 1.66 M
Explanation: Using M₁V₁ = M₂V₂
$5 \times 500=M_{2} \times 1500$
$M2=\frac{(5 \times 500)}{1500}$
$M_{2}=1.66 M$

Question:5

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?
(i) 4 g of He
(ii) 46g Na
(iii) 0.40 g of Ca
(iv) 12g He
Answer:

The answer is the option (iv) 12g He
Explanation:
Using the previous formula,
12 g He=3 moles Of He
Number of atoms of He $=3\times N_{A}$
Number of atoms of He $=3\times\left ( 6.022\times 10^{23} \right )$

Question:6

If the concentration of glucose $(C_{6}H_{12}O_{6})$ in blood is 0.9 g L^-1, what will be the molarity of glucose in blood?
(i) 5 M
(ii) 50 M
(iii) 0.005 M
(iv) 0.5 M

Answer:

The answer is the option (iii) 0.005 M
Explanation:
Given; Concentration of glucose- $(C_{6}H_{12}O_{6})$ in blood =0.9/L
Molar mass of glucose = 180 g/mol
Concentration/molar mass $=\frac{0.9}{180}$
$=\frac{1}{200}\; moles=0.005\; moles$
Therefore the Molarity of glucose in blood is 0.005 M.

Question:7

What will be the molality of the solution containing 18.25 g of HCI gas in 500 g of water?
(i) 0.1 m
(ii) 1M
(iii) 0.5 m
(iv) 1m

Answer:

The answer is the option (iv) 1m
Explanation:
Number. of moles of HCl $=\frac{18.25}{36.5}=0.5$
Molality (m) = Number of moles (in the solute)/ Mass of the solute (kg)
Molality $=\frac{0.5\times 1000}{500}$
Molality = 1m

Question:8

One mole of any Substance contains $6.022\; \; 10^{23}$ of Atoms/molecules of $H_{2}SO_{4}$ present in the 100 mL of 0.02M $H_{2}SO_{4}$ solution is
(i) $12.044 \times 10^{20}$ molecules
(ii) $6.022 \times 10^{23}$ molecules
(iii) $1 \times 10^{23}$ molecules
(iv) $12.044 \times 10^{23}$ molecules
Answer:

The answer is the option (i) $12.044\times 10^{20}$ molecules.
Explanation:
Given: Molarity = 0.02 M
Given; Volume of solution = 100ml = 0.1 L
Number of moles Of $H_{2}SO_{4}$ = Molarity × Volume ( litres)
Number of molecules of $H_{2}SO_{4}$ = $2 \times 10^{-3} \times 6.022 \times 10^{23}$
Number of molecules = $12.044\times 10^{20}$ Molecules

Question:9

What is the mass percent of carbon in carbon dioxide?
(i) 0.034%
(ii) 27.27%
(iii) 3.4%
(iv) 28.7%
Answer:

The answer is the option (ii) 27.27%

Explanation:
Molar mass of $CO_{2}$ $= (1\times 12) + (2 \times 16) = 44 g$
Therefore, 1 g of $CO_{2}$ contains 1 g of carbon
44 g of $CO_{2}$ therefore contains 12 g of carbon.
% of C in $CO_{2}$ = $=\frac{12}{44}\times 100=27.27\; ^{o}/_{o}$

Question:10

The empirical formula and molecular mass of a compound are $CH_{2}O$ and 180 g respectively. What will be the molecular formula of the compound?
(i) $C_{9}H_{18}O_{9}$
(ii) $CH_{2}O$
(iii) $C_{6}H_{12}O_{6}$
(iv) $C_{2}H_{4}O_{2}$

Answer:

The answer is the option (iii) $C_{6}H_{12}O_{6}$
Explanation:
Empirical formula of $CH_{2}O$ is $CH_{2}O$
Calculating empirical formula mass = $12+(1\times 2)+16=30g$
Given: Molecular mass= 180g
n = Molecular mass/ empirical formula mass
$n=\frac{180}{30}=6$
Therefore, Molecular formula = $n\times CH_{2}O=6\times CH_{2}O=C_{6}H_{12}O_{6}$

Question:11

If the density of a solution is 3.12 g mL^-1, the mass of 1.5 mL solution in significant figures is _______.
(i) 4.7g
(ii) 4680 × 10^-3 g
(iii) 4.680g
(iv) 46.80g
Answer:

The answer is the option (i) 4.7g
Explanation:
Given, the density of a solution = 3.12 g mL^-1
Volume = 1.5 mL
Mass = Volume × Density
= 1.5mL × 3.12g m/L
= 4.68 g which when rounded to 2 significant figures is 4.7 g.

