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Ncert Solutions of Chemistry Chapter 1:Some Basic Concepts of chemistry
This solution notes is the first chapter of chemistry for class 11th. This chapter is supposed to be the basic chapter of chemistry. In this solution notes, the questions have been asked from the topics which have been prescribed by ncert council. They are unit and its conversion into other units, Laws of chemical combinations, mole concepts, balancing of chemical equations, limiting reagents, concentration of solution like molarity, molality, mole fraction and ppm. Questions have also asked related empirical and molecular formulas, percentage of elements in a compound.
Ncert Solutions of Chemistry Chapter 1
Here are the answers of NCERT TEXT BOOK QUESTIONS:
Question 1. Calculate the molecular mass of the following:
(i) H20 (ii) C02 (iii) CH4
Ans: (i) Molecular mass of H2O = 2(1.008 amu) + 16.00 amu=18.016 amu
(ii) Molecular mass of CO2= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH4 = 12.01 amu + 4 (1.008 amu) = 16.042 amu
Ans:
Question 3. Determine the empirical formula of an oxide of Iron which has 69.9 % iron and 30.1 % dioxygen by mass.
Ans: % of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]
Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide will be Fe2O3.
Question 4. Calculate the amount of carbon dioxide that could be produced when (Ncert Solutions of Chemistry Chapter 1)
(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 of carbon in dioxygen / air is
C(s) + O2(g) → CO2 (g)
1 mole 1 mole (32 g) 1 mole(44 g)
(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.
(ii) Here, oxygen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxygen acts as a limiting reagent as 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.
Question 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 82.0245g mol-1.
Ans: 0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.
∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875
Molar mass of sodium acetate = 82.0245g mol-1
∴Mass of sodium acetate acquired = 0.1875×82.0245 g = 15.380g
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%.
Ans: Mass percent of 69% means that 100 g of nitric acid solution contains 69 g nitric acid by mass.
Molar mass of nitric acid (HNO3) = 1 + 14 + 48 = 68 gmol-1
Moles in 69 g HNO3 = 69 g/ 63 gmol-1 = 1.095 mole Volume of 100 g nitric acid solution = 100 g/ 1.41 gml-1 = 70.92 ml = 0.0792 L Conc. of HNO3 in moles per litre = 1.095 mole / 0.07092 L = 15.44 M (Ncert Solutions of Chemistry Chapter 1)
7. How much copper can be obtained from 100 g of copper sulphate (CuSO4 )?
Solution ?
Ans: 1 mole of CuSO4 contains 1mole (1g atom) of Cu and molar mass of CuSO4 = 63.5 + 32 + 4 × 16 = 159.5 gmol-1 Thus, Cu that can be obtained from 159.5 g of CuSO4 = 63.5 g.
∴ Cu that can be obtained from 100 g of CuSO4 = 63.5/159.5 ×100 g = 39.81 gram.
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
Ans: % of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given] Atomic mass of iron = 55.85 amu Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3 ∴ The empirical formula of the iron oxide will be Fe2O3. Now, Empirical formula mass of Fe2O3 = 2 × 55.85 + 3 × 16 = 159.7 gmol-1
∴ n = Molar mass/ Empirical formula mass = 159.8/159.7 = 1 Hence, molecular formula will be same as empirical formula Fe2O3
9. Calculate the atomic mass (average) of chlorine using the following data :
| Elements | Natural Abundance | Molar mass |
| 35Cl | 75.77 | 34.9689 |
| 37Cl | 24.23 | 36.9659 |
Ans: Fractional Abundance of 35Cl = 0.7577 and molar mass = 34.9689 Fractional Abundance of 37Cl = 0.2423 and molar mass = 36.9659
∴ Average atomic mass = (0.7577 × 34.9689) + (0.2423 × 36.9659) = 35.4527
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.
(Ncert Solutions of Chemistry Chapter 1)
Ans: (i) 1 mole of C2H6 contains 2 moles of carbon atoms ∴ 3 moles of C2H6 will contain C- atoms = 6 moles.
(ii) 1 mole of C2H6 contain 6 moles of hydrogen atoms ∴ 3 moles of C2H6 will contain H- atoms = 18 moles.
(iii) 1 mole of C2H6 contains ethane molecules = 6.022 × 1023 molecules. ∴ 3 mole of C2H6 will contain ethane molecules = 3 × 6.022 × 1023 = 18.06 × 1023 molecules.
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 × 12 + 22 × 1 + 11 × 16 = 342 gmol-1 NO. of moles in 20 g of sugar = 20 g/ 342 gmol-1 = 0.0585 mole.
Molar concentration = moles of solute/ volume of solution in L = 0.0585/ 2L = 0.0293 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?
Ans:- Molar mass of Methanol (CH3OH) = 32 gmol-1 = 0.032 kg mol-1.
Molarity of the given solution = 0.793 kg L-1/ 0.032 kg mol-1 = 24.78 mol L-1
Applying M1 × V1 (for given solution) = M2 × V2 (solution to be prepared)
24.78 × V1 = 0.25 × 2.5 L or V1 = 0.02522 L = 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.
