NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry
1.1. Calculate the molecular mass of the following: (i) H2O (ii) CO2 (iii) CH4.
Ans. (i) Molecular mass of water (H2O)
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= (2 × 1.008 u) + (1 × 16.00 u)
= 2.016 u + 16.00 u
= 18.02 u
(ii) 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) 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
1.2. Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
Ans. According to formula,
Mass percent of an element
$$=\frac{\text{Mass of that element in the compound}}{\text{Molar mass of the compoound}}×100$$
Molar mass of Na2SO4 = (2 × 23.0) + (1 × 32.0) + (4 × 16.0)
= 142 g
Elements present in given compound sodium sulphate is Na, S and O. Therefore,
$$\text{Mass percent of sodium (Na):}\bigg(\frac{46}{142}\bigg)×100\\=32.30\%\\\text{Mass percent of sulphur (S):}\bigg(\frac{32}{142}\bigg)×100\\=22.54\%\\\text{Mass percent of oxygen (O):}\bigg(\frac{64}{142}\bigg)×100\\=45.07\%$$
1.3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Ans. Given:
Percentage of iron by mass = 69.9%
Atomic mass of iron = 55.85 amu
Thus, relative moles of iron in iron oxide:
$$=\frac{\%\text{mass of iron by mass}}{\text{Atomic mass of iron}}\\=\frac{69.9}{55.85}$$
= 1.25
Now, Percentage of oxygen by mass = 30.1%
Atomic mass of oxygen = 16.00 amu
Thus, relative moles of oxygen in iron oxide
$$=\frac{\%\text{mass of oxygen by mass}}{\text{Atomic mass of oxygen}}\\=\frac{30.01}{16}$$
= 1.88
The simplest molar ratio of iron to oxygen will be,
$$= \frac{1.25}{1.25}:\frac{1.88}{1.25}\\=1:15\space\text{or}, 2:3$$
= 1 : 1.5 or, 2 : 3
Therefore, the empirical formula of the iron oxide is Fe2O3.
1.4. 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.
Ans. The balanced chemical equation for the combustion process of carbon can be represented as:
$$\underset{\text{1 mole}}{\text{C(s)}}+\underset{\text{1 mole}}{\text{O}_2(\text{g})}\xrightarrow{}\underset{\text{1 mole}}{\text{CO}_2(\text{g})}$$
(32 g) (44 g)
(i) According to the above equation, one mole of carbon burns in 1 mole of air or dioxygen and produce 1 mole of carbon dioxide.
(ii) When 16 g of dioxygen is taken i.e., half. Then the amount of carbon reacting will also be half. Thus, only 0.5 moles of carbon react with 16 g of dioxygen and give 22 g of carbon dioxide.
(iii) In the above reaction, oxygen acts as limiting reagent. Thus, only 0.5 moles of carbon can combine with 16 g of dioxygen and give 22 g of carbon dioxide.
1.5. 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.
Ans. Given,
0.375 M aqueous solution of sodium acetate means that 1000 mL of solution contain 0.375 moles of sodium acetate.
Molar mass of sodium acetate = 82.0245 g mol–1
Now,
No. of moles of sodium acetate in 500 mL
$$=\frac{0.375}{1000}×500\\=\frac{0.375}{2}$$
= 0.1875 mole
Therefore,
Mass of sodium acetate required
= 0.1875 mole × 82.0245 g mol–1
= 15.380 g
1.6. 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 percent of nitric acid in it being 69%.
Ans. Given,
Mass percent of nitric acid inthe sample = 69%
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Now,
Molar mass of nitric acid (HNO3)
= {(1 × Atomic mass of H) + (1 × atomic mass of N) + (3 × Atomic mass of Oxygen)}
= (1 × 1) + (1 × 14) + (3 × 16)
= (1 + 14 + 48)
= 63 g
And , number of moles in 69 g of HNO3
$$=\frac{69}{63}\space\text{moles}=1.095\space\text{moles}\\\text{Thus},\\\text{Volume of 100 g nitric acid solution}\\=\frac{\text{Mass of solution}}{\text{Density of solution}}\\\frac{100}{1.41}\text{g mL}^{-1}\\= 70.92 \text{mL}\\=\text{= 0.07092 L}\\\therefore\text{Conc. of HNO}_3\space\text{in moles per litre}=\frac{1.095}{0.07092}\\=15.44\space\text{mol/L}.$$
1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Ans. 1 mole of CuSO4 contains 1 mole of copper.