Question:12

Which of the following statements about a compound is incorrect?
(i) A molecule of a compound has atoms of different elements.
(ii) A compound cannot be separated into its constituent elements by physical methods of separation.
(iii) A compound retains the physical properties of its constituent elements.
(iv) The ratio of atoms of different elements in a compound is fixed.

Answer:

The answer is the option (iii) A compound retains the physical properties of its constituent elements.
Explanation:
The compound can be completely indifferent and in contrast to the physical properties of its constituent elements.

Question:13

Which of the following statements is correct about the reaction given below:
$4Fe(s) + 3O_{2} (g) \rightarrow 2Fe_{2}O_{3} (g)$
(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows Law of conservation of mass.
(ii) Total mass of reactants = total mass of product; therefore, Law of multiple proportions is followed.
(iii) Amount of $Fe_{2}O_{3}$ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(iv) Amount of $Fe_{2}O_{3}$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
Answer:

The answer is the option (i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows Law of conservation of mass.
Explanation:
Keeping the Law of conservation in mind, we evaluate both the sides of every equation. The mass of the reactants should be equal to the of the product, and this is only followed in equation (i), therefore it is correct.

Question:14

Which of the following reactions is not correct according to the law of conservation of mass.
(i) $2Mg(s) + O_{2} (g) \rightarrow 2MgO(s)$
(ii) $C_{2} H_{8} (g) + O_{2} (g) \rightarrow CO_{2} (g) + H_{2}O(g)$
(iii) $P_{4}(s) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s)$
(iv) $CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)$
Answer:

The answer is the option (ii) $C_{2} H_{8} (g) + O_{2} (g) \rightarrow CO_{2} (g) + H_{2}O(g)$
Explanation:
Again, following the Law of energy, the number of atoms on the reactant side should be equal to the product side. Option (ii) does not follow the rule.

Question:15

Which of the following statements indicates that Law of multiple proportion is being followed?
(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the simple ratio 1:2,
(ii) Carbon forms two oxides namely $CO_{2}$ and $CO$ . where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.
Answer:

The answer is the option (ii) Carbon forms two oxides namely $CO_{2}$ and $CO$ . where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
Explanation:
According to the Law of multiple proportions,

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)

Question:16

One mole of oxygen gas at STP is equal to ________
(i) $6.022\times 10^{23}$ molecules of oxygen
(ii) $6.022\times 10^{23}$ atoms of oxygen
(iii) 16 g of oxygen
(iv) 32 g of oxygen

Answer:

The answer are the options (i) and (iv)
Explanation:
1 mole of oxygen gas at STP = $6.022\times 10^{23}$ molecules of oxygen (which is the Avogadro number)
1 mole $O_{2}=32g \; of \; O_{2}$
Thus, one mole of $O_{2}$ is equal to the molecular weight of oxygen and the Avogadro number.

Question:17

Sulphuric acid reacts with sodium hydroxide as follows :
$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$
When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0. 1M sodium hydroxide solution, the amount of sodium sulphate and its molarity in the solution obtained is
(i) 0.1 mol L^-1
(ii) 7.10 g
(iii) 0.025 mol L^-1
(iv) 3.55 g
Answer:

The answer are the options (ii) and (iii)
$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$
Given, (0.1 M), Moles of H₂SO₄ = 0.1 moles
Given, (0.1M), Moles of NaOH = 0.1 moles
The ratio of H₂SO₄ and NaOH is 1:2, using the ratio, we figure for 0.1 mole of H₂SO₄ , 0.2 moles of NaOH is used, which is not available.
Therefore if 0.1 moles of NaOH reacts, then only $0.05(0.1/2)$ moles of H₂SO₄ is used. NaOH is the limiting reactant, and the reaction and products are evaluated according to the limiting reactant.
Therefore the number of moles formed of Na₂SO₄ will be 0.05 as well.
This is because 2 moles of NaOH produces 1 mole of H₂SO₄, therefore 1 mole of NaOH will logically produce 0.05 moles of Na₂SO₄.
Therefore, mass of Na₂SO₄ = 0.05 × (molar mass)
0.05 × (23 ×2 + 32 + 16× 4)
0.05 × 142 = 7.1 g
Molarity = Number of moles / Volume in L
Molarity = 0.05/ 2 L = 0.023 mol/L

