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 above. 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.

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 on calculating the molecular weight of compounds.
  • Numerical problems on calculating mass per cent 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 in 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. Find out the value of molecular weight of the given compounds:

(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. Sodium Sulphate (Na2SO4Na_{2}SO_{4}) has various elements, find out the mass percentage of each element.

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. Find out the empirical formula of an oxide of iron having 69.9% Fe and 30.1% O2 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. Find out the amount of CO2 that can be produced when

 (i) 1 mole carbon is burnt in air.

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

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

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. Find out the mass of CH3COONaCH_{3}COONa(sodium acetate) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of CH3COONaCH_{3}COONa is 82.0245  g  mol182.0245\;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

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

= 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 gram

 

Q6. A sample of HNO3 has a density of 1.41  g  mL11.41\;g\;mL^{-1} find the concentration of HNO3 in moles per litre and the mass percent of HNO3 in it is 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 Cu (Copper) can be obtained from 100 gram of CuSO4CuSO_{4}( copper sulphate)?

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 gram

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

Therefore, 100 gram 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 gram

 

Q8. The mass percent of iron and oxygen in an oxide of iron is 69.9 and 30.1 calculate the molecular formula of the oxide of iron. 159.69  g  mol1159.69\;g\;mol^{-1} is the given molar mass of an oxide.

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. Find out 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 3 moles of ethane (C2H6C_{2}H_{6}), calculate the given below:

(a) No. of moles of C- atoms

(b) No. of moles of H- atoms.

(c) No. of molecules of C2H6.

 

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 C12H22O11C_{12}H_{22}O_{11} in mol L1L^{-1} if its 20 gram are dissolved in enough H2O to make a final volume up to 2 Litre?

 

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. The density of CH3OH (methanol) is 0.793 kg L1L^{-1}. For making 2.5 Litre of its 0.25 M solution what volume is needed?

 

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 defined as force per unit area of the surface. Pascal, the SI unit of pressure is as given below:

1 Pa = 1 N m2m^{-2}

Assume that mass of air at the sea level is 1034 gcm-2. Find out 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. Write SI unit for mass. Also define mass.

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 prefixes with their multiples in the table given below:

 

  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 are 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 highly contaminated with CHCl3CHCl_{3}, chloroform, which is carcinogenic. 15 ppm (by mass) was the level of contamination.

(a) Express in terms of percent by mass.

(b) Calculate the molality of chloroform in the given 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 gram of the sample is having 1.5 ×10310^{-3}g of CHCl3CHCl_{3}.

1000 gram 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 gram mol1mol^{-1}

 

Therefore, molality of CHCl3CHCl_{3} I water

= 1.25 × 10410^{-4} m

 

Q18. Express the given number in scientific notation:

(a) 0.0047

(b) 235,000

(c)8009

(d)700.0

(e) 5.0013

 

Ans.

(a) 0.0047= 4.7 ×10310^{-3}

(b) 235,000 = 2.35 ×10510^{5}

(c) 8009= 8.009 ×10310^{3}

(d) 700.0 = 7.000 ×10210^{2}

(e) 5.0013 = 5.0013

 

Q19. Find the number of significant figures in the numbers given belowt.

(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 given numbers upto 3 significant numbers.

(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. When dioxygen and dinitrogen react together, they form various compounds. The information is given below:

  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

 

(1) In the data given above, which chemical combination law is obeyed? Also give the statement of the law.

(2) Convert the following:

            (a) 1 km =_____ mm = _____ pm

            (b) 1 mg = _____ kg = _____ ng

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

 

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 gram, 64 gram, 32 gram and 80 gram.

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. What is the distance covered by the light in 2 ns if the speed of light is 3 × 10810^{ 8 } ms1ms^{ -1 }

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 the reaction given below:

X  +  Y2    XY2X \; + \; Y_{ 2 } \; \rightarrow \; XY_{ 2 }

Find the limiting reagent if it is present in the  reactions given below:

(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. H2 and N2 react with each other to produce NH3 according to the given chemical equation

            N2  (g)   +  H2  (g)    2NH3  (g)N_{ 2 }\; (g)  \; + \; H_{ 2 }\; (g) \; \rightarrow \; 2NH_{ 3 }\;(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 6g28g  ×  2  ×  103\frac{ 6 g }{ 28 g } \; \times \; 2 \; \times \; 10^{ 3 } g H2

 

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 }

Therefore, N2N_{ 2 } is limiting reagent.

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

 

(b) N2N_{ 2 } is limiting reagent and 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. 0.50 mol Na2CO3Na_{ 2 }CO_{ 3 } and 0.50 M Na2CO3Na_{ 2 }CO_{ 3 } are different. How?

 

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 react with 5 volumes of dioxygen gas, how many volumes of vapour would be obtained?

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 given quantities 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 of the given below have the largest no. 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. What is the molarity of the solution of ethanol in water in which the mole fraction of ethanol is 0.040?

(Assume the density of water to be 1)

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. Calculate the mass of 1 12 C _{}^{ 12 }\textrm{ C } 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 numbers should be present in answer of the given 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. Calculate molar mass of Argon isotopes according to the data given in the table.

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. What is the number of atoms in the following compounds?

(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 hydrogen and carbon. If we burn a small sample, we get 3.38 g of carbon dioxide and 0.69 g of water. A volume of 10 L (at STP) of this welding gas weighs 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 and gives CaCl2CaCl_{ 2 } and CO2CO_{ 2 } according to the 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)

Calculate the mass of CaCO3CaCO_{ 3 } 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 = 71 g) react with 1 mol of CaCO3CaCO_{ 3 } (100 g)

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

= 10071  ×  0.6844  g\frac{ 100 }{ 71 } \; \times \; 0.6844 \; g

= 0.9639 g

 

Q36. Chlorine is prepared by adding manganese dioxide with hydrochloric acid acc. to the reaction.

4  HCl  (aq)  +  MnO2  (s)     2  H2O  (l)   +  MnCl2  (aq)  +  Cl2  (g)4 \; HCl\;(aq) \; + \; MnO_{ 2 }\;(s)  \; \rightarrow \; 2 \; H_{ 2 }O\; (l)  \; + \; MnCl_{ 2 }\; (aq) \; + \; Cl_{ 2 }\; (g)

How many grams of HCl react with 5 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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