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 on 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 CoolGyan’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 solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1
ncert solution for class 11 chemistry chapter 1

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×100frac{Mass;of;that;element;in;the;compound}{Molar;mass;of;the;compound} imes 100

 

Therefore, mass percent of the sodium element:

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

= 32.379

= 32.4%

Mass percent of the sulphur element:

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

= 22.57

= 22.6%

Mass percent of the oxygen element:

= 64.0g142.066g×100frac{64.0g}{142.066g} imes 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  ironfrac{percent;of;iron;by;mass}{Atomic;mass;of;iron}

= 69.955.85frac{69.9}{55.85}

= 1.25

 

Relative moles of O in iron oxide:

= percent  of  oxygen  by  massAtomic  mass  of  oxygenfrac{percent;of;oxygen;by;mass}{Atomic;mass;of;oxygen}

= 30.116.00frac{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} ightarrow 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 44×1632frac{44 imes 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

= 0.3751000×500frac{0.375}{1000} imes 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}:

= 69g63gmol1frac{69:g}{63:g:mol^{-1}}

= 1.095 mol

 

Volume of 100g HNO3 solution

= Mass  of  solutiondensity  of  solutionfrac{Mass;of;solution}{density;of;solution}

= 100g1.41g  mL1frac{100g}{1.41g;mL^{-1}}

= 70.92mL

= 70.92×103  L70.92 imes 10^{-3};L

 

Concentration of HNO3

= 1.095mole70.92×103Lfrac{1.095:mole}{70.92 imes 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 63.5×100g159.5frac{63.5 imes 100g}{159.5} of Cu.

= 63.5×100159.5frac{63.5 imes 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

= 69.9055.85frac{69.90}{55.85}

= 1.25

 

No. of moles of O present in oxide

= 30.116.0frac{30.1}{16.0}

=1.88

 

Ratio of Fe to O in oxide,

= 1.25: 1.88

= 1.251.25:1.881.25frac{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. 7g

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

Therefore n = Molar  massEmpirical  formula  mass=159.69  g159.7  gfrac{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 AbundanceMolar Mass
35Cl_{}^{35} extrm{Cl}75.7734.9689
37Cl_{}^{37} extrm{Cl}24.2336.9659

 

Ans.

Average atomic mass of Cl.

= [(Fractional abundance of 35Cl_{}^{35} extrm{Cl})(molar mass of 35Cl_{}^{35} extrm{Cl})+(fractional abundance of 37Cl_{}^{37} extrm{Cl} )(Molar mass of 37Cl_{}^{37} extrm{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 1 mole of ethane- 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  Litresfrac{Number;of;moles;of;solute}{Volume;of;solution;in;Litres}

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

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

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

= 0.0585  mol2  Lfrac{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  mol1frac{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 = FAfrac{F}{A}

= 1034  g  ×  9.8  ms2cm2×1  kg1000  g×(100)2  cm21m2frac{1034;g; imes ;9.8;ms^{-2}}{cm^{2}} imes frac{1;kg}{1000;g} imes 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.

The SI Unit of mass: 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:

 

 PrefixesMultiples
(a)femto10
(b)giga101510^{-15}
(c)mega10610^{-6}
(d)deca10910^{9}
(e)micro10610^{6}

 

Ans.

PrefixesMultiples
(a)femto101510^{-15}
(b)giga10910^{9}
(c)mega10610^{6}
(d)deca10
(e)micro10610^{-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.

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

Mass percent of 15 ppm chloroform in H2O

= 15106×100frac{15}{10^{6}} imes 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  CHCl3frac{1.5 ; imes 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 ×10010^{0}

 

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) 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: 2808

 

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

 Mass of dioxygenMass of dinitrogen
(i)16 g14 g
(ii)32 g14 g
(iii)32 g28 g
(iv)80 g28 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) If we fix the mass of N2 at 28 g, the masses of O2 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: 1: 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.”

(b) Convert:

(a) 1 km = ____ mm = ____ pm

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

 

  • 1 km = 1 km * 1000  m1  kmfrac{ 1000 ; m }{ 1 ; km } * 1  pm1012  mfrac{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  mgfrac{ 1 ; g }{ 1000 ; mg } * 1  kg1000  gfrac{ 1 ; kg }{ 1000 ; g }

1 mg = 10610^{ -6 } kg

 

  • 1 mg = 1 mg * 1  g1000  mgfrac{ 1 ; g }{ 1000 ; mg } * 1  ng109  gfrac{ 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  mLfrac{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  cmcm3frac{1 ; dm ; imes ; 1 ; dm ; imes ;1 ; dm }{10 ; cm ; imes ; 10 ; cm ; imes ; 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.

(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 limiting the amt. of products 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 agent.

 

(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 agent.

 

(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 agent.

 

(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 moles of B, so 2.5 moles of A is unused. Hence, B is the limiting agent.

 

(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 agent.

 

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 ; imes ;10^{ 3 } g N2 reacts with 1  ×  1031 ; imes ;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) Balance the given equation:

N2  (g)   +  3H2  (g)    2NH3  (g)N_{ 2 };(g)  ; + ; 3H_{ 2 } ;(g) ; ightarrow ; 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 ; imes ;10^{ 3 } g of N2 will react with 628   ×  2  ×  103frac{ 6}{ 28  } ; imes ; 2 ; imes ; 10^{ 3 } g NH3

 

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

 

Given:

Amt of H2 = 1  ×  1031 ; imes ;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  ×  2000frac{ 34 ; g }{ 28 ; g } ; imes ; 2000 g

= 2430 g of NH3NH_{ 3 }

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

(iii) Mass of H2 unreacted

= 1  ×  1031 ; imes ;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  molfrac{ 106 ; g }{ 1 ; mol } ; imes ; 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)  ; ightarrow ; 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) 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

25365 mg = 2.5365 × 10110^{ -1 } × 103  kg10^{ -3 } ; kg

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