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

NCERT Solutions Class 11 Chemistry Chapter 1 – Free PDF Download

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry are provided on this page for Class 11 Chemistry students. These solutions can also be downloaded in a PDF format for free by clicking the download button provided below. The free Class 11 Chemistry NCERT Solutions provided by BYJU’S have been prepared by seasoned academics and chemistry experts keeping the needs of Class 11 students in mind. The NCERT Solutions provided on this page contain step-by-step explanations in order to help students answer similar questions that might be asked in their board examinations.

The expert tutors have created NCERT Solutions for Class 11 in a simple and understandable language in accordance with the latest CBSE Syllabus 2022-23 and its guidelines. The basic concepts are covered in the solutions with the aim of providing a quality learning experience for Class 11 students. You can download the Class 11 Chemistry NCERT Solutions Chapter 1 PDF from BYJU’S to complete a part of the CBSE syllabus before the board exam.

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

“Some Basic Concepts of Chemistry” is the first chapter in the Class 11 Chemistry syllabus as prescribed by NCERT under the first term . 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.

The types of questions provided in the NCERT Class 11 Chemistry textbook for Chapter 1 include:

  • Numerical problems in calculating the molecular weight of compounds.
  • Numerical problems in calculating mass percent and concentration.
  • Problems on empirical and molecular formulae.
  • Problems on molarity and molality.
  • Other problems related to the mole concept (such as percentage composition and expressing concentration in parts per million).

The NCERT Solutions of Chemistry provided on this page for Class 11 Chapter 1 contains detailed explanations on the steps to be followed while solving the numerical value questions that are frequently asked in first term examinations. The subtopics covered under the chapter are listed below.

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

  1. Importance of Chemistry
  2. Nature of Matter
  3. Properties of Matter and Their Measurement  
  4. The International System of Units (SI)
  5. Mass and Weight
  6. Uncertainty in Measurement
  7. Scientific Notation
  8. Significant Figures
  9. Dimensional Analysis
  10. Laws of Chemical Combinations    
  11. Law of Conservation Of Mass    
  12. Law of Definite Proportions
  13. Law of Multiple Proportions  
  14. Gay Lussac’s Law of Gaseous Volumes  
  15. Avogadro’s Law
  16. Dalton’s Atomic Theory
  17. Atomic and Molecular Masses   
  18. Atomic Mass
  19. Average Atomic Mass
  20. Molecular Mass
  21. Formula Mass
  22. Mole Concept and Molar Masses
  23. Percentage Composition    
  24. Empirical Formula for Molecular Formula
  25. Stoichiometry and Stoichiometric Calculations                
  26. Limiting Reagent
  27. Reactions in Solutions

Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 1

Exercise

Q1. Calculate the molar mass of the following:

(i) CH4CH_{4}      (ii)H2OH_{2}O      (iii)CO2CO_{2}

Ans.

(i) CH4CH_{4} :

Molecular mass of CH4CH_{4} = Atomic mass of C + 4 x Atomic mass of H

= 12 + 4 x 1

= 16 u

(ii) H2OH_{2}O :

Molar mass of water H2OH_{2}O

Atomic mass of H = 1

Atomic mass of O = 16

H2O = 2×H+1×O

Molar mass of water = 2×1+16 = 18g/mol

(iii) CO2CO_{2} :

Molecular mass of CO2CO_{2} = Atomic mass of C + 2 x Atomic mass of O

= 12 + 2 × 16

= 44 u

Q2. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4Na_{2}SO_{4}) .

Ans.

Now for Na2SO4Na_{2}SO_{4}.

Molar mass of Na2SO4Na_{2}SO_{4}

= [(2 x 23.0) + (32.066) + 4(16.00)]

=142.066 g

Formula to calculate mass percent of an element = Mass  of  that  element  in  the  compoundMolar  mass  of  the  compound×100\frac{Mass\;of\;that\;element\;in\;the\;compound}{Molar\;mass\;of\;the\;compound}\times 100

 

Therefore, mass percent of the sodium element:

= 46.0g142.066g×100\frac{46.0g}{142.066g}\times 100

= 32.379

= 32.4%

Mass percent of the sulphur element:

= 32.066g142.066g×100\frac{32.066g}{142.066g}\times 100

= 22.57

= 22.6%

Mass percent of the oxygen element:

= 64.0g142.066g×100\frac{64.0g}{142.066g}\times 100

= 45.049

= 45.05%

 

Q3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Ans.

