Question

NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry – In the NCERT Solutions for the chapter Some Basic Concepts of Chemistry we have discussed all the concepts of the chapter. These NCERT solutions are made by chemistry experts. Our solution is the best way to learn the basic concepts of chemistry, prepare for board exams, and also for cracking tough competitive exams like JEE Mains, JEE Advance, and many more.

Our experts on the subject matter suggest one study the NCERT Solution Class 11 thoroughly to have a complete understanding of the Chemistry. We have discussed all the important topics of the chapter such as

• Some Basic Concepts of Chemistry – Here we have discussed some basic laws and concepts of chemistry such as Avogadro law, Avogadro number, Dalton’s atomic theory, and some basic chemical reactions.
• Importance of Chemistry – anything and everything we do in our day-to-day life, our life processes how we breathe, eat, walk, and digest our food are all part of chemistry. So it is very important to learn and have a common understanding of chemistry.
• Nature of Matter – all matter are made up of particle. The particle of matter is very small. Properties of matter are:
  • Matter is made up of small particles.
  • Matter has interparticle space among its constituting particle.
  • The constituting particle of all matter is very small.
  • The particle of the matter is not stationary and is always in constant motion.
• Properties of Matter and their Measurement – all matter exists in four forms. Four forms of matter are solid, liquid, gas, and plasma.
• Uncertainty in Measurement – the uncertainty of measurement is defined as the range of possible values in which the true values of the measurement lie.
• Laws of Chemical Combination – there are 5 laws of chemical combination.
  •  Avogadro’s law
  • Gay Lussac’s Law of Gaseous volume
  • Law of Multiple Proportions
  • Law of Definite Proportions.
  • Law of Conservation of Mass.
• Dalton’s Atomic Theory – Dalton’s atomic theory states that all matter is made up of atoms, which are inseparable from each other.
• Atomic and Molecular Masses – atomic mass is the mass of a single atom of any particular element.
• Mole Concept and Molar Masses – Molar masses are the mass of 1 mole of an element or the compound.

NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry is made by mentors at Sarthaks in a very concise manner to make it easy for students in their last-minute preparation.

Answer 1

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

1. Calculate the molecular mass of the following: 

(i) H₂O

(ii) CO₂

(iii) CH₄

Answer:

(i) H₂O:

The molecular mass of water, H₂O

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

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

= 2.016 u + 16.00 u

= 18.016

= 18.02 u

(ii) CO₂: 

The molecular mass of carbon dioxide, CO₂ 

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen) 

= [1(12.011 u) + 2 (16.00 u)] 

= 12.011 u + 32.00 u 

= 44.01 u 

(iii) CH₄: 

The molecular mass of methane, CH₄ 

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen) 

= [1(12.011 u) + 4 (1.008 u)] 

= 12.011 u + 4.032 u 

= 16.043 u

2. Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Answer:

The molecular formula of sodium sulphate is Na₂SO₄ .

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

= 142.066 g

Mass percent of an element = \(\frac{Mas\,of\, that\,element\,in\,the\,compound}{Molar\,mass\, of \,the\, compund}\times100\)

\(\therefore\) Mass percent of sodium:

= \(\frac{46.0 g}{142.066g}\times100\)

= 32.379

= 32.4 %

Mass percent of sulphur:

= \(\frac{32.066 g}{142.066g}\times100\)

= 22.57

= 22.6 %

Mass percent of oxygen:

= \(\frac{64.0 g}{142.066g}\times100\)

= 45.049

= 45.05 %

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

Answer:

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given]

 Relative moles of iron in iron oxide:

= \(\frac{\% \,of\,iron\,by\,mass}{Atomic\,mass\,of\,iron}\)

= \(\frac{69.9}{55.85}\)

= 1.25

Relative moles of oxygen in iron oxide:

= \(\frac{\% \,of\,iron\,by\,mass}{Atomic\,mass\,of\,oxygen}\)

= \(\frac{30.1}{16.00}\)

= 1.88

Simplest molar ratio of iron to oxygen: 

= 1.25: 1.88 

= 1: 1.5 

\(\simeq\)2: 3 

The empirical formula of the iron oxide is Fe₂O₃.