Question:18

Which of the following pairs have the same number of atoms ?
(i) 16 g of $O_{2}$ (g) and 4 g of $H_{2}$ (g)
(ii) 16 g of $O_{2}$ and 44 g of $CO_{2}$
(iii) 28 g of $N_{2}$ and 32 g of $O_{2}$
(iv) 12 g of C (s) and 23 g of Na(s)

Answer:

The answer are the options (iii) and (iv)
Explanation:
One mole is measured as the amount of a substance that equal number of atoms as contained in exactly 12 g of Carbon.
1 mole of C = 12g C
While 23 grams of Na (mass) = 1 mole of Na
28g N₂ = 1 mole of nitrogen = 2 × atoms of nitrogen
This is $2 \times (6.022 \times 10^{23})$ atoms of nitrogen
32g O₂ = 1 mole of oxygen = $2 \times (6.022 \times 10^{23})$ atoms of oxygen

Question:19

Which of the following solutions have the same concentration?
(i) 20 g of NaOH in 200 mL of solution
(ii) 0.5 mole of KCl in 200 mL of solution
(iii) 40 g of NaOH in 100 mL of solution
(iv) 20 g Of KOH in 200 mL of solution

Answer:

The answer are the options (i) and (ii)
Explanation: (i) 20 g NaOH in 200 mL solution.

$\frac{20}{40}=0.5$ mol of NaOH
(ii) Molar concentration Of NaOH
$\frac{0.5\; mol}{0.2\; L}=2.5 \; M$ as 200 mL = 0.2 L
Molar concentration of KCl = $\frac{0.5\; mol}{0.2\; L}=2.5 \; M$

Question:20

16 g of oxygen has same of molecules as in
(i) 16 g of CO
(ii) 28 g of $N_{2}$
(iii) 14 g of $N_{2}$
(iv)) 1.0 g of $H_{2}$
Answer:

The answer are the options (iii) and (iv)
Explanation: 1 mole or the number of molecules of oxygen in 16g Of oxygen is
$\frac{16}{32}\times \left ( 6.023 \times 10^{23} \right )$
$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
In 14g or N₂ and 1 g of H₂ contain 0.5 moles. Therefore they will also contain $0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
Number of molecules of N₂
$=\frac{14}{28}\times \left ( 6.023\times 10^{23} \right )$
$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules
Number of molecules of H₂

$=0.5\times \left ( 6.023\times 10^{23} \right )$ molecules

Question:21

Which of the following terms are unitless?
(i) Molality
(ii) Molarity
(iii) Mole fraction
(iv) Mass percent

Answer:

The answer are the options (iii) and (iv)
Explanation:
Mole Fraction: It is the particular ratio of the substance in the solution in comparison to the total components of the solution. It is the ratio of the number of moles of the particular component to the total number of moles in the solution.
While, Mass percent= $\frac{Mass\; of\; solute}{Mass\; of\; solution}\times 100$

Question:22

One of the Statements of Dalton 's atomic theory is given below; "Compounds are formed when atoms of different elements combine in a fixed ratio". Which of 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
(iv) Avogadro's Law

Answer:

The answer are the options (i) and (iv)
Explanation:
For (iv), Avogadro suggested that equal volumes of gases will contain equal number of molecules, if the temperature and pressure remain the same.
While the Law of conservation of mass states that matter can neither be created nor Destroyed but just transformed from one form to another.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Short Answer Type

Question:23

What will be the mass of one atom of C-12 in grams?
Answer:

Mass of $6.022\times 10^{23}$ atoms of C in gm =12g
Therefore, mass of one atom of $C=\frac{12}{6.023\times 10^{23}}$ , this is $1.992648\times 10^{-23}$ g

Question:24

How many significant figures should be present in the answer of the following calculations?

$\frac{2.5\times 1.25\times 3.5}{2.01}$

Answer:

$\frac{2.5\times 1.25\times 3.5}{2.01}$ has 2 significant figures.

