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

NCERT Solutions for Class 11 Chemistry: Chapter 1 (Some Basic Concepts of Chemistry) are provided in this page for the perusal of 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 chemistry 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 examinations.

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

NCERT Chemistry Class 11 Chapter 1 Subtopics (“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 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.

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 weight of methane, CH4CH_{4}

= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)

= [1(12.011 u) +4 (1.008u)]

= 12.011u + 4.032 u

= 16.043 u

 

(ii) H2OH_{2}O :

Molecular weight of water, H2OH_{2}O

= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)

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

= 2.016 u +16.00 u

= 18.016u

So approximately

= 18.02 u

 

(iii) CO2CO_{2} :

Molecular weight of carbon dioxide, CO2CO_{2}

= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)

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

= 12.011 u +32.00 u

= 44.011 u

So approximately

= 44.01u

 

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.

Percent of Fe by mass = 69.9 % [As given above]

Percent of O2 by mass = 30.1 % [As given above]

Relative moles of Fe 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 O 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

 

Simplest molar ratio of Fe to O:

= 1.25: 1.88

= 1: 1.5

\approx 2: 3

Therefore, empirical formula of 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.

Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O2 will react with 6g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18g of carbon (1.5 mol) will not undergo combustion.

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 Maqueous 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, mass that is required 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 per cent 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)] gr

= 159.69 g

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 and the empirical formula.

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

 

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.

Average atomic mass of Cl.

= [(Fractional abundance of 35Cl_{}^{35}\textrm{Cl})(molar mass of 35Cl_{}^{35}\textrm{Cl})+(fractional abundance of 37Cl_{}^{37}\textrm{Cl} )(Molar mass of 37Cl_{}^{37}\textrm{Cl} )]

 

= [{(75.77100(34.9689u)\frac{75.77}{100}(34.9689u) } + {(24.23100(34.9659  u)\frac{24.23}{100}(34.9659\;u) }]

 

= 26.4959 + 8.9568

 

= 35.4527 u

Therefore, the average atomic mass of Cl = 35.4527 u

 

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.

(a) 1 mole 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 * 3

= 6

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

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

= 3 * 6

= 18

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

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

= 3 * 6.023 * 102310^{23}

= 18.069 * 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 CH3OHCH_{3}OH

= (1 * 12) + (4 * 1) + (1 * 16)

= 32 g mol1mol^{-1}

= 0.032 kg mol1mol^{-1}

 

Molarity of the solution

= 0.793  kg  L10.032  kg  mol1\frac{0.793\;kg\;L^{-1}}{0.032\;kg\;mol^{-1} }

= 24.78 molL1L^{-1}

 

(From the definition of density)

M1V1=M2V2M_{1}V_{1} = M_{2}V_{2} (24.78 molL1L^{-1}) V1V_{1} =  (2.5 L) (0.25 molL1L^{-1})

V1V_{1} = 0.0252 Litre

V1V_{1} = 25.22 Millilitre

 

 

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.

As per definition, pressure is force per unit area of the surface.

P = FA\frac{F}{A}

= 1034  g  ×  9.8  ms2cm2×1  kg1000  g×(100)2  cm21m2\frac{1034\;g\;\times \;9.8\;ms^{-2}}{cm^{2}}\times \frac{1\;kg}{1000\;g}\times \frac{(100)^{2}\;cm^{2}}{1 m^{2}}

= 1.01332 × 10510^{5} kg m1s2m^{-1} s^{-2}

 

Now,

1 N = 1 kg ms2s^{-2}

Then,

1 Pa = 1 Nm2Nm^{-2}

= 1 kgm2kgm^{-2}s2s^{-2}

Pa   = 1 kgm1kgm^{-1}s2s^{-2} Pressure  (P) = 1.01332 × 10510^{5} Pa

 

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

Ans.

Si Unit: Kilogram (kg)

Mass:

“The mass equal to the mass of the international prototype of kilogram is known as mass.”

 

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 the 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.

(a) 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} %

 

(b) 100 grams of the sample is having 1.5 ×10310^{-3}g of CHCl3CHCl_{3}.