Given there is an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass:

Relative moles of iron in iron oxide:

= percent  of  iron  by  massAtomic  mass  of  iron\frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}

= 69.955.85\frac{69.9}{55.85}

= 1.25

 

Relative moles of oxygen in iron oxide:

= percent  of  oxygen  by  massAtomic  mass  of  oxygen\frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}

= 30.116.00\frac{30.1}{16.00}

= 1.88

 

The simplest molar ratio of iron to oxygen:

 1.25: 1.88 ⇒ 1: 1.5 ⇒ 2: 3

Therefore, the empirical formula of the iron oxide is Fe2O3Fe_{2}O_{3}.

 

Q4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Ans.

(i) 1 mole of carbon is burnt in air.

C+O2CO2C+O_{2}\rightarrow CO_{2}

1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2.

Amount of CO2CO_{2} produced = 44 g

 

(ii) 1 mole of carbon is burnt in 16 g of O2.

1 mole of carbon burnt in 32 grams of O2 it forms 44 grams of CO2CO_{2}.

Therefore, 16 grams of O2 will form 44×1632\frac{44\times 16}{32}

= 22 grams of CO2CO_{2}

 

(iii) 2 moles of carbon are burnt in 16 g of O2.

Here again, dioxygen is the limiting reactant. 16g of dioxygen can combine only with 0.5mol of carbon. CO2 produced again is equal to 22g.

Q5. Calculate the mass of sodium acetate (CH3COONa)(CH_{3}COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.

 

Ans.

0.375 M aqueous solution of CH3COONaCH_{3}COONa

= 1000 mL of solution containing 0.375 moles of CH3COONaCH_{3}COONa

Therefore, no. of moles of CH3COONaCH_{3}COONa in 500 mL

= 0.3751000×500\frac{0.375}{1000}\times 500

= 0.1875 mole

Molar mass of sodium acetate = 82.0245  g  mol182.0245\;g\;mol^{-1}

Therefore, the mass of CH3COONaCH_{3}COONa

= (82.0245  g  mol1)(0.1875  mole)(82.0245\;g\;mol^{-1})(0.1875\;mole)

= 15.38 grams

 

Q6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%.

Ans.

Mass percent of HNO3 in sample is 69 %

Thus, 100 g of HNO3 contains 69 g of HNO3 by mass.

Molar mass of HNO3

= { 1 + 14 + 3(16)} g.mol1g.mol^{-1}

= 1 + 14 + 48

=63g  mol1= 63g\;mol^{-1}

 

Now, no. of moles in 69 g of HNO3HNO_{3}:

= 69g63gmol1\frac{69\:g}{63\:g\:mol^{-1}}

= 1.095 mol

 

Volume of 100g HNO3 solution

= Mass  of  solutiondensity  of  solution\frac{Mass\;of\;solution}{density\;of\;solution}

= 100g1.41g  mL1\frac{100g}{1.41g\;mL^{-1}}

= 70.92mL

= 70.92×103  L70.92\times 10^{-3}\;L

 

Concentration of HNO3

= 1.095mole70.92×103L\frac{1.095\:mole}{70.92\times 10^{-3}L}

= 15.44mol/L

Therefore,

Concentration of HNO3 = 15.44 mol/L

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

Ans.

1 mole of CuSO4CuSO_{4} contains 1 mole of Cu.

Molar mass of CuSO4CuSO_{4}

= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 grams

159.5 grams of CuSO4CuSO_{4} contains 63.5 grams of Cu.

Therefore, 100 grams of CuSO4CuSO_{4} will contain 63.5×100g159.5\frac{63.5\times 100g}{159.5} of Cu.

= 63.5×100159.5\frac{63.5\times 100}{159.5}

=39.81 grams

 

Q8. 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. 

Ans.