Question:25

What is the symbol for SI unit of mole? How is the mole defined?

Answer:

The symbol for Sl unit of mole is 'mol'. One mole is measured as the amount of a substance that equal number of atoms as contained in exactly 12 g of Carbon.

Question:26

What is the difference between molality and molarity?

Answer:

Molarity: It is the number of moles of solute present in 1 litre of the solution. It is denoted by M. It is dependent on temperature.
Molarity (M) = $\frac{Number\; of\; moles\; of\; solute}{Volume\; of \; solution\; in\; litres}$
Molality: It is the number of moles of solute present in 1 kg of solvent. It is denoted by m. It is independent on temperature.
Thus, Molality (m)= $\frac{Number\; of\; moles\; of\; the\; solute}{Mass\; of\; Solvent\; in\; kg}$

Question:27

Calculate the mass percent of calcium, phosphorus and oxygen in calcium Phosphate; $Ca_{3} (PO_{4}��)_{2}$

Answer:

Molecular mass of $Ca_{3} (PO_{4}��)_{2}$
$=\left ( 3\times 40 \right )+\left [ 2\left ( 31+16\times 4 \right ) \right ]=310$
Mass percent of Calcium = $\frac{3x \; Atomic \; mass \; of \; calcium}{Total\; molecular \; mass\; of \; calcium \; Phosphate}\times 100$
$=\frac{3\times 40}{310}\times 100= 38.71\; ^{o}/_{o}$
Likewise,
Mass percent of phosphorus = $\frac{2x \; Atomic \; mass \; of \; Phosphorous}{Total\; molecular \; mass\; of \; calcium\; Phosphate}\times 100$
$=\frac{2\times 31}{310}\times 100= 20\; ^{o}/_{o}$
Mass percent of Oxygen= $\frac{2x \; Atomic \; mass \; of \; Oxygen}{Total\; molecular \; mass\; of \; calcium\; Phosphate}\times 100$
$=\frac{2\times 64}{310}\times 100= 41.29\; ^{o}/_{o}$

Question:28

45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 454 L of nitrous oxide was formed. The reaction is given below. $2N_{2}(g)+O_{2}(g)\rightarrow 2N_{2}O(g)$ Which Law is being obeyed in this experiment? Write the statement of the Law?

Answer:

The Law being obeyed in this experiment is Gay Lussac's Law of Gaseous Volumes. The Law states, gases react with each other in a simple ratio by volume, provided the gases are maintained at a constant temperature and pressure.

Some basic concepts of Chemistry Excercise: 1.2

Question:29

If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number
(a) Is this statement true?
(b) If yes, according to Which Law?
(c) Give one example related to this Law.

Answer:

(a) Yes, the statement is true.
(b) The statement is true due to the Law of multiple proportions.
(c) Citing an example; carbon combines with oxygen to form ; CO and CO2
The mass of oxygen, when reacted with a fixed mass of carbon (12 g) have a simple ratio i.e.,16:32 = 1:2

Question:30

Calculate the average atomic mass of hydrogen using the following data:

Isotope	% Natural Abundance	Molar Mass
¹H	99.985	1
²H	0.015	2

Answer:

Average atomic mass = $\frac{(Natural\; Abudance\; of\; ^{1}H \times Molar\; mass)+(Natural\; Abudance\; of\; ^{2}H \times Atomic\; mass)}{100}$
$=\frac{(99.985\times 1)+(0.015\times 2)}{100}$
$=\frac{100.015}{100}=1.00015\; \upsilon$

Question:31

Hydrogen gas is prepared in the laboratory by reacting dilute HCI with granulated zinc. Following reaction takes place.
$Zn + 2HCl -ZnCl_{2}+ H_{2}$
Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts With HCI. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u.
Answer:

Given that 65.3 g of Zn reacts with HCI to form 22.7 of H₂(g)
Therefore, 32.65 g of Zn at STP reacts with HCI to form $=\frac{22.7\times 32.65}{65.3}=11.35L$

Question:32

The density of 3 molal solutions of NaOH is 1.110 g mL^-1 . Calculate the molarity of the solution.
Answer:

3 molal solution Of NaOH has 3 mols Of NaOH which are dissolved in 1000 g of solvent.
Therefore, the Mass of Solution-= Mass of SoIvent + Mass of Solute
$=1000g+(3\times 40\; g)$
$=1120\; g$
Volume of Solution = $\frac{MASS}{DENSITY}$
$=\frac{1120}{1.110\; mL}=100.90\; mL$
Given the density of solution =1.110 g m/L^-1
Since 1009 mL solution has 3 mols Of NaOH
Molarity (M) = $\frac{Number \; of\; moles\; of\; solute}{Volume\; of\; solution\; in\; litres}$
$\frac{3\; mol}{1009.00}=2.97\; M$

Question:33

Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer.

