Table of Contents

## Ncert Solutions For Class 11 Chemistry Chapter 1 in English

Adda247 provides NCERT solutions for class 11 chemistry chapter 1. The NCERT Solutions provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers.

These NCERT Solutions Class 11 chemistry chapter 1 are presented in a very simple language so that you can understand the basic of chemistry with ease. These NCERT Solutions class 11 Chemistry cover chapters 1 to 14 with all important questions and answers explained in a detailed way.

Students can download the Class 111 Chemistry NCERT Solutions, which they want to study with the comfort of their house.

The solutions available are in depth and simplest way. Thus will help the students beyond examination marks. This will help them develop a core understanding of the subject. Because this subject demands to understand rather than just memorizing solutions of Class 11 Chemistry. Here below we are providing you with the overview of all the chemistry Class 11 Chapters that are there in the NCERT textbook.

At Adda247, students can access chapter wise solutions to get their doubts clarified instantly. The faculty had provided both online and offline mode of Solutions which can be used free of cost.

**Benefits of Solutions of NCERT class 11 Chemistry:**

- NCERT Solutions for Class 11 is helpful to solve questions from other reference books too.
- NCERT Solutions for class 11 Chemistry will assist students to cross check answers and prepare for the exams in a strategic way.

The students can access the solutions anywhere while browsing web easily. The solutions are very precise and accurate.

## NCERT Solutions for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry

Chemistry is the science of molecules and their transformations. It is the science not so much of the one hundred elements but of the infinite variety of molecules that may be built from them.

Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it’s the study of everything.

Adda247 provides you with Class 11 Chemistry NCERT Solutions Chapter 1. Our experts of Chemistry explain the solutions of all questions as per the NCERT pattern. Some basic concepts of Chemistry Class11 NCERT Solutions are given to make your study simplistic and enjoyable at Adda247. You can get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from Adda247 website to assist you through the complete syllabus properly and obtain the best marks in your examinations.

The Chapter touches upon topics such as the importance of Chemistry, atomic mass, and molecular mass. Some basic laws and theories in Chemistry such as Dalton’s atomic theory, Avogadro’s law and the law of conservation of mass are also discussed in this chapter.

#### Download your free content now!

#### Download success!

Thanks for downloading the guide. For similar guides, free study material, quizzes, videos and job alerts you can download the Adda247 app from play store.

**NCERT Solutions for Class 11 Chemistry Chapter 1 Subtopics **

**Importance of Chemistry.****Nature of Matter****Properties of Matter and Their Measurement.****The International System of Units.****Mass And Weight****Uncertainty in Measurement.****Scientific notation****Significant Figures****Dimensional Analysis****Laws of Chemical Combinations****Laws of Conservation of Mass****Law of Definite proportions****Law of Multiple Proportions****Gay Lussac’s Law of Gaseous Volumes****Avogadro Law****Dalton’s Atomic Theory****Atomic and Molecular Masses****Atomic mass****Average Atomic Mass****Molecular Mass****Formula Mass****Mole concept And Molar Masses****Percentage Composition****Empirical Formula for Molecular Formula****Stoichiometry and Stoichiometry Calculations****Limiting Reagent****Reactions In Solutions**

** Read Ncert Solutions Class 12 Chemistry**

### Key features of NCERT solutions for class 11 Chemistry chapter 1: Some Basic Concepts of Chemistry

- The NCERT solution provide clear and precise answer.
- The columns are used wherever necessary.

** **

## Important questions of NCERT Solutions of Chemistry Class 11 Chapter 1

**Question: 1 Calculate the molar mass of the following:**

**H**_{2}O**CO**_{2}**CH**_{4}

Answer:

- H
_{2}O

The molecular mass of water

= (2 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)

= [2(1.0084) + 1(16.00u)]

= 2.016 u + 16.00u

= 18.016 u

= 18.02 u

- CO
_{2}

The molecular mass of carbon dioxide

= (1 x atomic mass of carbon) + (2 x atomic mass of oxygen)

= [1(12.011u) + 2(16.00 u)