4. 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.

Answer:

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

(i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide. 

(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant. 

(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. 

Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^–1

Answer:

0.375 M aqueous solution of sodium acetate

≡ 1000 mL of solution containing 0.375 moles of sodium acetate 

\(\therefore\) Number of moles of sodium acetate in 500 mL

= \(\frac{0.375}{1000}\times500\)

= 0.187 mole

Molar mass of sodium acetate = 82.0245 g mole^–1 (Given) 

\(\therefore\) Required mass of sodium acetate = (82.0245 g mol^–1) (0.1875 mole) 

= 15.38 g

6. 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%.

Answer:

Mass percent of nitric acid in the sample = 69 % [Given] 

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass. 

Molar mass of nitric acid (HNO₃)

= {1 + 14 + 3(16)} g mol^–1 

= 1 + 14 + 48 

= 63 g mol^–1 

\(\therefore\) Number of moles in 69 g of HNO₃

= \(\frac{69 g}{63g\,mol^{-1}}\)

= 1.095 mol

Volume of 100g of nitric acid solution

\(= \frac{Mass\,of\,solution}{density\,of\,solution}\)

= \(\frac{100g}{1.41g\,mL^{-1}}\)

= 70.92 mL ≡ 70.92 x 10^-3 L

Concentration of nitric acid

= \(\frac{1.095\,mole}{70.92\times10^{-3}L}\)

= 15.44 mol/L

\(\therefore\) Concentration of nitric acid = 15.44 mol/L

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

Answer:

1 mole of CuSO₄ contains 1 mole of copper. 

Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00) 

= 63.5 + 32.00 + 64.00 

= 159.5 g 

159.5 g of CuSO₄ contains 63.5 g of copper.

⇒ 100 g of CuSO₄ will contain \(\frac{63.5\times100g}{159.5}\) of copper.

\(\therefore\) Amount of copper that can be obtained from 100 g CuSO₄ = \(\frac{63.5\times100}{159.5}\)

= 39.81 g

Answer 2

8. 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. Given that the molar mass of the oxide is 159.69 g mol^–1.

Answer:

Mass percent of iron (Fe) = 69.9% (Given) 

Mass percent of oxygen (O) = 30.1% (Given) 

Number of moles of iron present in the oxide = \(\frac{69.90}{55.85}\)

= 1.25

Number of moles of oxygen present in the oxide = \(\frac{30.1}{16.0}\)

= 1.88

Ratio of iron to oxygen in the oxide,

= 1.25 : 1.88

= \(\frac{1.25}{1.25}:\frac{1.88}{1.25}\)

= 1 : 1.5 

= 2 : 3

\(\therefore\) The empirical formula of the oxide is Fe₂O₃. 

Empirical formula mass of Fe₂O₃ = [2(55.85) + 3(16.00)] g 

Molar mass of Fe₂O₃ = 159.69 g

\(\therefore\) n = \(\frac{Molar \,mass}{Emprical\,formula\,mas}\) = \(\frac{159.69g}{159.7g}\)

= 0.999

= 1 (approx)

Molecular formula of a compound is obtained by multiplying the empirical formula with n. 

Thus, the empirical formula of the given oxide is Fe₂O₃ and n is 1. 

Hence, the molecular formula of the oxide is Fe₂O₃.

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

	% Natural Abundance	Molar Mass
\(^{35}\, CL\)	75.77	34.9689
\(^{37} \, CL\)	24.23	36.9659

Answer:

The average atomic mass of chlorine

= 26.4959 + 8.9568 

= 35.4527 u 

\(\therefore\)The average atomic mass of chlorine = 35.4527 u

10. In three moles of ethane (C₂H₆), calculate the following: 

(i) Number of moles of carbon atoms. 

(ii) Number of moles of hydrogen atoms. 

(iii) Number of molecules of ethane.

Answer:

(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.

\(\therefore\) Number of moles of carbon atoms in 3 moles of C₂H₆ 

= 2 × 3 = 6 

(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.