Answer:

No, it will not. The molality of any solution is indifferent to temperature changes as molality is expressed in mass and mass is unaffected by temperature fluctuations.
Molality (m) = $\frac{number\; of\; moles\; of\; the\; solute}{Mass\; of\; Solvent\; in\; kg}$

Question:34

If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL^-1).

Answer:

As per information given in the question, Mass of NaOH = 4g
Therefore, Number of moles of NaOH= $\frac{4}{40}=0.1\; mol$
Number of moles of water $=\frac{36}{18}=2\; mols$
Mole fraction of water = $\frac{Number\; of\; moles\; of\; water}{Total\; number\; of\; moles(water+NaOH)}\frac{2}{2.1}=0.952$
Mole fraction of NaOH = $\frac{Number\; of\; moles\; of\; NaOH}{Total\; number\; of\; moles(water+NaOH)}\frac{0.1}{2.1}=0.047$
Now, mass of solution = Mass of water + Mass of NaOH= $36+4=40 \; g$
Volume $=\frac{40}{1}=40\; mL$
Molarity in solution = $\frac{Number\; of\; moles\; of\; solute}{Volume\; of\; solute\; in\; Litre}=\frac{0.1\; mol\; NaOH}{0.04L}=2.5 M$

Question:35

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 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:

$2A + 4B \rightarrow 3C + 4D$
A and B are in the ratio- 2:4, 2 mols of 'A' require 4 mols of 'B' .
(i) For 5 mols of A, 10 mols of 'B' will be required using the ratio of 2:4.
i.e. 5 mol of $A\times \frac{4\; mol\; of\; B}{2\; mol\; of\; A}=$ 10 mol of
But only 6 mols of B is available therefore it is the limiting reagent
(ii), Amount of 'C' formed is depends on 'B' ,
4 mols of 'B' give 3 mols of C, the ratio is 4:3.
Therefore, 6 mols of B will give, 6 mol of $B\times \frac{3\; mol\; of\; C}{4\; mol\; of\; C}=4.5\; moles\; of\; C$

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type

Question:36

Match the following:

Column I	Column II
A. 88 g of CO2	a. 0.25 mol
B. 6.022 x 10²³ molecules of H2O	b. 2 mol
C. 5.6 litres of O₂ at STP	c. 1 mol
D. 96 g of O₂	d. 6.022 x 10²³ molecules
E. 1 mole of any gas	e. 3 mol

Answer:

i) – b

ii) – c

iii) – a

iv) - e

v) – d

Question:37

Match the following physical quantities with units

Column I (Physical)	Column II (Unit)
(i) Molarity	a. g mL^-1
(ii) Mole fraction	b. mol
(iii) Mole	c. Pascal
(iv) Molality	d. Unitless
(v) Pressure	e. mol L^-1
(vi) Luminous intensity	f. Candela
(vii) Density.	g. mol kg^-1
(viii) Mass	h. N m^-1
	i. kg

Answer:

(i) – e
(ii) – d
(iii) – b
(iv) – g
(v) – c
(vi) – f
(vii) – a
(viii) - i

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Assertion and Reason Type

Question:38

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The empirical mass of ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
(i) Both A and R are true, and R is 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:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: Molecular formula = n × Empirical formula mass
Empirical formula of ethene = $CH_{2}$
Empirical formula mass of ethene = 14 amu
$=\frac{1}{2}\times mass\; of\; ethene$
Empirical formula shows that ethene has (C : H)= 1:2

Question:39

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): One atomic mass unit is 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 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:

The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: Carbon has isotopes and Carbon-12 is one of the isotopes is represented as 12C.
12C has a mass of exactly 12 amu –(atomic mass units) and mass of all other elements are calculated with respect to this. One amu is the mass exactly of one twelfth the mass of one 12C atom.