Ans:- As we know that pressure is the force (weight) acting per unit area. But weight = mg (where ‘g’ is gravity due to acceleration)
∴ pressure = weight per unit area = 1034 g × 9.8 ms-2 / cm2 = 1034 g × 9.8 ms-2 / cm2 × 1kg /1000 g × 100 cm/1 m × 100 cm/ 1 m × 1 N / kg ms-2 × 1Pa / 1 Nm-2 = 1.0132 × 105 Pa.
14. What is the SI unit of mass? How is it defined?
Ans:- S.I. unit of mass is kilogram (kg). The kilogram is the mass of platinum- iridium cylinder that is stored in an airtight jar at international Bureau of weights and measure in France.
15. Match the following prefixes with their multiples:
| S.NO. | prefixes | Multiples |
| A | femto | 10 |
| B | giga | 10-15 |
| C | mega | 10-6 |
| D | deca | 109 |
| E | micro | 106 |
Ans : micro = 10-6 , deca = 10, mega = 106 , giga = 109 , femto = 10-15
16. What do you mean by significant figures ?
Ans:- The total number of digits in a number including the last digit whose value is uncertain is called the number of significant figures. For example, 14.56 has four significant figures. Similarly 2.5 has two significant figures and 0.024 has two significant figures.
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) 15 ppm means 15 parts in million (106) parts. ∴ % by mass = 15/106 ×100 = 15 × 10-4 = 1.5 × 10 -3%
(ii) Molar mass of Chloroform (CHCl3) = 12 + 1 + 3 × 35.5 = 118.5 g mol-1. 100 g of the sample contains Chloroform = 1.5 × 10-3 g.
∴ 1000 g (1kg) of the sample will contain Chloroform = 1.5 × 10-2 g = 1.5 × 10-2 g/ 118.5 g mol-1 = 1.266 × 10-4 mole ∴ molality = 1.266 × 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
Ans:- (i) 4.8 × 10-3 (ii) 2.34 × 105 (iii) 8.008 × 103 (iv) 5.000 × 102 (v) 6.0012 × 100
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) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5
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) 0.0460 (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
Ans:- (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These are in the ratio 1 : 2 : 1 : 5 which is a simple whole number ratio. Hence, the given data obey the law of multiple proportions.
(b) (i) 1 km = 1 km × 1000 m/1km × 100 cm/ 1 m × 10 mm/1 cm = 106 mm
1 km = 1 km ×1000 m/1 km × 1pm/10-12 m = 1015 pm
(ii) 1 mg = 1 mg × 1 g /1000 mg × 1 kg / 1000 g = 10-6 kg
1 mg = 1 mg × 1 g/ 1000 mg × 1ng / 10-9 g = 106 mg
(iii) 1 ml = 1ml × 1L/1000ml = 10-3 L
1 ml = 1 cm3 = 1 cm3 × 1dm/10 cm × 1dm/10 cm × 1dm/10 cm = 10-3 dm3
22. If the speed of light is 3.0 × 108ms-1, calculate the distance covered by light in 2.00 ns.
Ans:- Distance covered = speed × time = 3.0 × 108 ms-1 × 2.00 ns = 3.0 × 108 ms-1 × 2.00 ns × 10-9 s /1ns = 6.00 × 10-1 m = 0.6 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
Ans:- (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. ∴ 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B is the limiting reagent and A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B. ∴ 2 mol of A will react with 2 mol of B. Hence, A is the limiting reagent.
(iii) No limiting reagent
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.
24. 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 ammonia produced if 2.00×103g dinitrogen reacts with 1.00×103g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
(i) 1 mol of N2 that is 28 g react with 3 mol of H2, i.e., 6 g of H2.
∴ 200 g of N2 will react with H2 = 6/28 × 200 g = 428.6 g. Thus N2 is the limiting reagent while H2 is the excess reagent. 2 mol of N2, i.e.,28 g of N2 produces NH3 = 2 mol = 34 g NH3. Hence, 200 g N2 will produce NH3 = 34/28 × 2000 g = 2428.57 g
(ii) H2 will remain unreacted.
(iii) Mass left unreacted = 1000 — 428.6 g = 571.4 g.
25. How are 0.50 mol Na2CO3 and 0.50 M Na2 CO3 different?
Ans:- Molar mass of Na2CO3 = 2 × 23 + 12 + 3 × 16 = 106 gmol-1
0.50 mol of Na2CO3 means 0.50 × 106 g = 53 g Na2CO3 and 0.50 M Na2CO3 means 0.50 mol or 53 g of Na2CO3 are present in 1 litre of the solution.
26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Ans:- H2 and O2 react according to the equation
2H2 (g) + O2(g) —— 2H2O(g)
Thus, 2 Volumes of H2 react with 1 volume of O2 to produce 2 volumes of water vapour. Hence, 10 Volumes of H2 will react completely with 5 Volumes of O2 to produce 10 volumes of water vapour.
27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 μs
(iii) 25365 mg
Ans:- (i) 28.7 pm = 28.7 pm × 10-12 m/ 1pm = 2.87 × 10-11 m
(ii) 15.15 μs = 10.15 μs × 10-6 s/1μs = 1.515 × 10-5 s.