Here,

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

 

No. of moles of Fe present in oxide

= 69.9055.85\frac{69.90}{55.85}

= 1.25

 

No. of moles of O present in oxide

= 30.116.0\frac{30.1}{16.0}

=1.88

 

Ratio of Fe to O in oxide,

= 1.25: 1.88

= 1.251.25:1.881.25\frac{1.25}{1.25}:\frac{1.88}{1.25}

=1:1.51:1.5

= 2:32:3

Therefore, the empirical formula of oxide is Fe2O3Fe_{2}O_{3}

 

Empirical formula mass of Fe2O3Fe_{2}O_{3}

= [2(55.85) + 3(16.00)] g

= 159. 7g

The molar mass of Fe2O3Fe_{2}O_{3} = 159.69g

Therefore n = Molar  massEmpirical  formula  mass=159.69  g159.7  g\frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}

= 0.999

= 1(approx)

 

The molecular formula of a compound can be obtained by multiplying n with the empirical formula.

Thus, the empirical of the given oxide is Fe2O3Fe_{2}O_{3} and n is 1.

Therefore, the molecular formula of the oxide is Fe2O3Fe_{2}O_{3}

 

Q9. Calculate the atomic mass (average) of chlorine using the following data:

Percentage Natural Abundance Molar Mass
35Cl_{}^{35}\textrm{Cl} 75.77 34.9689
37Cl_{}^{37}\textrm{Cl} 24.23 36.9659

 

Ans.

Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689  

Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659 

Average Atomic mass = (0.7577 x 34.9689)amu + (0.2423 x 36.9659) 

= 26.4959 + 8.9568 = 35.4527

 

Q10. In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atom

(iii) Number of molecules of ethane

Ans.

(i) 1 mole of C2H6C_{2}H_{6} contains two moles of C- atoms.

No. of moles of C- atoms in 3 moles of C2H6C_{2}H_{6}.

= 2 x 3

= 6

(ii) 1 mole of C2H6C_{2}H_{6} contains six moles of H- atoms.

No. of moles of H- atoms in 3 moles of C2H6C_{2}H_{6}.

= 3 x 6

= 18

(iii) 1 mole of C2H6C_{2}H_{6} contains 1 mole of ethane- atoms.

No. of molecules in 3 moles of C2H6C_{2}H_{6}.

= 3 x 6.023 x 102310^{23}

= 18.069 x 102310^{23}

 

Q11. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?

 

Ans.

Molarity (M) is as given by,

= Number  of  moles  of  soluteVolume  of  solution  in  Litres\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}

= Mass  of  sugarMolar  mass  of  sugar2  L\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}

= 20  g[(12  ×  12)  +  (1  ×  22)  +  (11  ×  16)]g]2  L\frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}

= 20  g342  g2  L\frac{\frac{20\;g}{342\;g}}{2\;L}

= 0.0585  mol2  L\frac{0.0585\;mol}{2\;L}

= 0.02925 molL1L^{-1}

Therefore, Molar concentration = 0.02925 molL1L^{-1}

 

Q12. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

 

Ans.)

Molar mass of methanol (CH3OH)

= 32 gmol-1 = 0.032 kgmol-1

molarity of the given solution

=W2inkgMw2×V(sol)L=dsol(kgL1)Mw2(kg)=0.793kgL10.032kgmol1=24.78MApplyingM1×V1(Givensolution)=M2V2(solutiontobeprepared)=\frac{W_{2}in kg}{M_{w_{2}}\times V_{(sol)}L}=\frac{d_{sol}(kgL^{-1})}{Mw_{2}(kg)}\\=\frac{0.793kgL^{-1}}{0.032kgmol^{-1}}= 24.78 M\\\underset{(Given solution)}{Applying M_{1}\times V_{1}}= \underset{(solution to be prepared)}{M_{2}V_{2}}

24.78 x V1 = 0.25 x 2.5 L

or V1 = 0.02522L = 25.22mL

 

Q13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal

Ans.

Pressure is the force (i.e., weight) acting  per unit area

But weight = mg

∴ Pressure = Weight per unit area

=1034g×9.8ms2cm2=1034g×9.8ms2cm2×1kg1000g×100cm×100cm1m×1m×1Nkgms2×1Pa1Nm2=1.01332×105Pa=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\\=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\times \frac{1kg}{1000g}\times \frac{100cm\times 100cm}{1m\times 1m}\times \frac{1N}{kgms^{-2}}\times \frac{1Pa}{1Nm^{-2}}\\= 1.01332\times 10^{^{5}}Pa

 

Q14. What is the SI unit of mass? How is it defined?

Ans.

The SI unit of mass is kilogram (kg). A kilogram is equal to the mass of a platinum-iridium cylinder kept at the International Bureau of Weights and Measures at Sèvres, France.