Question:40

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1.
Reason (R): Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
(i) Both A and R are true and R is correct explanation of A.
(ii) Both A and R are true but R is not a correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:

The answer is the option (iii) A is true but R is false.
Explanation:
Zero's at the end of a number which does not have a decimal point, depends on the measurement.

Question:41

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Combustion Of 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:

The answer is the option (iii) A is false but R is true.
Explanation:
$CH_{4} (g) + 2O_{2} (g) \rightarrow CO_{2} �(g) + 2H_{2}O (g)$
16g of CH4 on complete combustion will give 36g of water.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type

Question:42

A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(i) volume of the new vessel.
(ii) number of molecules of dioxygen.

Answer:

$P_{1}=1 \; atm$
$T_{1}=273\; K$
We will be calculating V₁ now,
We know that 32 g of oxygen occupies 22.4 L of volume at STP⁺
Hence, 1.6 g Of O₂ will occupy,
$1.6\; g\; of\; Oxygen\times \frac{22.4\; L}{32\; g\; of\; oxygen}=1.12\; L$
$Hence\; V_{1}=1.12\; L$
We know that, $p_{2}=p_{1}/2=\frac{1}{2}=0.5 \; atm$
Now we must calculate $V_{2}$
As per Boyle's law, $P_{1}V_{1}=P_{2}V_{2}$
Therefore, $V_{2}=\frac{P_{1}V_{1}}{2}$
$\frac{1\; atm\times 1.12\; L}{2}=0.5\; atm\times v_{2}$
Therefore, $V_{2}=2.24L$
(iii) Number of molecules of O₂that are present in the vessel,
$= 6.023\times 10^{23}\times \frac{1.6}{32}=3.011\times 10^{22}$

Question:43

Calcium carbonate reacts with aqueous HCl to give $CaCl_{2}$ and $CO_{2}$ according to the reaction given below:
$CaCO_{3} (s) + 2HCl (aq) \rightarrow CaCl_{2}(aq) + CO_{2}(g) + H_{2}O (l)$
What mass of $CaCl_{2}$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $CaCO_{3}$ ? Name the limiting reagent. Calculate the number of moles of $CaCl_{2}$ formed in the reaction.

Answer:

Number of moles of HCI= $250\; mL\times \frac{0.76\; M}{1000}=0.19\; mol$
Given: Mass of $CaCO_{3}$ = 1000 g
Number of moles of $CaCO_{3}$ = $\frac{1000\; g}{100\; g}=10\; \; mol$
Referring to the equation, 1 mol of $CaCO_{3}$ (s) needs 2 mol of HCI (aq), the ratio is 1:2.
Hence, for 10 mol of $CaCO_{3}$ (s), the number of moles of HCI required would be:
$10\; Mol\; CaCO_{3}\times \frac{2\; Mol\; HCl}{1\; Mol\; CaCO_{3}}=20\; Mol\; HCl (aq)$
But only 0.19 mol of HCI (aq) is available and therefore, HCI is limiting reagent.
Amount of $CaCl_{2}$ will depend on the amount of HCl present. 2 mol of HCl (aq), gets 1 mol of $CaCl_{2}$ , the ratio is 2:1.
Thus, 0.19 mol of HCl will get= $0.19\; Mol\; HCl\times \frac{1\; mol\; CaCl_{2}}{2\; mol\; HCl}=0.095\; mol\; of\; CaCl_{2}$
0.095 ×molar mass of $CaCl_{2}$ = 0.095 ×111 = 10.54 g

Question:44

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

Answer:

The Law of multiple proportions states that "If two elements can combine to form more than one compound, the mass of element combines with the fixed mass of the other element and is in the ratio of small whole numbers." For example, hydrogen reacts with oxygen to form water and
hydrogen peroxide.
Hydrogen + Oxygen → Water
2g 16g 18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g 32g 34g
Here, the masses of oxygen combine with a fixed mass of hydrogen and have a simple ratio of 16:32 or 1:2. This Law reflects that reactants always combine in fixed proportions.

Question:45

A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB₂, A₂B and A₂B₃ and show that Law of multiple proportions is applicable.