Now,
Molar mass of CuSO4
= (1 × atomic mass of Copper + 1 × atomic mass
of Sulphur + 4 × atomic mass of Oxygen)
= (1 × 63.5 + 1 × 32 + 4 × 16)
= 63.5 + 32.00 + 64.00 = 159.5 g
Thus, Cu that can be obtained from 159.5 g of CuSO4 is 63.5 g
Cu that can be obtained from 100 g CuSO4 is
$$=\frac{63.5}{159.5}×100=39.81\space\text{g}$$
1.8. 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. To determine molecular formula, first empirical formula is determined.
Percentage of iron by mass = 69.9 % (Given)
Atomic mass of iron = 55.85 amu
Relative moles of iron in iron oxide
$$=\frac{\%\text{mass of iron by mass}}{\text{Atomic mass of iron}}\\\frac{69.9}{55.85}=1.25$$
Percentage of oxygen by mass = 30.1 % (Given)
Atomic mass of oxygen = 16.00 amu
Relative moles of oxygen in iron oxide
$$=\frac{\%\text{mass of oxygen by mass}}{\text{Atomic mass of oxygen}}\\=\frac{30.1}{16}=1.88$$
$$\text{Simplest molar ratio =}\frac{1.25}{1.25}:\frac{1.88}{1.25}\\=1:1.5=2:3\\\therefore\text{The empirical formula of an oxide of iron is Fe}_2\text{O}_3.\\\text{Empirical formula mass of Fe}_2\text{O}_3\\=(2 × 55.85) + (3 × 16.00) = 159.7 \text{g mol}^{–1}\\\text{According to the formula,}\\\ n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\=\frac{159.7}{159.6}=1$$
Thus, molecular formula is same as empirical formula i.e. Fe2O3.
1.9. Calculate the atomic mass (average) of chlorine using the following data :
| % Natural Abundance | Molar Mass | |
| 35Cl | 75.77 | 34.9689 |
| 37Cl | 24.23 | 36.9659 |
Ans. Given,
Fractional abundance of 35Cl = 0.7577
Molar mass of 35Cl = 34.9689
Fractional abundance of 37Cl = 0.2423
Molar mass 37Cl = 36.9659
According to the average atomic mass formula,
Average atomic mass
= (Fractional abundance of 35Cl × molar mass of 35Cl) + (Fractional abundance of 37Cl × molar mass of 37Cl)
= (0.7577 × 34.9689) u + (0.2423 × 36.9659) u
= 26.4959 u + 8.9568 u = 35.4527u
1.10. 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.
Ans. (i) 1 mole of C2H6 contains 2 moles of carbon atoms
∴ 3 moles of of C2H6 will contain (2 × 3) i.e, 6 moles of carbon atoms
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms
∴ 3 moles of of C2H6 will contain (3 × 6) i.e, 18 moles of hydrogen atoms
(iii) 1 mole of C2H6 contains Avogadro’s number = 6.02 × 1023 molecules
∴ 3 moles of of C2H6 will contain = 3 × 6.02 × 1023 = 18.06 × 1023 ethane molecules.
1.11. 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 2L?
Ans. Molar mass of sugar (C12H22O11)
= (12 × atomic mass of carbon) + (22 × atomic mass of hydrogen) + (11 × atomic mass of oxygen)
= (12 × 12) + (22 × 1) + (11 × 16)
= 342 g mol–1
Number of moles in 20 g of sugar
= 0.0585 mole
Given that,
Volume of Solution = 2L According to the formula,
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?
Ans. Molar mass of methanol (CH3OH)
= (1 × Atomic mass of carbon) + (4 × atomic mass of hydrogen) + (1 × atomic mass of oxygen)
= (1 × 12) + (4 × 1) + (1 × 16)
= 32 g mol–1
= 0.032 kg mol–1
Now,
$$\text{Molarity of the solution}= \frac{0.793\space\text{kg L}^{\normalsize -1}}{0.032 \space\text{kg}\space\text{mol}^{\normalsize -1}}$$
= 24.78 mol L–1
Applying the formula,
M1V1 = M2V2 (Here, M1 and V1 are molarity and volume of given solution and M2 and V2 are molarity and volume of solution to be prepared, respectively.)
24.78 × V1 = 0.25 × 2.5 L
V1 = 0.02522 L = 25.22 mL
1.13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, is pascal as shown below :
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.