1000 grams of the sample is having 1.5 ×10210^{-2}g of CHCl3CHCl_{3}.

Molality of CHCl3CHCl_{3} in water

= 1.5  ×102  gMolar  mass  of  CHCl3\frac{1.5 \;\times 10^{-2} \;g}{Molar \; mass \; of \; CHCl_{3}}

Molar mass (CHCl3CHCl_{3})

= 12 + 1 + 3 (35.5)

= 119.5 grams mol1mol^{-1}

 

Therefore, molality of CHCl3CHCl_{3} I water

= 1.25 × 10410^{-4} m

 

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

 

Ans.

(a) 0.0048= 4.8 ×10310^{-3}

(b) 234,000 = 2.34 ×10510^{5}

(c) 8008= 8.008 ×10310^{3}

(d) 500.0 = 5.000 ×10210^{2}

(e) 6.0012 = 6.0012

 

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

(a) 0.0027

(b) 209

(c)6005

(d)136,000

(e) 900.0

(f)2.0035

Ans.

(i) 0.0027: 2 significant numbers.

(ii) 209: 3 significant numbers.

(iii)6005: 4 significant numbers.

(iv)136,000:3 significant numbers.

(v) 900.0: 4 significant numbers.

(vi)2.0035: 5 significant numbers.

 

Q20. Round up the following upto three significant figures:

(a) 35.217

(b) 11.4108

(c)0.05577

(d)2806

 

Ans.

(a) The number after round up is: 35.2

(b) The number after round up is: 11.4

(c)The number after round up is: 0.0560

(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.

(1) If we fix the mass of N2 at 28 g, the masses of N2 that will combine with the fixed mass of N2 are 32 grams, 64 grams, 32 grams and 80 grams.

The mass of O2 bear whole no. ratio of 1: 2: 2: 5. Therefore, the given information obeys the law of multiple proportions.

The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”

(2) Convert:

(a) 1 km = ____ mm = ____ pm

  • 1 km = 1 km * 1000  m1  km\frac{ 1000 \; m }{ 1 \; km } × 100  cm1  m\frac{ 100 \; cm }{ 1 \; m } * 10  mm1  cm\frac{ 10 \; mm }{ 1 \; cm }
1 km = 10610^{ 6 } mm

 

  • 1 km = 1 km * 1000  m1  km\frac{ 1000 \; m }{ 1 \; km } * 1  pm1012  m\frac{1 \; pm}{10^{ -12 } \; m}
1 km = 101510^{ 15 } pm

Therefore, 1 km = 10610^{ 6 } mm = 101510^{ 15 } pm

 

(b) 1 mg = ____ kg = ____ ng

  • 1 mg = 1 mg * 1  g1000  mg\frac{ 1 \; g }{ 1000 \; mg } * 1  kg1000  g\frac{ 1 \; kg }{ 1000 \; g }

1 mg = 10610^{ -6 } kg

 

  • 1 mg = 1 mg * 1  g1000  mg\frac{ 1 \; g }{ 1000 \; mg } * 1  ng109  g\frac{ 1 \; ng }{ 10^{ -9 } \; g }

1 mg = 10610^{ 6 } ng

Therefore, 1 mg = 10610^{ -6 } kg = 10610^{ 6 } ng

 

(c) 1 mL = ____ L = ____ dm3dm^{ 3 }

  • 1 mL = 1 mL * 1  L1000  mL\frac{1 \; L}{1000 \; mL}

1 mL = 10310^{ -3 } L

 

  • 1 mL = 1 cm3cm^{ 3 } = 1 * 1  dm  ×  1  dm  ×  1  dm10  cm  ×  10  cm  ×  10  cmcm3\frac{1 \; dm \; \times \; 1 \; dm \; \times \;1 \; dm }{10 \; cm \; \times \; 10 \; cm \; \times \; 10 \; cm } cm^{ 3 }

1 mL = 103dm310^{ -3 } dm^{ 3 }

Therefore, 1 mL = 10310^{ -3 }L = 10310^{ -3 } dm3dm^{ 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 }

So,

Distance travelled in 2 ns = speed of light * 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.