 

Q15. Match the following prefixes with their multiples:

 

  Prefixes Multiples
(a) femto 10
(b) giga 101510^{-15}
(c) mega 10610^{-6}
(d) deca 10910^{9}
(e) micro 10610^{6}

 

Ans.

Prefixes Multiples
(a) femto 101510^{-15}
(b) giga 10910^{9}
(c) mega 10610^{6}
(d) deca 10
(e) micro 10610^{-6}

 

 

Q16. What do you mean by significant figures?

Ans.

Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in experimented value.

e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3.

Therefore, “the total number of digits in a number with the last digit that shows the uncertainty of the result is known as significant figures.”

 

Q17. 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 per cent by mass.
(ii) Determine the molality of chloroform in the water sample.

Ans.

(i) 1 ppm = 1 part out of 1 million parts.

Mass percent of 15 ppm chloroform in H2O

= 15106×100\frac{15}{10^{6}} \times 100

= \approx 1.5 ×10310^{-3} %

 

(ii) Molarity=15/119.5106×103=1.25×104Molarity = \frac{15/119.5}{10^{6}\times 10^{-3}}= 1.25 \times 10^{-4}

 

Q18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

 

Ans.

(i) 0.0048= 4.8 ×10310^{-3}

(ii) 234,000 = 2.34 ×10510^{5}

(iii) 8008= 8.008 ×10310^{3}

(iv) 500.0 = 5.000 ×10210^{2}

(v) 6.0012 = 6.0012 ×10010^{0}

 

Q19. How many significant figures are present in the following?

(a) 0.0025

(b) 208

(c) 5005

(d) 126,000

(e) 500.0

(f) 2.0034

Ans.

(a) 0.0025: 2 significant numbers.

(b) 208: 3 significant numbers.

(c) 5005: 4 significant numbers.

(d) 126,000:3 significant numbers.

(e) 500.0: 4 significant numbers.

(f) 2.0034: 5 significant numbers.

 

Q20. Round up the following upto three significant figures:

(a) 34.216

(b) 10.4107

(c)0.04597

(d)2808

 

Ans.

(a) The number after round up is: 34.2

(b) The number after round up is: 10.4

(c)The number after round up is: 0.0460

(d)The number after round up is: 2810

 

Q21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

  Mass of dioxygen Mass of dinitrogen
(i) 16 g 14 g
(ii) 32 g 14 g
(iii) 32 g 28 g
(iv) 80 g 28 g

 

(a) Which law of chemical combination is obeyed by the above experimental data?
Give its 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

 

Ans.

(a)

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

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.

(b)

i. 1km=1km×1000m1km×100cm1m×10mm1cm=106mm1km=1km×1000m1km×1pm1012m=1015pm1 km = 1 km \times \frac{1000m}{1km}\times \frac{100cm}{1m}\times \frac{10mm}{1cm}= 10^{6}mm\\1km = 1 km \times \frac{1000m}{1km}\times \frac{1pm}{10^{-12}m}= 10^{^{15}}pm

ii. 1mg=1mg×1g1000mg×1kg1000g=106kg1mg=1mg×1g1000mg×1ng109g=106ng1 mg = 1 mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}= 10^{-6}kg\\1mg = 1mg\times \frac{1g}{1000mg}\times \frac{1ng}{10^{-9}g}=10^{6}ng

iii. 1mL=1mL×1L1000mL=103L1mL=1cm3=1cm3×1dm×1dm×1dm10cm×10cm×10cm=103dm31mL = 1mL\times \frac{1L}{1000mL}=10^{-3}L\\1mL = 1cm^{3} \\ =1cm^{3}\times \frac{1dm\times 1dm\times 1dm}{10cm\times 10cm\times 10cm}= 10^{-3}dm^{3}

 

Q22. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns.

Ans.

Time taken = 2 ns

= 2 × 10910^{ -9 } s

Now,

Speed of light = 3 × 10810^{ 8 } ms1ms^{ -1 }

We know that,

Distance = Speed x Time

So,

Distance travelled in 2 ns = speed of light x time taken

= (3 × 10810^{ 8 })(2 × 10910^{ -9 })

= 6 × 10110^{ -1 } m

= 0.6 m

 

Q23. 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 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

 

Ans.