Ans. Pressure is given as the force acting per unit area i.e.,
$$\text{Pressure}=\frac{\text{Force}}{\text{Area}}\\=\frac{\text{Mass×Acceleration due to gravity}}{\text{Area}}$$
Given:
Mass of air over 1 cm2 area at sea level
= 1034 g = 1034 × 10–3kg
Area = 1 cm2 = 1 × (10–2 m)2
= 1 × 10–4 m2
Acceleration due to gravity = g = 9.8 ms–2
$$\therefore\space\text{Pressure}=\frac{(1034×10^{\normalsize-3}\text{kg})(9.8\space\text{ms}^{\normalsize-2})}{1×10^{\normalsize-4}\text{m}^{2}}\\= 1.01332 × 10^{5} \text{kg m}^{\normalsize–1}\text{s}^{\normalsize–2}\\= 1.01332 × 10^{5} \text{Pa}$$Thus, the pressure in pascal is 1.01332 Pa.
1.14. What is the SI unit of mass? How is it defined?
Ans. The SI unit of mass is kilogram. It is represented as kg. It can be defined as the mass equal to the mass of the international prototype of kilogram.
1.15. Match the following prefixes with their multiples:
| Prefixes | Multiples |
| (1) micro | 106 |
| (2) deca | 109 |
| (3) mega | 10–6 |
| (4) giga | 10–15 |
| (5) femto | 10 |
Ans.
| Prefixes | Multiples |
| (1) micro | 10–6 |
| (2) deca | 10 |
| (3) mega | 106 |
| (4) giga | 109 |
| (5) femto | 10-15 |
1.16. What do you mean by significant figures ?
Ans. Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures.
The uncertainity is indicated by writing the certain digits and the last uncertain digit.
For example; if the value of an experiment is 235.3, then , 235 is certain and 3 is uncertain. But total number of significant figures will be 4. Thus, significant figures also include last uncertain figure.
1.17. 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.
Ans. (i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
15 ppm means 15 parts in million (106) parts.
$$\therefore\text{percentage by mass}=\frac{15}{10^{6}}×100$$
= 15 × 10-4
= 1.5 × 10-3 %
(ii) Molar mass of chloroform (CHCl3)
= (1 × atomic mass of Carbon + 1 x atomic mass of hydrogen + 3 atomic mass of chlorine)
= 12 + 1 + (3 × 35.5) = 119.5 g mol–1
100 g of the sample contain chloroform
= 1.5 × 10–3 g
∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5 × 10–2 g$$\text{molality\space=}\space\frac{1.5×10^{\normalsize -2}}{119.5}\text{mole}$$
= 1.265 × 10–4 mole
∴ Molality = 1.265 × 10–4 m.
1.18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Ans. Scientific notations can be given as:
(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 × 100
1.19. 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
Ans. (i) 0.0025 – Two significant figures.
(ii) 208 – Three significant figures.
(iii) 5005 – Four significant figures
(iv) 126,000 – Three significant figures
(v) 500.0 – Four significant figures
(vi) 2.0034 – Five significant figures
1.20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Ans. (i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810
1.21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
| Mass of dinitrogen | Mass of dioxygen |
| (1) 14 g | 16 g |
| (2) 14 g | 32 g |
| (3) 28 g | 32 g |
| (4) 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
Ans. (a) The given experimental data obeys the law of multiple proportions because if the mass of dinitrogen is fixed to 28 g, then the mass of dioxygen mass of dioxygen combined is 32, 64, 32 and 80 g. The corresponding ratio is 2 : 4 : 2 : 5. It is a simple whole number ratio.
The law states that “ If two elements combine to form more than one compound, the masses of one element that combines with a fixed mass of the other element bears a simple whole number ratio.”
(b) (i) We know,
1 km = 1000 m
or 1 m = 1000 mm
Therefore,
1 km = 1000 × 1000 mm
= 106 mm
$$\text{1 km = 1 km}×\frac{1000\space\text{m}}{1\text{km}}×\frac{1 \text{pm}}{10^{-12}\space\text{m}}$$1 km = 1015 pm
Hence, 1 km = 106mm = 1015 pm
(ii) We know
1 kg = 1000 g
or 1 g = 1000 mg
Therefore,
1 kg = 1000 × 1000 mg
or 106 mg = 1 kg
or
1 mg = 10–6 kg
$$\text{1 mg = 1 mg ×}\frac{1g}{1000\text{mg}}×\frac{1 \text{ng}}{10^{\normalsize-9}g}$$
1 mg = 106 ng
1 mg = 10–6 kg = 106 ng
(iii) We know,
100 ml = 1L
$$\text{or \space 1 mL =}\frac{1}{1000}×1=0.001\text{L}\\1 \text{mL} = 1 \space\text{cm}^3 = 1 \space\text{cm}^3×\frac{1 \space\text{dm}×1 \space\text{dm}×1 \space\text{dm}}{10 \space\text{cm}×10 \space\text{cm}×10 \space\text{cm}}\\\Rarr1 \text{mL} = 10^{–3} \text{dm}^{3}\\ 1 \text{mL} = 10^{–3} \text{L} = 10^{–3} \text{dm}^{3}$$