(iii) 25365 mg = 25365 mg × 1 g/ 1000 mg × 1 kg/1000 g = 2.5365 × 10-2 kg.
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:- (i) 1 g Au (s) = 1/197 mol = 1/197 × 6.022 × 1023 atoms.
(ii) 1 g Na (s) = 1/23 mol = 1/23 × 6.022 × 1023 atoms.
(iii) 1 g Li (s) = 1/7 mol = 1/7 × 6.022 × 1023 atoms.
(iv) 1 g of Cl2 (g) = 1/71 mol = 1/71 × 6.022 × 1023 molecules = 2/71 × 6.022 × 1023 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).
Ans:- χ C2H5OH = n(C2H5OH) / n(C2H5OH) + n(H2O) = 0.040 (given). Now our aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1 L of water (because solution is dilute).
∴ No. of moles in 1L of water = 1000 g/ 18 gmol-1 = 55.55 moles.
Now, χ C2H5OH = n(C2H5OH) / n(C2H5OH) + 55.55 = 0.040
Or, 0.96 n (C2H5OH) = 55.55 × 0.040 or n (C2H5OH) = 2.31 mol. Hence, molarity of the solution = 2.31 M.
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.
∴ 1 atom of 12C will have mass = 12/ 6.022 × 1023 g = 1.9927 × 10-23 g.
31. How many significant figures should be present in the answer of the following calculations?
(i) 0.02856 × 298.15 × 0.112 /0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Ans:- (i) The least precise term has 3 significant figures (in 0.112). Hence, the answer should have 3 significant figures.
(ii) Leaving the exact number (5), the second term has 4 significant figures. Hence, the answer should have 4 significant figures.
(iii) In the given addition, the least number of decimal places in the term is 4, Hence, the answer should have 4 significant figures.
32. Use the data given in the following table to calculate the molar mass of naturally occuring 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 Ar = 35.96755 × 0.00337 + 37.96272 × 0.00063 + 39.96924 × 0.99600 = 39.948 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.
Ans:- (i) 1 mole of He = 6.022 × 1023 atoms. ∴ 52 moles of He = 52 × 6.022 × 1023 atoms = 3.131 × 1025 atoms.
(ii) 1 atom of He = 4 u of He that means 4 u of He = 1 atom of He
∴ 52 u of He = 1/4 ×52 atoms = 13 atoms.
(iii) 1 mole of He = 4 g = 6.022 × 1023 atoms ∴ 52 g of He = 6.022 × 1023 /4 × 52 atoms = 7.8286 × 1024 atoms.
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.
Amount of carbon in 3.38 g CO2 = 12/44 × 3.38 g = 0.9218 g.
Amount of hydrogen in 0.690 g H2O = 2/18 × 0.690 g = 0.0767 g. As compound contains only C and H, therefore, total mass of the compound = 0.9218 + 0.0767 g = 0.9985 g
% of C in the compound = 0.9218/ 0.9985 × 100 = 92.32.
% of H in the compound = 0.0767/ 0.9985 × 100 = 7.68
Calculation of Empirical Formula
Moles of C = 92.32/12 = 7.69 and moles of H = 7.68/1 = 7.68. Simplest molar ratio of C and H = 7.69/7.68 : 7.68/7.68 = 1 : 1 hence, Empirical formula = CH.
10.0 L of the gas at STP weigh = 11.6 g. Hence, 22.4 L of the gas will weigh = 11.6/10.0 × 22.4 L = 26 g. ∴ Molar mass of the gas = 26 gmol-1 and empirical formula mass of CH = 12 + 1 = 13 g
∴ n = Molecular mass/empirical formula mass = 26/13 = 2, hence, Molecular formula = 2 × CH = C2H2
35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Ans:- Step 1. To calculate mass of HCl in 25 ml of 0.75 M HCl.
1000 ml of 0.75 M HCl contain HCl = 0.75 × 36.5 g = 24.375 g. ∴ 25 ml of 0.75 M HCl contain HCl = 24.375 / 1000 × 25 g = 0.6844 g
Step 2. To calculate mass of CaCO3 reacting completely with 0.6844 g of HCl CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2(g) + H2O (l)
According to the equation, 2 mol of HCl, i.e. , 2 × 36.5 g = 73 g HCl react completely with CaCO3 = 1 mol = 100 g ∴ 0.6844 g HCl will react completely with CaCO3 = 100/73 × 0.6844 g = 0.938 g
36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction
4HCl (aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Ans:- 1 mole of MnO2, i.e. , 55 + 32 = 87 g MnO2 react with 4 moles of HCl, i.e. , 4 × 36.5 g = 146 g of HCl. ∴ 5.0 g of MnO2 will react with HCl = 146/ 87 × 5.0 g = 8.40 g
Conclusion of Ncert Solutions of Chemistry Chapter 1
Ncert Solutions of Chemistry Chapter 1 is completed well. We have included all the questions which have been prescribed according to the ncert chemistry 11 book. We have used simple and step wise solution of each questions. Proper calculation has been done. There are 36 questions in this chapter. All the questions have been solved properly.
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