\(\therefore\) Number of moles of carbon atoms in 3 moles of C₂H₆ =

 3 × 6 = 18 

(iii) 1 mole of C₂H₆ contains 6.023 × 10²³ molecules of ethane.

\(\therefore\) Number of molecules in 3 moles of C₂H₆ 

= 3 × 6.023 × 10²³ 

= 18.069 × 10²³

11. 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 2 L?

Answer:

Molarity (M) of a solution is given by,

= 0.02925 mol L^–1

\(\therefore\) Molar concentration of sugar = 0.02925 mol L^–1

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

Answer:

Molar mass of methanol (CH₃OH) = (1 × 12) + (4 × 1) + (1 × 16) 

= 32 g mol^–1

= 0.032 kg mol^–1 

Molarity of methanol solution = \(\frac{0.793\,kg\,L^{-1}}{0.032\,kg\,mol^{-1}}\)

= 24.78 mol L^–1 

(Since density is mass per unit volume) 

Applying, 

M₁V₁ = M₂V₂ 

(Given solution) (Solution to be prepared)

(24.78 mol L^–1) V₁ = (2.5 L) (0.25 mol L^–1) 

V₁ = 0.0252 L 

V₁ = 25.22 mL

13. 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.

Answer:

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

P = \(\frac{F}{A}\)

= 1.01332 × 10⁵ kg m^–1s^–2

We know, 

1 N = 1 kg ms^–2 

Then, 

1 Pa = 1 Nm^–2 = 1 kg m^–2s^–2 1 

Pa = 1 kg m^–1s^–2 

\(\therefore\) Pressure = 1.01332 × 10⁵ Pa

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

Answer:

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

Answer 3

15. Match the following prefixes with their multiples:

	Prefixes	Multiples
(i)	micro	106
(ii)	deca	109
(iii)	mega	10–6
(iv)	giga	10–15
(v)	femto	10

Answer:

	Prefixes	Multiples
(i)	micro	10–6
(ii)	deca	10
(iii)	mega	106
(iv)	giga	109
(v)	femto	10–15

16. What do you mean by significant figures?

Answer:

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

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

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

(i) Express this in percent by mass. 

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

Answer:

(i) 1 ppm is equivalent to 1 part out of 1 million (10⁶) parts. 

Mass percent of 15 ppm chloroform in water = \(\frac{15}{10^6}\times100\)

= 1.5 x 10^-1 %

(ii) 100 g of the sample contains 1.5 × 10^–3 g of CHCl₃. 

⇒ 1000 g of the sample contains 1.5 × 10^–2 g of CHCI₃. 

\(\therefore\) Molality of chloroform in water

= \(\frac{1.5\times10^{-2}g}{Molar\,mass \,of\,CHCl_3}\)

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

= 119.5 g mol^–1 

\(\therefore\) Molality of chloroform in water = 0.0125 × 10^–2 m 

= 1.25 × 10^–4 m

18. Express the following in the scientific notation: 

(i) 0.0048 

(ii) 234,000 

(iii) 8008

(iv) 500.0 

(v) 6.0012

Answer:

(i) 0.0048 = 4.8× 10^–3 

(ii) 234, 000 = 2.34 ×10⁵ 

(iii) 8008 = 8.008 ×10³ 

(iv) 500.0 = 5.000 × 10² 

(v) 6.0012 = 6.0012

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

(i) 0.0025 

(ii) 208 

(iii) 5005 

(iv) 126,000 

(v) 500.0

(vi) 2.0034

Answer:

(i) 0.0025 

There are 2 significant figures. 

(ii) 208 

There are 3 significant figures. 

(iii) 5005 

There are 4 significant figures. 

(iv) 126,000 

There are 3 significant figures. 

(v) 500.0 

There are 4 significant figures. 

(vi) 2.0034 

There are 5 significant figures.