= 12.011 u + 32.00 u

= 44.01 u

- CH
_{4}

= The molecular mass of methane

= (1 x atomic mass of carbon) + (4 x atomic mass of hydrogen)

= [1(12.011 u ) + 4(1.0084 u)

= 12.011 u + 4.032 u

= 16.043 u

**Question: 2 Calculate the mass percent of different elements present in sodium sulphate (Na _{2}SO_{4}).**

Answer:

The molecular formula of the sodium sulphate is Na2SO4

Molar mass of Na2SO4 = [(2 x 23.0) + (32.066) + 4(16.00)]

= 142.066 g

Mass percentage = Mass of that element in the compound x 100 / Molar mass of the compound

:Mass percent of the sodium:

= 46.0g x 100 / 142.066g

= 32.37 g

= 32.4%

: Mass percent of sulphur:

= 32.066 g x 100 / 142.066g

= 22.57

= 22. 6%

Mass percent of oxygen:

= 64.0 g x 2100 / 142.066 g

= 45.049

= 45.05%

** **

**Question: 3 Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.**

Answer:

% of iron by mass 69.9%

% 0f oxygen by mass 30.1%

Relative moles of iron in iron oxide:

= % of iron by mass / Atomic mass of iron

= 69.9 / 55.85

= 1.25

Relative moles of oxygen in iron oxide:

= % of oxygen by mass / Atomic mass of oxygen

= 30.1 / 16.00

= 1.88

Simplest molar ratio of iron to oxygen:

= 1.25 : 1.88

= 1 : 1.5

= 2 : 3

: The empirical formula of iron oxide is Fe2O3.

**Question: 4 Calculate the amount of carbon dioxide that could be produced when**

**1 mole of carbon is burnt in air.****1 mole of carbon is burnt in 16g of dioxygen.****2 moles of carbon are burnt in 16g of dioxygen.**

Answer:

The balanced reaction of combustion of carbon can be written as:

C + O_{2} à CO_{2}

- As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen to produce 1 mole of carbon dioxide.
- According to the equation, only 16 g dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reagent.
- According to the question, only 16 g dioxygen is available. It is a limiting reagent. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

**Question: 5 Calculate the mass of sodium acetate (CH _{3}COONa) required to make 500mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^{-1}.**

Answer:

0.375 M aqueous solution of sodium acetate

= 1000 mL of solution containing 0.375 moles of sodium acetate

: Number of moles of sodium acetate in 500 Ml

= 0.375 x 500 / 1000

= 0.1875 mole

Molar mass of sodium acetate = 82.0245 g mole^{-1} (given)

: Required mass of sodium acetate = 82.0245 g mole^{-1}(0.1875 mole)

= 15.38 g

**Question: 6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density. 1.41 g mol ^{-1} and the mass percent of nitric acid in it being 69%.**

Answer:

Mass percent of nitric acid is the sample = 69%

The 100g 0f nitric acid contains 60 g of nitric acid by mass

Molar mass of nitric acid (HNO3)

= {1 + 14 + 3(16)} gmol^{-1}

= 1 + 14 + 18

= 63 g mol^{-1}

: Number of moles in 69 g of HNO3

= 69 g / 63 g mol^{-1}

= 1.095 mol

Volume of 100g of nitric acid solution

= Mass of solution / density of solution

= 100g / 1.41 gmL^{-1}

= 70.92 Ml = 70.92 x 10^{-3}

= 15.44mol/L

Concentration of nitric acid = 15.44 mol/L

**Question: 7 How much copper can be obtained from 100g of copper sulphate (CuSO _{4}).**

Answer:

1 mole of CuSO_{4} contains 1 mole of copper

Molar mass of CuSO_{4} = (63.5) + (32.00) + 4(16)

= 159.5 g

159.5 g of CuSO_{4} contains 63.5 g of copper

= 100 g of CuSO_{4} will contains 63.5 x 100g / 159.5 of copper

= Amount of copper that can be obtained from 100g of CuSO_{4} = 63.5 x 100 / 159.5

= 39.81 g

**Question: 8 Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.**

Answer:

From the available data Percentage of iron = 69.9

Percentage of oxygen= 30.1

Total percentage of iron & oxygen= 69.9+30.1= 100%

Step 1 calculation of simplest whole number ratios of the elements

Element | Percentage | Atomic mass | Atomic ratio | Simplest ratio | Simplest whole ratio |

F = 1e | 69.9 | 55.84 | 69.9 / 55.84 = 1.25 | 1.25 | 2 |

O | 30.1 | 16 | 30.1 / 16 = 1.88 | 1.88 = 1.5 | 3 |

Step 2 Writing the empirical formula of the compound

The empirical formula of the compound = Fe2 O3

Step 3 determination of molecular formula of the compound

Empirical formula mass = 2 X69.9 + 3 X16=187.8 amu

Molecular mass of oxide= 159.69g/mol(given)

Now we know molecular formula = n x Empirical formula

And n= molecular mass / empirical formula mass= 159.69/187.8 = 0.85 = approx 1

Therefore molecular formula = n x empirical formula

=1 x(Fe2O3) = Fe2O3

The molecular formula of the oxide is Fe2O3

** **

**Question: 9 Calculate the atomic mass (average) of chlorine using the following data:**

** % Natural abundance Molar Mass**

^{35}**Cl 75.77 34.9689**

^{37}**Cl 24.23 36.9659**

Answer:

The average atomic mass of chlorine

= (Fractional abundance of ^{35}Cl) + (Molar mass of ^{36}Cl) + (Fractional abundance of ^{37}Cl)

= [{75.77 / 100}(34.9689 u)} + {(24.23 / 100)(36.9659)}[]

= 35.4527 u.

**Question: 10 In three moles of ethane (C _{2}H_{6}), calculate the following:**

**Number of moles of carbon atoms.****Number of moles of hydrogen atoms.****Number of molecules of ethane.**

Answer:

- 1 mole of C2H6 contains 2 moles of carbon atoms

: Number of moles of carbon atoms in 3 moles of C2H6

= 3 x 6 = 18

- 1 mole of C2H6 contains 6.023 x 10
^{23}molecules of ethane

:Number of molecules in 3 moles of C2H6

= 3 x 6.022 x 10^{23} = 18.069 x 10^{23}

**Question: 11 What is the concentration of sugar in molL-1. If its 20 g are dissolved in enough water to make a final volume up to 2L?**

Answer:

Molarity of a solution is given by:

= Number of moles of solute / Volume of solution in Litres

= Mass of sugar / molar mass of sugar

2 L

= 20 g / [(12 x 12) + (1 x 22) + (11 x 16)]g

2 L

= 20g / 342 g

2L

= 0.0585 mol / 2L

= : Molar concentration of sugar = 0.02925 mol /L

**Question:12 If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of its 0.25M solution?**

Answer:

Molar mass of methanol(CH2OH) = (1 x 12) + (4 x 1) + (1 x 16)

= 32 g mol-1

= 0.032 kg mol-1

Molarity of the methanol solution = 0.793kg L-1 / 0.032 kg mol-1

= 24.78 mol L-1

(Since density is mass per unit volume)

Applying, M1V1 = M2V2

(Given solution)(Solution to be prepared)

V1 = 0.0252 L

V1 = 25.22Ml

**Question:13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:**

**1 Pa = 1 N m-2**

**If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.**

Answer:

Pressure is defined as force acting per unit area of the surface.

P = F / A

= 1034g x 9.8ms-1 x 1kg x (100)2cm2

Cm2 1000g 1 m2

= 1.01332 x 106kg m-1s-2

We know,

1N = 1kg ms-2

Then,

1 Pa = 1Nm-2 1Kg m-1s-2

: Pressure = 1.01332 x 106 Pa

**Question:14 What is the SI unit of mass? How is it defined?**

Answer:

The SI unit of mass is kilogram (kg). 1 kilogram is defined as the mass equal to the mass of international prototype of kilogram.