Answer:

Applying the Law of multiple proportions
For the combination, AB
1 g of A combines with $\frac{5}{2}g\; of\; B=2.5g\; of\; B$
For AB₂
1 g of A combines $\frac{10}{2}g\; of\; B=5g\; of\; B$
For A₂B
1 g of A combines with $\frac{5}{4}g\; of\; B=1.25g\; of\; B$
For A₂B₂
1 g of A combines with $\frac{15}{4}g\; of\; B=3.75g\; of\; B$
Thus, it is proved that Law of multiple proportions is applicable.

Students preparing for their examinations can make use of the NCERT Exemplar Class 11 Chemistry solutions chapter 1 PDF download feature to download a PDF to prepare for their exams, as it covers many important topics which are asked in the competitive exams. NCERT Exemplar solutions for Class 11 Chemistry chapter 1 includes the following topics:

Main subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 1

Importance of Chemistry
Nature of matter
Properties of matter and their measurement
The International System of Units (S.I)
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
Avogadro law
Dalton Atomic Theory
Atomic and Molecular masses
Average atomic Mass
Formula mass
Mole Concept and Molar Mass
Percentage composition
Empirical Formula
Stoichiometry and stoichiometric calculations
Limiting Reagents
Reactions in solutions

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What Will Students Learn?

Through NCERT Exemplar Class 11 Chemistry solutions chapter 1 Some Basic Concepts of Chemistry students will learn the basic properties of matter and also about the various S.I units that can be used. All the laws and theories such as Dalton’s Atomic Theory and Avogadro Law among others have been explained in a detailed way to form a good base for the students. The students will also get to learn and solve numerical based on topics such as molecular weight, mass concentration, and mole concept among others which will be helpful if they are preparing for competitive exams and also for their class 12.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 7 Equilibrium

Chapter 8 Redox Reactions

Chapter 9 Hydrogen

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Important Topics To Cover For Exams From NCERT Exemplar Class 11 Chemistry Solutions Chapter 1

Here are the important topics that the students must specifically learn in this chapter:

The students will get to learn about the various parameters to denote the strength of the solution such as Molarity, Normality, Molality and their important relations. The related numerical and their solutions are given in Class 11 Chemistry NCERT Exemplar solutions chapter 1.
NCERT Exemplar Class 11 Chemistry solutions chapter 1 also introduces important concepts such as Dilution law, Equivalence concept, and Stoichiometry and has exercises for the revision of students.
Students will study various laws of chemical combination like Law of Multiple Proportion, Law of Definite proportion.

Check Solutions of Textbook Chapters

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Mathematics Solutions

NCERT Exemplar Class 11 Physics Solutions

Read more NCERT Solution subject wise -

Also, Read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Are the concepts given in NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 easy to understand?

Yes, the concepts are given in an easy and detailed way for the readers to understand better.

2. Is the chapter helpful in the preparation of JEE/NEET?

It is an important chapter from the competition point of view. The questions regarding molarity, molality etc. are frequently asked.

3. Do we have to solve all the questions given after each chapter?

For a better understanding of any concept/ chapter, one should solve all the exercises given. It helps in self-evaluation.

4. Does it cover all the important topics?

Yes, NCERT Exemplar Solutions for Class 11 Chemistry Chapter 1 covers all the important topics.

Also Read

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

NCERT Class 11 Biology Chapter 22 Notes Chemical Coordination And Integration- Download PDF Notes

Some basic concepts of Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 1 Notes - Download PDF

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Books for class 11 Hindi 2024 - Download Free PDF

NCERT Book for class 11 Biology 2024 - Download Free PDF

NCERT Books for class 11 Chemistry 2024 - Download Free Pdf

NCERT Books for class 11 Maths 2024 - Download Free PDF

NCERT Books for Class 11 Physics 2024 - Download PDF

NCERT Books for Class 11 English 2024 - Download Free PDF

NCERT Syllabus for Class 11 2024 - Download All Subjects Pdf Here

NCERT Syllabus for Class 11 English 2024 - Download PDF here

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

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NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 1)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: MCQ (Type 2)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Short Answer Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Matching Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Assertion and Reason Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1: Long Answer Type

Main subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 1

Importance of Chemistry

What Will Students Learn?

NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 - Chapter-Wise

Important Topics To Cover For Exams From NCERT Exemplar Class 11 Chemistry Solutions Chapter 1

Check Solutions of Textbook Chapters

NCERT Exemplar Class 11 Solutions

Frequently Asked Question (FAQs)

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