Limiting reagent:

It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limits the amount of product formed.

 

(i) 300 atoms of A + 200 molecules of B

1 atom of A reacts with 1 molecule of B. Similarly, 200 atoms of A reacts with 200 molecules of B, so 100 atoms of A are unused. Hence, B is the limiting reagent.

 

(ii) 2 mol A + 3 mol B

1 mole of A reacts with 1 mole of B. Similarly, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused. Hence, A is the limiting reagent.

 

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

1 atom of A reacts with 1 molecule of Y. Similarly, 100 atoms of A reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting reagent.

 

(iv) 5 mol A + 2.5 mol B

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

 

(v) 2.5 mol A + 5 mol B

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

 

Q24. 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 NH3NH_{ 3 } produced if 2  ×  1032 \; \times \;10^{ 3 } g N2 reacts with 1  ×  1031 \; \times \;10^{ 3 } g of H2?

 (ii) Will any of the two reactants remain unreacted?

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

 

Ans.

(i) 1 mol of N2 i.e., 28 g reacts with 3 moles of H2 i.e., 6 g of H2

∴ 2000 g of N2 will react with H2628×200g=428.6g\frac{6}{28}\times 200g = 428.6g

Thus, N2 is the limiting reagent while H2 is the excess reagent

2 mol of N2 i.e., 28 g of N2 produces NH3 = 2 mol

= 34 g

Therefore, 2000 g will produces NH33428×2000g\frac{34}{28}\times 2000 g

= 2428.57 g

(ii) H2 will remain unreacted

(iii) Mass left unreacted = 1000g – 428.6g = 571.4g

 

Q25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Ans.

Molar mass of Na2CO3Na_{ 2 }CO_{ 3 }:

= (2 × 23) + 12 + (3 × 16)

= 106 g mol1mol^{ -1 }

1 mole of Na2CO3Na_{ 2 }CO_{ 3 } means 106 g of Na2CO3Na_{ 2 }CO_{ 3 }

Therefore, 0.5 mol of Na2CO3Na_{ 2 }CO_{ 3 }

= 106  g1  mol  ×  0.5  mol\frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol Na2CO3Na_{ 2 }CO_{ 3 }

= 53 g of Na2CO3Na_{ 2 }CO_{ 3 }

0.5 M of Na2CO3Na_{ 2 }CO_{ 3 } = 0.5 mol/L Na2CO3Na_{ 2 }CO_{ 3 }

Hence, 0.5 mol of Na2CO3Na_{ 2 }CO_{ 3 } is in 1 L of water or 53 g of Na2CO3Na_{ 2 }CO_{ 3 } is in 1 L of water.

 

Q26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Ans.

Reaction:

2H2  (g)   +  O2  (g)     2H2O  (g)2H_{ 2 }\;(g)  \; + \; O_{ 2 }\; (g)  \; \rightarrow \; 2H_{ 2 }O\; (g)

 

2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of water vapour.

Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of water vapour.

 

Q27. Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

 

Ans.

(i) 28.7 pm

1 pm = 1012  m10^{ -12 } \; m

28.7 pm = 28.7 × 1012  m10^{ -12 } \; m

= 2.87 × 1011  m10^{ -11 } \; m

 

(ii) 15.15 pm

1 pm = 1012  m10^{ -12 } \; m

15.15 pm = 15.15 × 1012  m10^{ -12 } \; m

= 1.515 × 1011  m10^{ -11 } \; m

 

(iii) 25365 mg

1 mg = 103  g10^{ -3 } \; g

1 mg = 10-6 kg

25365 mg = 25365 x 10-6 kg

25365 mg = 2.5365 × 102  kg10^{ -2 } \; kg

 

Q28. Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2Cl_{ 2 } (g)

 

Ans.