**Question:15 Match the following prefixes with their multiples:**

**Prefixes Multiples**

**i.) Micro 106**

**ii.) Deca 109**

**iii.) Mega 10-6**

**iv.) Giga 10-15**

**v.) femto 10**

Answer:

Prefixes

Multiples

micro

10-6

Deca

10

Mega

106

Giga

109

Femto

10-15

**Question:16 What do you mean by significant figures?**

Answer:

Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, tyhen 15 is certain while 6 is uncertain and the total number of significant figures are 3.

Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainity of the result.

**Question:17 A sample of drinking water was found to be severely contaminated with chloroform. CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).**

**i.) Express this in percent by mass.**

**ii.) Determine the molality of chloroform in the water sample.**

Answer:

i.) 1 ppm is equivalent ti 1 part out of 1 million (1060 parts.

: Mass percent of 15 ppm chloroform in water

= 15 x 100 / 106

= 15 x 10-3%

ii.) 100 g of the sample contains 1.5 x 10-3g of CHCl3

= 1000 g of the sample contains 1.5 x 10-2 g of CHCl3

Molality of chloroform in water

= 1.5 x 10-2 g / Molar mass of CHCl3

Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5)

= 119.5 g / mol

Molality of chjloroform in water = 0.0125 x 10-2m

= 1.25 x 10-4m

**Question:18 Express the following in scientific notation:**

**i.) 0.0048**

**ii.) 234,000**

**iii.) 8008**

**iv.) 500.0**

**v.) 6.0012**

Answer:

i.) 0.0048 = 4.8 x 10-3

ii.) 234,000 = 2.34 x 105

iii.) 8008 = 8.008 x 103

iv.) 500.0 = 5.000 x 102

v.) 6.0012 = 6.0012 x 100

**Question:19 How many significant figures are present in the following:**

**i.) 0.0025**

**ii.) 208**

**iii.) 5005**

**iv.) 126,000**

**v.) 500.0**

**vi.) 2.0034**

Answer:

i.) 0.0025

There are two significant figures.

ii.) 208

There are 3 significant figures

iii.) 5005

There are 4 significant figures

iv.) 126.000

There are 3 significant figures

v.) 500.0

There are 4 significant figures

vi.) 2.0034

There are 5 significant figures.

**Question:20 Round up the following upto three significant figures:**

**i.) 34.216**

**ii.) 10.4107**

**iii.) 0.04597**

**iv.) 2808**

Answer:

i.) 34.216

ii.) 10.4107

iii.) 0.04597

iv.) 2808

**Question:21 The following data obtained when dinitrogen and dioxygen react together to form different compound:**

**Mass of dinitrogen**

**Mass of dioxygen**

**14 g**

**16 g**

**14 g**

**32 g**

**28 g**

**32 g**

**28 g**

**80 g**

**a.) Which law of chemical combination is obeyed by the above experimental data? Give statement.**

**b.) Fill in the blanks in the following conversions:**

**i.) 1 km = ———mm = ………….. pm**

**ii.) 1 mg = ………..kg = ……………ng**

**iii.) 1 ml = …………L = ……………..dm3**

Answer:

Let us fix 14 parts by weight of nitrogen as fixed weight.

Now let us calculate the weights of oxygen which combine with 14 parts by weight of nitrogen

S No

No. of parts by weight of nitrogen

No. of parts by weight of oxygen

14 parts of nitrogen as fixed weight

No. of parts by weight of oxygen which combine with 14 parts by weight of nitrogen

1

14 g

16 g

14 g

16

2

14 g

32 g

14 g

32

3

28 g

32 g

14 g

32

4

28 g

80 g

14 g

80

(a) If we fix the mass of dinitrogen at 14 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16 g, 32 g, 32 g, and 80 g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions.

This law was given by Dalton in 1804. The law states that if two elements combine to form 2 or more compound, then the weight of one element which combines a fixed weight of other element in these compounds, bears a simple whole number ratio by weight.