(a) 2 mol X + 3 mol Y

(b) 100 atoms of X + 100 molecules of Y

(c) 300 atoms of X + 200 molecules of Y

(d) 2.5 mol X + 5 mol Y

(e) 5 mol X + 2.5 mol Y

 

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 limiting the amt. of products formed.

 

(a) 2 mol X + 3 mol Y

1 mole of X reacts with 1 mole of Y. Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. Hence, X is limiting agent.

 

(b) 100 atoms of X + 100 molecules of Y

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

 

(c) 300 atoms of X + 200 molecules of Y

1 atom of X reacts with 1 molecule of Y. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. Hence, Y is limiting agent.

 

(d) 2.5 mol X + 5 mol Y

1 mole of X reacts with 1 mole of Y. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. Hence, X is limiting agent.

 

(e) 5 mol X + 2.5 mol Y

1 mole of X reacts with 1 mole of Y. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. Hence, Y is limiting agent.

 

Q24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + H2(g)→ 2NH3 (g)

(a) What is the mass of NH3NH_{ 3 } produced if 2  ×  1032 \; \times \;10^{ 3 } g N2 reacts with 1  ×  1031 \; \times \;10^{ 3 } g of H2?

 (b) Will the reactants N2 or H2 remain unreacted?

(c) If any, then which one and give it’s mass.

 

Ans.

(a) Balance the given equation:

N2  (g)   +  3H2  (g)    2NH3  (g)N_{ 2 }\;(g)  \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g)

 

Thus, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of NH3NH_{ 3 }.

 

2  ×  1032 \; \times \;10^{ 3 } g of N2 will react with 628   ×  2  ×  103\frac{ 6}{ 28  } \; \times \; 2 \; \times \; 10^{ 3 } g NH3

 

2  ×  1032 \; \times \;10^{ 3 } g  of N2 will react with 428.6 g of H2.

 

Given:

Amt of H2 = 1  ×  1031 \; \times \;10^{ 3 }

28 g of N2N_{ 2 } produces 34 g of NH3NH_{ 3 }

Therefore, mass of NH3NH_{ 3 } produced by 2000 g of N2N_{ 2 }

= 34  g28  g  ×  2000\frac{ 34 \; g }{ 28 \; g } \; \times \; 2000 g

= 2430 g of NH3NH_{ 3 }

(b) H2H_{ 2 } is the excess reagent. Therefore, H2H_{ 2 } will not react.

(c) Mass of H2 unreacted

= 1  ×  1031 \; \times \;10^{ 3 } – 428.6 g

= 571.4 g

 

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 vapour.

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

 

Q27. Convert the following into basic units:

(i) 29.7 pm

(ii) 16.15 pm

(iii) 25366 mg

 

Ans.

(i) 29.7 pm

1 pm = 1012  m10^{ -12 } \; m

29.7 pm = 29.7 × 1012  m10^{ -12 } \; m

= 2.97 × 1011  m10^{ -11 } \; m

 

(ii) 16.15 pm

1 pm = 1012  m10^{ -12 } \; m

16.15 pm = 16.15 × 1012  m10^{ -12 } \; m

= 1.615 × 1011  m10^{ -11 } \; m

 

(iii) 25366 mg

1 mg = 103  g10^{ -3 } \; g

25366 mg = 2.5366 × 10110^{ -1 } × 103  kg10^{ -3 } \; kg

25366 mg = 2.5366 × 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 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 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 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 12 C _{}^{ 12 }\textrm{ C } atom

= 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 × 5.365

(iii) 0.012 + 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 no. of calculation = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in the least precise no.

= 3

 

(ii) 5 × 5.365

Least precise no. of calculation = 5.365

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.365

= 4

(iii) 0.012 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4, 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 occuring argon isotopes:

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

= 7.8286  ×  10247.8286 \; \times \; 10^{ 24 } atoms of He

 

Q34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of itin oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. Avolume 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

= 0.9217 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, and

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 is (CH)n(CH)_{ n } that is 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 contins 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 mol of MnO2MnO_{2} = 55 + 2 × 16 = 87 g

4 mol of HCl = 4 × 36.5 = 146 g

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

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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