20. Round up the following upto three significant figures: 

(i) 34.216 

(ii) 10.4107 

(iii) 0.04597 

(iv) 2808

Answer:

(i) 34.2 

(ii) 10.4 

(iii) 0.0460 

(iv) 2810

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

Mass of dinitrogen   Mass of dioxygen 

(i) 14 g                              16 g 

(ii) 14 g                             32 g 

(iii) 28 g                            32 g 

(iv) 28 g                            80 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

Answer:

(a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g. 

The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.

Answer 4

(b)

(i) 1 km = 1 km × \(\frac{1000\,m}{1 \,km}\times\frac{100\,cm}{1\,m}\times\frac{10\,mm}{1\,cm}\)

\(\therefore\) 1 km = 10⁶ mm

1 km = 1 km × \(\frac{1000\,m}{1 \,km}\times\frac{1\,pm}{10^{-12}m}\)

\(\therefore\) 1 km = 10¹⁵ pm 

Hence, 1 km = 10⁶ mm = 10¹⁵ pm

(ii) 1 mg = 1 mg × \(\frac{1\,g}{1000\,mg}\times\frac{1\,kg}{1000\,g}\)

⇒ 1 mg = 10^–6 kg 

1 mg = 1 mg × \(\frac{1\,g}{1000\,mg}\times\frac{1\,ng}{10^{-9}\,g}\)

⇒ 1 mg = 10⁶ ng 

1 mg = 10^–6 kg = 10⁶ ng

(iii) 1 mL = 1 mL × \(\frac{1\,L}{1000\,mL}\)

⇒ 1 mL = 10^–3 L 

1 mL = 1 cm³ = 1 \(\frac{1\,dm\times1\,dm\times1\,dm}{10\,cm\times10\,cm\times10\,cm}\) cm³

⇒ 1 mL = 10^–3 dm³ 

\(\therefore\) 1 mL = 10^–3 L = 10^–3 dm³

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

Answer:

According to the question: 

Time taken to cover the distance 

= 2.00 ns = 2.00 × 10^–9 s 

Speed of light = 3.0 × 10⁸ ms^–1 

Distance travelled by light in 2.00 ns 

= Speed of light × Time taken

= (3.0 × 10⁸ ms ^–1) (2.00 × 10^–9 s) 

= 6.00 × 10^–1 m 

= 0.600 m

23. In a reaction 

A + B₂ → AB₂ 

Identify the limiting reagent, if any, in the following reaction mixtures. 

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

(ii) 2 mol A + 3 mol B 

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

(iv) 5 mol A + 2.5 mol B 

(v) 2.5 mol A + 5 mol B

Answer:

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

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B.

Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent. 

(ii) According to the reaction, 1 mole of A reacts with 1 mole of B.

Thus, 2 mole of A will react with only 2 mole of B. As a result, 1 mole of A will not be consumed. Hence, A is the limiting reagent. 

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B.

Thus, all 100 atoms of A will combine with all 100 molecules of B.

Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mole of atom A combines with 1 mole of molecule B.

Thus, 2.5 mole of B will combine with only 2.5 mole of A. As a result, 2.5 mole of A will be left as such. 

Hence, B is the limiting reagent.

(v) According to the reaction, 1 mole of atom A combines with 1 mole of molecule B.

Thus, 2.5 mole of A will combine with only 2.5 mole of B and the remaining 2.5 mole of B will be left as such. Hence, A is the limiting reagent.

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

N_2(g) + H_2(g) → 2NH_3(g)

(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen. 

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

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

Answer:

(i) Balancing the given chemical equation,

N_2(g) + 3H_2(g) → 2NH_3(g)

From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.

⇒ 2.00 × 10³ g of dinitrogen will react with \(\frac{6\,g}{28\,g}\times2.00\times10^3 g\) dihydrogen 

i.e., 2.00 × 10³ g of dinitrogen will react with 428.6 g of dihydrogen. 

Given, 

Amount of dihydrogen = 1.00 × 10³ g 

Hence,

N₂ is the limiting reagent. 

\(\therefore\) 28 g of N₂ produces 34 g of NH₃. 

Hence, mass of ammonia produced by 2000 g of N₂ = \(\frac{34\,g}{28\,g}\times2000g\) = 2428.57 g

(ii) N₂ is the limiting reagent and H₂ is the excess reagent. 