(i) 1 g of Au (s)

= 1197\frac{ 1 }{ 197 } mol of Au (s)

= 6.022  ×  1023197\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 } atoms of Au (s)

= 3.06 ×  1021\times \; 10^{ 21 } atoms of Au (s)

 

(ii) 1 g of Na (s)

= 123\frac{ 1 }{ 23 } mol of Na (s)

= 6.022  ×  102323\frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 } atoms of Na (s)

= 0.262 ×  1023\times \; 10^{ 23 } atoms of Na (s)

= 26.2 ×  1021\times \; 10^{ 21 } atoms of Na (s)

 

(iii) 1 g of Li (s)

= 17\frac{ 1 }{ 7 } mol of Li (s)

= 6.022  ×  10237\frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 } atoms of Li (s)

= 0.86 ×  1023\times \; 10^{ 23 } atoms of Li (s)

= 86.0 ×  1021\times \; 10^{ 21 } atoms of Li (s)

 

(iv)1 g of Cl2Cl_{ 2 } (g)

= 171\frac{ 1 }{ 71 } mol of Cl2Cl_{ 2 } (g)

(Molar mass of Cl2Cl_{ 2 } molecule = 35.5 × 2 = 71 g mol1mol^{ -1 })

= 6.022  ×  102371\frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 } atoms of Cl2Cl_{ 2 } (g)

= 0.0848 ×  1023\times \; 10^{ 23 } atoms of Cl2Cl_{ 2 } (g)

= 8.48 ×  1021\times \; 10^{ 21 } atoms of Cl2Cl_{ 2 } (g)

 

Therefore, 1 g of Li (s) will have the largest no. of atoms.

 

Q29. 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).

Ans.

Mole fraction of C2H5OHC_{ 2 }H_{ 5 }OH

= Number  of  moles  of  C2H5OHNumber  of  moles  of  solution\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}

0.040 = nC2H5OHnC2H5OH  +  nH2O\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}} ——(1)

 

No. of moles present in 1 L water:

nH2O  =  1000  g18  g  mol1n_{ H_{ 2 }O} \; = \; \frac{ 1000 \; g}{18 \; g \; mol^{ -1 }} nH2On_{ H_{ 2 }O} = 55.55 mol

 

Substituting the value of nH2On_{ H_{ 2 }O} in eq (1),

nC2H5OHnC2H5OH  +  55.55\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55} = 0.040 nC2H5OHn_{C_{ 2 }H_{ 5 }OH} = 0.040nC2H5OHn_{C_{ 2 }H_{ 5 }OH} + (0.040)(55.55)

0.96nC2H5OHn_{C_{ 2 }H_{ 5 }OH} = 2.222 mol

nC2H5OHn_{C_{ 2 }H_{ 5 }OH} = 2.2220.96  mol\frac{ 2.222 }{ 0.96 } \; mol nC2H5OHn_{C_{ 2 }H_{ 5 }OH} = 2.314 mol

 

Therefore, molarity of solution

= 2.314  mol1  L\frac{ 2.314 \; mol }{ 1 \; L }

= 2.314 M

 

Q30. What will be the mass of one 12C atom in g?

Ans.

1 mole of carbon atoms

= 6.023  ×  10236.023 \; \times \; 10^{ 23 } atoms of carbon

= 12 g of carbon

Therefore, mass of 1 atom of 12 C _{}^{ 12 }\textrm{ C }

= 12  g6.022  ×  1023\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}

= 1.993  ×  1023g1.993 \; \times \; 10^{ -23 } g

 

Q31. How many significant figures should be present in the answer of the following calculations?

(i) 0.02856  ×  298.15  ×  0.1120.5785\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

 

Ans.

(i) 0.02856  ×  298.15  ×  0.1120.5785\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }

Least precise number = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 0.112

= 3

 

(ii) 5 × 5.364

Least precise number = 5.364

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.364

= 4

 

(iii) 0.0125 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4. Hence, the no. of significant numbers in the answer is also 4.

 

Q32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope Molar mass Abundance
36Ar^{36}Ar 35.96755 g  mol1g \; mol^{ -1 } 0.337 %
38Ar^{38}Ar 37.96272 g  mol1g \; mol^{ -1 } 0.063 %
40Ar^{40}Ar 39.9624 g  mol1g \; mol^{ -1 } 99.600 %

 

Ans.

Molar mass of Argon:

= [(35.96755  ×  0.337100)( 35.96755 \; \times \; \frac{ 0.337 }{ 100 }) + (37.96272  ×  0.063100)( 37.96272 \; \times \; \frac{ 0.063 }{ 100 }) + (39.9624  ×  99.600100)( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })]

= [0.121 + 0.024 + 39.802] g  mol1g \; mol^{ -1 }

= 39.947 g  mol1g \; mol^{ -1 }

 

Q33. Calculate the number of atoms in each of the following

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He

 

Ans.