1.22. If the speed of light is 3.0 × 108ms–1, calculate the distance covered by light in 2.00 ns.
Ans. According to the formula,
Distance covered = Speed × Time
= 3.0 × 108 ms–1 × 2.00 ns
$$= 3.0 × 10^8 \text{ms}^{\normalsize–1} × 2.00 \text{ns}×\frac{10^{\normalsize-9}}{1\text{ns}}\\= 6.00 × 10^{\normalsize–1} \text{m} = 0.600 \text{m}$$
1.23. In a reaction :
$$\textbf{A}+\textbf{B}_2\xrightarrow{}\textbf{AB}_2$$
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 The reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent. When it gets consumed, the reaction will not proceed irrespective of the amount of other reactant present in the reaction. In other words, it determines the extent of reaction.
According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus,
(i) 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) 2 mol of A will react with only 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, A is the limiting reagent.
(iii) 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) 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.
1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$$\textbf{N}_2\textbf{(g)}+\textbf{H}_2\textbf{(g)}\xrightarrow{}2\textbf{NH}_3\textbf{(g)}$$
(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?
Ans. (i) Balanced chemical equation for the formation of ammonia can be given as:
$$\text{N}_2\text{(g)}+\text3{\text{H}}_2\text{(g)}\xrightarrow{}2\text{NH}_3\text{(g)}$$
Thus, according to 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 × 103g of N2 will react with H2
$$=\frac{6}{28}×2.00×10^3 \text{g}\\= 428.6 \space\text{g}.\\ \text{Thus}, 2.00 × 10^{3}\text{g of N}_2 \text{will react with}\space 426.8 \space\text{g of H}_2. \space\text{Here, N}_2 \space\text{is the limiting reagent while H}_2 \space\text{is in excess.}\\\text{As,}\space28 \text{g of N}_2\space\text{produce}\space 34 \text{g of NH}_3.\\\therefore \space2.00×10^{3}\text{g of N}_2\text{will produce}\\=\frac{34}{28}×2.00×10^{3}\text{g}\\= 2428.57 \text{g of NH}_3.$$
(ii) N2 is the limiting reagent and H2 is the excess reagent. Therefore, H2 will remain unreacted.
(iii) Mass of dihydrogen left unreacted
= 1000 g – 428.6g
= 571.4 g
1.25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
Ans. Molar mass of Na2CO3
= (2 × atomic mass of Na + 1 × atomic mass of C+ 3 × atomic mass of O)
= (2 × 23) + 12.00 + (3 × 16) = 106 g mol–1
∴ 0.50 mol Na2CO3 means 0.50 × 106 g mol–1
= 53 g
0.50 M Na2CO3 means 0.50 mol/liter of Na2CO3 i.e. 53 g of Na2CO3 are present in 1 litre of the solution.
1.26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Ans. The balanced chemical equation for reaction of dihydrogen gas with dioxygen gas can be given as ,
$$\text{2H}_2(\text{g})+\text{O}_2(\text{g})\xrightarrow{} \text{2H}_2\text{O(g)}$$Here, two volumes of dihydrogen react with one volume of dioxygen to produce two volumes of water vapour. Therefore, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.