Hence, H₂ will remain unreacted.

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

= 571.4 g

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

Answer:

Molar mass of Na₂CO₃ = (2 × 23) + 12.00 + (3× 16) = 106 g mol^–1 

Now, 1 mole of Na₂CO₃ means 106 g of Na₂CO₃.

\(\therefore\) 0.5 mol of Na₂CO₃ = \(\frac{106\,g}{1\,mole}\times0.5\) mol Na₂CO₃

= 53 g Na₂CO₃ 

⇒ 0.50 M of Na₂CO₃ = 0.50 mol/L Na₂CO₃ 

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

26. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer:

Reaction of dihydrogen with dioxygen can be written as:

\(2H_{2(g)}+O_{2(g)}\longrightarrow 2H_2O_{(g)}\)

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

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

27. Convert the following into basic units: 

(i) 28.7 pm 

(ii) 15.15 pm 

(iii) 25365 mg

Answer:

(i) 28.7 pm: 

1 pm = 10^–12 m 

\(\therefore\) 28.7 pm = 28.7 × 10^–12 m 

= 2.87 × 10^–11 m

(ii) 15.15 pm: 

1 pm = 10^–12 m 

\(\therefore\) 15.15 pm = 15.15 × 10^–12 m 

= 1.515 × 10^–12 m

(iii) 25365 mg: 

1 mg = 10^–3 g 

25365 mg = 2.5365 × 10⁴ × 10^–3 g 

Since, 

1 g = 10^–3 kg 

2.5365 × 10¹ g = 2.5365 × 10^–1 × 10^–3 kg 

\(\therefore\) 25365 mg = 2.5365 × 10^–2 kg

Answer 5

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

(i) 1 g Au (s) 

(ii) 1 g Na (s) 

(iii)1 g Li (s) 

(iv)1 g of Cl₂(g)

Answer:

(i). 1 g of Au (s) = \(\frac{1}{197}\) mol of Au (s)

= \(\frac{6.022\times10^{23}}{197}\) atoms of Au (s)

= 3.06 × 10²¹ atoms of Au (s)

(ii). 1 g of Na (s) = \(\frac{1}{23}\) mol of Na (s)

= \(\frac{6.022\times10^{23}}{23}\) atoms of Na (s)

= 0.262 × 10²³ atoms of Na (s) 

= 26.2 × 10²¹ atoms of Na (s)

(iii). 1 g of Li (s) = \(\frac{1}{7}\) mol of Li (s)

= \(\frac{6.022\times10^{23}}{7}\) atoms of Li (s)

= 0.86 × 10²³ atoms of Li (s) 

= 86.0 × 10²¹ atoms of Li (s)

(iv). 1 g of Cl₂ (g) = \(\frac{1}{71}\) mol of Cl₂ (g)

(Molar mass of Cl₂ molecule = 35.5 × 2 = 71 g mol^–1)

= \(\frac{6.022\times10^{23}}{71}\) atoms of Cl₂ (g)

= 0.0848 × 10²³ atoms of Cl₂ (g) 

= 8.48 × 10²¹ atoms of Cl₂ (g) 

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

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

Answer:

Mole fraction of C₂H₅OH = \(\frac{Number\,of\,moles\,of\,C_2H_5OH}{Number\,of\,moles\,of\,solution}\)

0.040 = \(\frac{n_{C_2H_5OH}}{n_{C_2H_5OH}\,+\,n_{H_2O}}\) ........(1)

Number of moles present in 1 L water:

\(n_{H_2O} = \frac{1000g}{18\,g\,mol^{-1}}\)

\(n_{H_2O} = 55.55 \,mol\)

Substituting the value of \(n_{H_2O}\) in equation (1),

\(\therefore\) Molarity of solution = \(\frac{2.314\,mol}{1\,L}\)

= 2.314 M

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

Answer:

1 mole of carbon atoms = 6.023 × 10²³ atoms of carbon 

= 12 g of carbon 

Mass of one ¹²C atom = \(\frac{12\,g}{6.022\times10^{23}}\)