(i) 52 moles of Ar

1 mole of Ar = 6.023  ×  10236.023 \; \times \; 10^{ 23 } atoms of Ar

Therefore, 52 moles of Ar = 52 × 6.023  ×  10236.023 \; \times \; 10^{ 23 } atoms of Ar

= 3.131  ×  10253.131 \; \times \; 10^{ 25 } atoms of Ar

 

(ii) 52 u of He

1 atom of He = 4 u of He

OR

4 u of He = 1 atom of He

1 u of He = 14\frac{ 1 }{ 4 } atom of He

52 u of He = 524\frac{ 52 }{ 4 } atom of He

= 13 atoms of He

 

(iii) 52 g of He

4 g of He = 6.023  ×  10236.023 \; \times \; 10^{ 23 } atoms of He

52 g of He = 6.023  ×  1023  ×  524\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 } atoms of He

= 7.829  ×  10247.829 \; \times \; 10^{ 24 } atoms of He

 

Q34. 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.  Find:

(i) Empirical formula

(ii) Molar mass of the gas, and

(iii) Molecular formula

 

Ans.

(i) Empirical formula

1 mole of CO2CO_{ 2 } contains 12 g of carbon

Therefore, 3.38 g of CO2CO_{ 2 } will contain carbon

= 12  g44  g  ×3.38  g\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g

= 0.9218 g

 

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

= 2  g18  g  ×0.690\frac{ 2 \; g }{ 18 \; g } \; \times 0.690

= 0.0767 g

 

As hydrogen and carbon are the only elements of the compound. Now, the total mass is:

= 0.9217 g + 0.0767 g

= 0.9984 g

 

Therefore, % of C in the compound

= 0.9217  g0.9984  g  ×100\frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 100

= 92.32 %

% of H in the compound

= 0.0767  g0.9984  g  ×100\frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 100

= 7.68 %

 

Moles of C in the compound,

= 92.3212.00\frac{ 92.32 }{ 12.00 }

= 7.69

 

Moles of H in the compound,

= 7.681\frac{ 7.68 }{ 1 }

= 7.68

 

Therefore, the ratio of carbon to hydrogen is,

7.69: 7.68

1: 1

 

Therefore, the empirical formula is CH.

 

(ii) Molar mass of the gas

Weight of 10 L of gas at STP = 11.6 g

Therefore, weight of 22.4 L of gas at STP

= 11.6  g10  L  ×  22.4  L\frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L

= 25.984 g

\approx 26 g

 

(iii) Molecular formula

Empirical formula mass:

CH = 12 + 1

= 13 g

n = Molar  mass  of  gasEmpirical  formula  mass  of  gas\frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}

= 26  g13  g\frac{ 26 \; g }{ 13 \; g}

= 2

Therefore, molecular formula = 2 x CH = C2H2C_{ 2 }H_{ 2 }.

 

Q35. 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?

Ans.

0.75 M of HCl

≡ 0.75 mol of HCl are present in 1 L of water

≡ [(0.75 mol) × (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 contains 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

= 27.375  g1000  mL  ×  25  mL\frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL

= 0.6844 g

 

Given chemical reaction,

CaCO3  (s)  +  2  HCl  (aq)     CaCl2  (aq)   +  CO2  (g)   +  H2O  (l)CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq)  \; \rightarrow \; CaCl_{ 2 }\;(aq)  \; + \; CO_{ 2 }\; (g)  \; + \; H_{ 2 }O\; (l)

 

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO3CaCO_{ 3 } (100 g)

Therefore, amt of CaCO3CaCO_{ 3 } that will react with 0.6844 g

= 10073  ×  0.6844  g\frac{ 100 }{ 73 } \; \times \; 0.6844 \; g

= 0.9375 g

 

Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:

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

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

Ans.

1 mole of MnO2MnO_{2} = 55 + 2 × 16 = 87 g

4 mole of HCl = 4 × 36.5 = 146 g

1 mole of MnO2MnO_{2} reacts with 4 mol of HCl

Hence,

5 g of MnO2MnO_{ 2 }will react with:

= 146  g87  g  ×  5  g\frac{146 \; g}{87 \; g} \; \times \; 5 \; g HCl

= 8.4 g HCl

Therefore, 8.4 g of HCl will react with 5 g of MnO2MnO_{2}.

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