1.27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Ans. (i) 28.7 pm
1 pm = 10–12 m
28.7 pm = 28.7 × 10–12 m
= 2.87 × 10–11 m
(ii) 15.15 pm
1 pm = 10–12 m
15.15 pm = 15.15 × 10–12 m
= 1.515 × 10–11 m
(iii) 25365 mg
1 mg = 10–3 g
25365 mg = 2.5365 × 104×10–3 g
= 2.5365 × 101
Also,
1 g = 10–3 kg
2.5365 × 101 g = 2.5365 × 101 ×10–3 kg
∴ 25365 mg = 2.5365 × 10–2 kg
1.28. 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)
Ans. Number of atoms = Number of moles × Avogadro’s number
(i) 1 g Au = 1/197 mol atom of Au (atomic mass of Au = 197)
Number of atoms
$$=\frac{1}{197}×6.022 × 10^{23} \text{atoms of Au}\\\text{(ii) 1 g Na (s) (atomic mass of Na = 23)}\\=\frac{1}{23}\text{mol atom of Na}\\\text{Number of atoms}=\frac{1}{23}×6.022×10^{23}\text{atoms of Na}\\\text{(iii) 1 g Li = 1/7 mol atom of Li (atomic mass of Li = 7)}\\\text{Number of atoms}=\space\frac{1}{7}×6.022×10^{23}\text{atoms of Li}\\\text{(iv) 1 g Cl}_2 = 1/71 \text{mol molecules of Cl}_2 \text{(atomic mass of Cl = 35.5)}\\=\frac{1}{71}×6.022×10^{23}\text{molecules of Cl}_2$$
Thus, 1 g of Li has the largest number of atoms.
1.29. 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.)
Ans. Given :
Mole fraction of ethanol = 0.040.
Mole fraction of C2H5OH$$=\frac{\text{No. of moles of C}_2\text{H}_5\text{OH}}{\text{No. of moles of solution}}\\ \text{x}_{\text{C}_2\text{H}_5\text{OH}}=\frac{\text{n}(\text{C}_2\text{H}_5\text{OH})}{\text{n}(\text{C}_2\text{H}_5\text{OH})+\text{n}(\text{H}_2\text{0})}\\=0.040(\text{Given})$$
The objective is to find number of moles of ethanol in 1L of the solution which is nearly equal to 1L of water as solution is dilute.
Now, calculating number of moles of water,
$$\text{No. of moles in 1L of water =}\frac{1000\text{g}}{18\text{g}}\text{mol}^{-1}\\=55.55\text{moles (molar mass of water = 18 g) Substituting n(H}_2\text{O}) = 55.55 \space\text{in equation (i)}\\\frac{n(\text{C}_2\text{H}_5\text{OH})}{n(\text{C}_2\text{H}_5\text{OH})+55..55}=0.040\\0.96\text{n(C}_2\text{H}_5\text{OH)} = 55.55 × 0.040\\\text{or}\space \text{n(C}_2\text{H}_5\text{OH})=2.31\space\text{mol}$$Hence, molarity of the solution is 2.31 M
1.30. What will be the mass of one 12C atom in g ?
Ans. 1 mol of 12C atoms = 6.022 × 1023 atoms = 12 g
Thus, 6.022 × 1023 atoms of 12C have mass = 12 g$$\text{1 atom}\space^{12}\text{C}\space \text{will have mass =}\frac{12}{6.022×10^{23}}\text{g}\\= 1.993 × 10^{–23} \text{g}$$
1.31. How many significant figures should be present in the answer of the following calculations?$$(i)\space0.02856 × 298.15 ×\frac{0.112}{0.5785}\\(ii)\space5 × 5.364 \\(iii)\space0.0125 + 0.7864 + 0.0215$$
Ans. (i) Least precise term in the calculation is 0.112. It is having 3 significant digits. Therefore, the answer should have 3 significant figures.
(ii) Least precise term in the calculation is 5.364. It is having 4 significant figures. Therefore, the answer should have 4 significant figures.
(iii) In the given addition, least number of decimal places in each term is 4. Therefore, the answer should have 4 significant figures.
1.32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
| Isotope | Isotopic molar mass | Abundance |
| 36Ar | 35.96755 g mol–1 | 0.337% |
| 38Ar | 37.96272 g mol–1 | 0.063% |
| 40Ar | 39.9624 g mol–1 | 99.600% |
Ans. Molar mass of Argon = Sum of molar mass of all isotopes + Abundance percentage of all isotopes
$$=\bigg(3596755×\frac{0.337}{100}\bigg)+\bigg(37.96272×\frac{0.063}{100}\bigg)+\bigg(39.9624×\frac{99.600}{100}\bigg)$$
= (35.96755 × 0.00337) + (37.96272 × 0.00063) + (39.9624 ×0.99600)
= 39.948 g mol–1
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.