= 1.993 × 10^–23 g

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

(i) \(\frac{0.02856\times298.15\times0.112}{0.5785}\) 

(ii) 5 × 5.364 

(iii) 0.0125 + 0.7864 + 0.0215

Answer:

(i) \(\frac{0.02856\times298.15\times0.112}{0.5785}\) 

Least precise number of calculation = 0.112 

\(\therefore\) Number of significant figures in the answer = Number of significant figures in the least precise number 

= 3

(ii) 5 × 5.364 

Least precise number of calculation = 5.364 

\(\therefore\) Number of significant figures in the answer = Number of significant figures in 5.364

= 4

(iii) 0.0125 + 0.7864 + 0.0215 

Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4.

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

Isotope	Isotopic molar mass	Abundance
36 Ar	35.96755 gmol–1	0.337%
38 Ar	37.96272 gmol–1	0.063%
40 Ar	39.9624 gmol–1	99.600%

Answer:

Molar mass of argon

= 39.947 gmol^–1

33. 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.

Answer:

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

\(\therefore\) 52 mol of Ar = 52 × 6.022 × 10²³ atoms of Ar

= 3.131 × 10²⁵ atoms of Ar

(ii) 1 atom of He = 4 u of He 

Or, 

4 u of He = 1 atom of He

1 u of He = 1/4 atom of He

52u of He = 52/4 atom of He

= 13 atoms of He

(iii) 4 g of He = 6.022 × 10²³ atoms of He

\(\therefore\) 52 g of He = \(\frac{6.022\times10^{23}\times52}{4}\) atoms of He

= 7.8286 × 10²⁴ atoms of He

Answer 6

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

(i) empirical formula,

(ii) molar mass of the gas, and

(iii) molecular formula.

Answer:

(i) 1 mole (44 g) of CO₂ contains 12 g of carbon.

\(\therefore\) 3.38 g of CO₂ will contain carbon = \(\frac{12g}{44g}\times3.38g\)

= 0.9217 g 

18 g of water contains 2 g of hydrogen.

\(\therefore\) 0.690 g of water will contain hydrogen = \(\frac{2g}{18g}\times0.690\)

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is: 

= 0.9217 g + 0.0767 g 

= 0.9984 g

\(\therefore\) Percent of C in the compound = \(\frac{0.9217\,g}{0.9984\,g}\times100\)

= 92.32%

Percent of H in the compound = \(\frac{0.0767g}{0.9984g}\times100\)

= 7.68%

 Moles of carbon in the compound = \(\frac{92.32}{12.00}\)

= 7.69 

Moles of hydrogen in the compound = \(\frac{7.68}{1}\)

= 7.68

\(\therefore\) Ratio of carbon to hydrogen in the compound = 7.69: 7.68 = 1: 1

Hence, the empirical formula of the gas is CH.

(ii) Given, 

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

\(\therefore\) Weight of 22.4 L of gas at STP = \(\frac{11.6\,g}{10.0\,L}\times22.4L\)

 = 25.984 g 

≈ 26 g

Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g

n = \(\frac{Molar\,mass\,of\,gas}{Empircial\,formula\,mass\,of\,gas}\)

= \(\frac{26g}{13g}\)

n = 2 

\(\therefore\) Molecular formula of gas = (CH)_n 

= C₂H₂

35. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, 

CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l) 

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

Answer:

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

≡ [(0.75 mol) × (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.

\(\therefore\) Amount of HCl present in 25 mL of solution

= \(\frac{27.375\,g}{1000\,mL}\times25\, mL\)

= 0.6844 g 

From the given chemical equation,

CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l) 

2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO₃ (100 g).

\(\therefore\) Amount of CaCO₃ that will react with 0.6844 g = \(\frac{100}{71}\times0.6844g\)

= 0.9639 g

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

4HCl_(aq) + MnO_2(s) → 2H₂O_(l) + MnCl_2(aq) + Cl_2(g)

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

Answer:

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

5.0 g of MnO₂ will react with 

= \(\frac{146g}{87g}\times5.0g\) of HCl

= 8.4 g of HCl 

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