(b) (i) We know 1km=1000m

Or 1m = 1000 mm

Therefore 1km = 1000x 1000mm= 106 mm

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

1 km = 1015 pm

Hence, 1 km = 106 mm = 1015 pm

(ii) We know 1kg = 1000mg

Or 1000mg= 1kg

Or 1mg= 1/1000* 1= 0.01 kg

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

⇒ 1 mg = 106 ng

1 mg = 10-6 kg = 106 ng

(iii) We know 1000 ml=l L

Or 1ml=1/1000*1= 0.01L

1 mL = 1 cm3 = 1 cm3

⇒ 1 mL = 10-3 dm3

1 mL = 10-3 L = 10-3 dm3

**Question:22 If the speed of light is 3.0 x 108m s-1. Calculate the distance covered by light in 2.00ns.**

Answer:

According to the equation:

Time taken to cover the distance = 2.00 ns

= 2.00 x 10-9s

Speed of light = 3 x 108ms-1

Distance travelled by light in 2.00 ns

= Speed of light x time taken

= (3.0 x 108 ms-1) (2.00 x 10-9)

= 6.00 x 10-1 m

0.600 m

**Question:23 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 A + 3 mol B**

**iii.) 100 atom of A + 100 molecules of B**

**iv.) 5 mol A + 2.5 mol B**

**v.) 2.5 mol A + 5 mol B**

Answer:

A limiting reagent determines the extentof a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting tha amount of product formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent. Here atom B is in lesser amount(200).

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent because B is less as compared to A

(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent

**Question:24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:**

**N2(g) + H2(g) 2NH3 (g)**

**i.) Calculate the mass of ammonia produced if 2.00 x 103g dinitrogen reacts with 1.00 x 103 g of hydrogen.**

**ii.) Will any of the reactants remain unreacted?**

**iii.) If yes, which one and what would be its mass?**

Answere:

(i) Balancing the given chemical equation,

N2(g) + H2(g) 2NH3 (g)

Total mass of Ammonia = 2((14) +3(1)) = 34 g

From the chemical equation, we can write

28gm of N2 reacts with 6gm of H2 to produce ammonia= 34g

Or 1 gm of N2 reacts with 1gm of H2 to produce ammonia= 34/28*1

Or when 2.00 x 103 g of N2 reacts with 1.00 x 103 gm of H2 to produce ammonia

=34/28 *2.00 x 103 = 2428.57g

Hence 2.00 × 103 g of dinitrogen will react with 1.00 x 103 g of dihydrogen to give 2428.57 g of ammonia

Given, Amount of dihydrogen = 1.00 × 103 g

Hence, N2 is the limiting reagent.

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g

= 571.4 g

**Question:25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?**

Answer:

Molar mass of Na2CO3 = (2 x 23) + 12.00 + (3 x 6)

= 106 g mol-1

Now, 1 mole of Na2CO3 means 106 g of Na2CO3

: 0.5 mol of Na2CO3 = 106 g x 0.5 mol / 1 mole

= 53 g Na2CO3

= 0.50 M of Na2CO3 = 0.50 mol / L Na2CO3

Hence, 0.50 mol of Na2CO3 is present in 1 L of water or 53 g Na2CO3 is present in 1 L of water.

**Question: 26 If ten volumes of dihydrogen gas reacts with five volume of dioxygen gas, how many volumes of water vapour would be produced?**

Answer:

Reaction of dihydrogen with dioxygen can be written as:

2H2(g) + O29g) 2H2O(g)

Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour

Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

**Question:27 Convert the following in basic units:**

**i.) 28.7 pm**

**ii.) 15.15 pm**

**iii.) 25365 mg**

Answer:

(i) 28.7 pm:

1 pm = 10-12 m

28.7 pm = 28.7 × 10-12 m

= 2.87 × 10-11 m

(ii) 15.15 pm:

1 pm = 10-12 m

15.15 pm = 15.15 × 10-12 m

= 1.515 × 10-12 m

(iii) 25365 mg:

1 mg = 0.01 kg

Therefore 25365 mg = 0.01/1 x 25365= 253.65 kg

**Question:28 Which one of the following will have largest no. of atoms:**

**i.) 1 g Au (s)**

**ii.) 1 g Na (s)**

**iii.) 1 g Li (s)**

**iv.) 1 g Cl2 (g)**

Answer:

(i) Gram atomic mass of Au= 197 g

Or

197g of Au contains = 6.022 x 1023

Therefore 1gm of Au contains = 6.022 x 1023/197*1 = 3.06 x 1021 atoms

(ii) Gram atomic mass of Na = 23 g

Or

23 g of Na contains atoms = 6.022x 1023

Or

1gm of Na contains atoms = 6.022×1023 /23 *1 = 26.2 x1021 atoms

(iii) Gram atomic mass of Li = 7

Or

7g of Li contains atoms = 6.022 x 1023

Or

1g of Li contains atoms = 6.022 x 1023/7 *1= 86.0 x 1021 atoms

(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022×1023

Or

1 g of Cl contains atoms = 6.022×1023 /71 * 1= 8.48 x 1021 atoms

Hence, 1 g of Li (s) will have the largest number of atoms

**Question:29 Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).**

Answer:

Mole fraction of C2H5OH= Number of moles of C2H5OH / Number of moles of solution

Let the moles of C2H5OH= X

Now density of water = 1 (given)

And the weight of 1000ml of water = volume * density (from density = mass/volume)

= 1000 x 1=1000g

Therefore moles of water = 1000/18 = 55.55 mol (18g is molecular mass of water)

Also mole fraction of C2H5OH= 0.040 (given)

Putting the values in equation 1,we get

= 0.040 = X/X + 55.55

=0.040X + 2.222 = X

OR

X= 2.3145 mol

Molarity of solution = 2.314 M

**Question:30 What will be the mass of one 12C atom in g?**

Answer:

1 mole of carbon atoms = 6.022 x 1023 atoms of carbon

= 12 g of carbon

: Mass of one 12C atom = 12g/ 6.022 x 1023

= 1.993 x 10-23g

**Question:31 How many significant figures should be present in the answer of the following calculations?**

i.) 0.02856 x 298.15 x 0.112

0.5785

ii.) 5 x 5.364

iii.) 0.0125 + 0.7864 + 0.0215

Answer:

(i) Least precise term i.e. 0.112 is having 3 significant digits.

∴ There will be 3 significant figures in the calculation.

(ii) 5.364 is having 4 significant figures.

∴ There will be 4 significant figures in the calculation.

(iii) Least number of decimal places in each term is 4.

∴ There will be 4 significant figures

**Question:32 Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:**

**Isotope**

**Isotopic molar mass**

**Abundance**

**36Ar**

**35.96755 g mol-1**

**0.337%**

**38Ar**

**37.96272 g mol-1**

**0.063%**

**40Ar**

**39.9624 g mol-1**

**99.600%**

Answer:

Molar mass of argon

= [ (35.96755 x 0.337/100) + (37.96272 x 0.063/100) + (39.9624 x 90.60/100) ] g mol-1

= [ 0.121 + 0.0024 + 39.802 ] gmol-1

= 39.947 gmol-1

**Question:33 Calculate the no. of atoms in each of the following :**

**i.) 52 moles of Ar**

**ii.) 52 u of He**

**iii.) 52 g of He**

Answer:

(i) 1 mole of Ar = 6.022 × 1023 atoms of Ar

52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar

= 3.131 × 1025 atoms of Ar

(ii) Atomic mass of He = 4amu

Or

4amu is the mass of He atoms = 1

Therefore 52 amu is the mass of He atoms= ¼*52 = 13 atoms of He

(iii) Gram atomic mass of He = 4g

Or

4g of He contains = 6.022x 1023 atoms

Therefore 52 g of He contains = 6.022x 1023 /4 * 52 = 7.83 x 1024 atoms

**Question:34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate:**

**i.) Empirical formula**

**ii.) Molar mass of the gas**

**iii.) Molecular formula**

Answer:

(i) Mass of carbon in 3.38 g of CO2 = 3.38 g/44 x 12 = 0.922 g

Mass of hydrogen in 0.690 g of H2O = 0.690 g/18 x 2 = 0.077 g

Total mass of the sample burnt = 0.922 g + 0.077 g = 0.999 g

Percentage of carbon in the fuel = {0.922}/{0.999 g} x 100 = 92.29%

Percentage of hydrogen in the fuel = {0.077 g}/{0.999 g} x 100 = 7.71%

Element

Mass percent

Atomic mass

Relative no. of atoms

Simple atomic ratio

Carbon (C)