Ans. (i) One mole of an element contains 6.022 × 1023 atoms i.e.,
1 mol of Ar
= 6.022 × 1023 atoms
∴ 52 mol of Ar = 52 × 6.022 × 1023 atoms
= 3.131 × 1025 atoms
(ii) 4 u of He = 1 Atom of He $$\text{1 u of He =}\space\frac{1}{4}\text{atom of He}\\\therefore 52 \space\text{u of He} =\frac{1}{4}×52=13\space\text{atoms}\\\text{(iii) 1 mol of He = 4 g}= 6.022×10^{23}\text{atoms}\\\therefore\space52 \space\text{g of He}=\bigg(\frac{6.022×10^{23}}{4}\bigg)×52\space\text{atoms}\\= 7.8286 × 10^{24}\space\text{atoms}$$
1.34. 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.
Ans. Calculation of the amount of carbon in 3.38 g of CO2
One mole of carbon dioxide contains 12 g of carbon $$\therefore 3.38\space\text{g of CO}_2\space\text{will contain carbon}=\frac{12}{44}×3.38\text{g}\\=0.9218 \space\text{g}$$
Now, amount of hydrogen in 0.690 g H2O
18 g of water contains 2 g of hydrogen
∴ 0.690 g of water will contain hydrogen $$=\frac{2}{18}×0.690\space\text{g}=0.0767\text{g}$$
Since, the compound contains only C and H, therefore,
Total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
Now,
Percentage (%) of C in the compound $$=\frac{0.9218}{0.9985}×100=92.32\\\text{Percentage} (\%) \text{of H in the compound}\\=\frac{0.0767}{0.9985}×100=7.68$$(i) Calculation of empirical formula,$$\text{Moles of carbon in the compound =}\frac{92.32}{12}\\\text{= 7.69 (Atomic mass of carbon is 12)}\\\text{Moles of hydrogen in the compound =}\frac{7.68}{1}\\\text{= 7.68 (Atomic mass of Hydrogen is 1)}\\\text{Simplest molar ratio = 7.69 : 7.68 = 1(approx)}\\\therefore\text{Empirical formula will be CH}\\\text{(ii) 10.0 L of the gas at STP will weight = 11.6 g}\\\therefore 22.4 \text{L of the gas at STP will weight}\\=\frac{11.6}{10.0}×22.4\\=25.984 g ≅ 26 g\\\text{Therefore, the molar mass of gas is 26 g.}\\\text{(iii) Calculation of molecular formula}\\ \text{Empirical formula Mass (CH) = 12 + 1 = 13}\\\therefore\space n=\frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}\\=\frac{26}{13}=2$$ ∴ Molecular Formula = n × empirical formula
= 2 × (CH) = C2H2
1.35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
$$\textbf{CaCO}_3 (\textbf{s}) + 2\textbf{HCl (aq)}\xrightarrow{} \textbf{CaCl}_2(\textbf{aq}) + \textbf{CO}_2(\textbf{g}) + \textbf{H}_2\textbf{O(l)}.$$What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Ans. 1000 mL of 0.75 M HCl have 0.75 mol of HCl
= 0.75 × 36.5 g = 24.375 g
(Molar mass of HCl is 36.5)
∴ Mass of HCl in 25 mL of 0.75 M HCl will be $$=\frac{24.375}{1000}×25\space\text{g}\\= 0.6844 \space\text{g}\\\text{According to the given chemical equation,}\\ \text{CaCO}_3 (\text{s}) + 2\text{HCl (aq)}\xrightarrow{} \text{CaCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}2\text{O}(\text{l})\\\text{2 mol of HCl i.e., 73 g HCl react completely with 1 mol of CaCO}_3\space\text{i.e., 100 g}\\\therefore\text{0.6844 g HCl reacts completely with CaCO}_3\\=\frac{100}{73}×0.6844 \space\text{g}.=0.938\text(g)$$
1.36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction
$$\textbf{4HCl (aq) + MnO}_2(\textbf{s})\xrightarrow{}2\textbf{H}_2\textbf{O}(1)+\textbf{MnCl}_2\textbf{aq}+\textbf{Cl}_2\textbf{(g)}$$ How many grams of HCl react with 5.0 g of manganese dioxide?
Ans. 1 mol of MnO2 = (Atomic mass of Mn + 2 × atomic mass of oxygen) = 55 + 32 g = 87 g
87 g of MnO2 react with 4 moles of HCl i.e. 4 × 36.5 g = 146 g of HCl. (Molar mass of HCl is 36.5)
∴ 5.0 g of MnO2 will react with HCl $$=\frac{146}{87}×5.0\space\text{g}=8.40\space\text{g}.$$
NCERT Solutions for Class 11 Chemistry Chapter 1 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