92.29

12.0

92.29 / 12.0 = 7.69

7.69 / 7.69 = 1

Hydrogen

7.71

1.0

7.71 / 1.0 = 7.71

7.71 / 7.69 = 1

Therefore, empirical formula of the compound = CH

(ii) Volume of the gaseous fuel = 11.6 g

Molar mass of the fuel = {11.6 g}/{10.0 L} x 22.4 L/ml = 26.0 g mol-1

(iii) Empirical formula mass of the fuel = (12 + 1) g mol-1 = 13 g mol-1

Molar mass of the fuel = 26.0 g mol-1 n = {26.0 g mol-1}/{13 g mol-1} = 2

Molecular formula of the fuel = 2 x Empirical formula = 2 x CH = C2H2

**Question:35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) CaCl2(aq) + CO2(g) + H2O(l)**

**What mass of CaCO3 is required to react completely with 25 ml of 0.75 M HCl?**

Answer:

0.75 M of HCl = 0.75 mol of HCl are present in 1 L of water

= [(0.75 mol ) x (36.5 g mol-1)] HCl is present in 1 L of water

= 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution

= 27.375g x 25mL / 1000 Ml

= 0.6844 g

From the given chemical equation

CaCO2(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

2 mol of HCl (2 x 36.5 = 71 g) react wuth 1 mol of CaCO3 (100g).

: Amount of CaCO3 that will react with 0.6844 g = 100 x 0.6844 g / 71

= 0.9639 g

**Question:36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction**

**4HCl(aq) + MnO2(s) 2H2O(l) + MnCl2(aq) + Cl2(g)**

**How many grams of HCl react with 5.0 g of manganese dioxide?**

Answer:

1 mol [55 + 2 x 16 = 87 g] MnO2 reacts completely with 4 mol[4 x 36.5 = 146 g] of HCl

: 50 g of MnO2 will react with

= 146g x 5.0 g / 87 g

= 8.4 g of HCl

Hence, 8.4 g of HCl will react with 5.0 g of manganese dioxide.

## FAQs on NCERT solution of Chemistry Class 11 Chapter 1

**Carbon generates two oxides, namely CO and CO2, where masses of carbon which combine with Mass of oxygen are in the ratio 1:2. State how the law of multiple proportions is being followed.**

Ans. The element carbon, combines with oxygen to form two different chemical compounds, which are, carbon dioxide and carbon monoxide. In CO2, 12 parts of carbon mass combine with 32 parts of oxygen mass while in CO, 12 parts of carbon mass combine with 16 parts of oxygen mass. Therefore, the masses of oxygen mix with a definite mass of 12 parts of carbon in CO2 and CO are 32 and 16 respectively. These oxygen masses bear a simple ratio of 32:16 or 2:1 to each other and this is a classic example of the law of multiple proportions.

**Define the law of multiple proportions. Explain it with an example. How does this law prove the existence of atoms?**

Ans. Law of multiple proportions: When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another, e.g., carbon combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide.

The masses of oxygen which combine with a fixed mass of carbon in CO2 and CO are 32 and 16 respectively. These masses of oxygen bear a simple ratio of 32 : 16 or 2 : 1 to each other. For example, sulphur combines with oxygen to form two compounds, namely, sulphur trioxide and sulphur dioxide.

The masses of oxygen which combine with a fixed mass of sulphur in SO3 and SO2 are 48 and 32 respectively. These masses of oxygen bear a simple ratio of 48 : 32 or 3 : 2 to each other. This law shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine to form molecules.

**Give an overview of questions present in NCERT Solutions for Class 11 Chemistry Chapter 1.**Ans. NCERT Solutions for Class 11 Chemistry Chapter 1 has 3 exercises. The concepts of this chapter are listed below.- Numerical problems in calculating the molecular weight of compounds.
- Numerical problems in calculating mass percentage and concentration.
- Problems on empirical and molecular formulae.
- Problems on molarity and molality.
- Other problems related to the mole concept.