01 Some Basic Concepts of Chemistry
Exercise1
Question:
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 Cl_{2} (g)
Answer:
Step 1: Determine the atomic mass of each element. Au = 197 g/mol, Na = 23 g/mol, Li = 6.94 g/mol, Cl_{2} = 70.9 g/mol.
Step 2: Determine the number of moles of each element. 1 g Au (s) = 0.005 moles, 1 g Na (s) = 0.043 moles, 1 g Li (s) = 0.144 moles, 1 g of Cl_{2} (g) = 0.014 moles.
Step 3: Determine the number of atoms of each element. 1 g Au (s) = 0.005 moles x 6.02x10^{23} atoms/mole = 3.01x10^22 atoms, 1 g Na (s) = 0.043 moles x 6.02x10^{23} atoms/mole = 2.59x10^{23} atoms, 1 g Li (s) = 0.144 moles x 6.02x10^{23} atoms/mole = 8.67x10^{22} atoms, 1 g of Cl_{2} (g) = 0.014 moles x 6.02x10^{23} atoms/mole = 8.43x10^{22} atoms.
Step 4: Compare the number of atoms of each element. The element with the largest number of atoms is Na (s) with 2.59x10^{23} atoms.
Question:
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) Empirical Formula:
 Calculate the moles of carbon dioxide (CO_{2}) and water (H_{2}O) produced from the burning of the welding fuel gas:
Moles of CO2 = 3.38 g/44.01 g/mol = 0.0765 mol Moles of H2O = 0.690 g/18.02 g/mol = 0.0383 mol
 Calculate the moles of carbon (C) and hydrogen (H) present in the welding fuel gas:
Moles of C = 0.0765 mol/2 = 0.0383 mol Moles of H = 0.0383 mol/2 = 0.0192 mol
 Calculate the empirical formula of the welding fuel gas:
Empirical Formula = CH_{2}
(ii) Molar Mass of the Gas:
 Calculate the molar mass of the welding fuel gas using the empirical formula:
Molar Mass = (1 mol C x 12.01 g/mol) + (2 mol H x 1.01 g/mol) = 14.04 g/mol
(iii) Molecular Formula:
 Calculate the moles of the welding fuel gas present in 10.0 L (measured at STP):
Moles of Gas = 11.6 g/14.04 g/mol = 0.827 mol
 Calculate the molar mass of the molecular formula using the moles of the welding fuel gas:
Molar Mass = 0.827 mol x 14.04 g/mol = 11.6 g/mol
 Calculate the molecular formula of the welding fuel gas:
Molecular Formula = CH_{2}
Question:
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:
Step 1: Calculate the mass of ethanol in the solution.
Mass of ethanol = 0.040 x density of water x volume of solution
Step 2: Calculate the moles of ethanol in the solution.
Moles of ethanol = mass of ethanol / molecular weight of ethanol
Step 3: Calculate the molarity of the solution.
Molarity = moles of ethanol / volume of solution
Question:
In a reaction A+B_{2}→AB_{2}, identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B_{2} (ii) 2 mol A + 3 mol B_{2} (iii) 100 atoms of A + 100 molecules of B_{2} (iv) 5 mol A + 2.5 mol B_{2} (v) 2.5 mol A + 5 mol B_{2}
Answer:
(i) 300 atoms of A + 200 molecules of B_{2}
There are 300 atoms of A and 200 molecules of B_{2}, which is equivalent to 600 atoms of B_{2}. Since there are more atoms of B2 than A, B_{2} is the limiting reagent.
(ii) 2 mol A + 3 mol B2
There are 2 moles of A and 3 moles of B_{2}. Since there are more moles of B_{2} than A, B_{2} is the limiting reagent.
(iii) 100 atoms of A + 100 molecules of B_{2}
There are 100 atoms of A and 100 molecules of B_{2}, which is equivalent to 200 atoms of B_{2}. Since there are equal amounts of A and B_{2}, neither is the limiting reagent.
(iv) 5 mol A + 2.5 mol B_{2}
There are 5 moles of A and 2.5 moles of B_{2}. Since there are more moles of A than B_{2}, B_{2} is the limiting reagent.
(v) 2.5 mol A + 5 mol B_{2}
There are 2.5 moles of A and 5 moles of B_{2}. Since there are more moles of B_{2} than A, B_{2} is the limiting reagent.
Question:
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) 0.046 (iv) 2810
Question:
How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,00 (v) 500.0 (vi) 2.0034
Answer:
(i) 3 significant figures (ii) 3 significant figures (iii) 4 significant figures (iv) 5 significant figures (v) 4 significant figures (vi) 5 significant figures
Question:
How much copper can be obtained from 100 g of copper sulphate (CuSO_{4})?
Answer:
Step 1: Determine the molecular weight of copper sulphate (CuSO_{4}).
The molecular weight of copper sulphate (CuSO_{4}) is 159.609 g/mol.
Step 2: Calculate the amount of copper present in 100 g of copper sulphate (CuSO_{4}).
The amount of copper present in 100 g of copper sulphate (CuSO_{4}) is 100 g x (1 mol/159.609 g) = 0.6265 mol.
Step 3: Calculate the amount of copper obtained from 100 g of copper sulphate (CuSO_{4}).
The amount of copper obtained from 100 g of copper sulphate (CuSO_{4}) is 0.6265 mol x (63.546 g/mol) = 39.87 g.
Question:
Calculate the molar mass of CO_{2}.
Answer:
Step 1: Identify the elements in CO_{2}. Answer: Carbon and Oxygen
Step 2: Look up the atomic masses of the elements. Answer: Carbon = 12.01 g/mol; Oxygen = 16.00 g/mol
Step 3: Calculate the molar mass of CO_{2}. Answer: Molar mass of CO_{2} = 12.01 g/mol + (2 x 16.00 g/mol) = 44.01 g/mol
Question:
Calculate the mass percent of O present in sodium sulphate (Na_{2}SO_{4}).
Answer:
Step 1: Calculate the molar mass of sodium sulphate. Molar mass of Na2SO4 = (2 x 23) + (32 + 4 x 16) = 142 g/mol
Step 2: Calculate the mass of oxygen present in sodium sulphate. Mass of oxygen = (4 x 16) = 64 g
Step 3: Calculate the mass percent of oxygen present in sodium sulphate. Mass percent of oxygen = (64/142) x 100 = 45%
Question:
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
Answer:
(i) 5 significant figures; (ii) 4 significant figures; (iii) 4 significant figures.
Question:
If the density of methanol is 0.793kgL^{1}, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:

Calculate the mass of 2.5 L of 0.25 M methanol solution: Molarity (M) = moles of solute/liters of solution Moles of solute = Molarity x Liters of solution Moles of solute = 0.25 M x 2.5 L = 0.625 moles Mass of solute = moles x molar mass of solute Molar mass of methanol = 32 g/mol Mass of methanol = 0.625 moles x 32 g/mol = 20 g

Calculate the volume of methanol needed to make 2.5 L of 0.25 M methanol solution: Density (ρ) = mass/volume Volume = mass/density Volume = 20 g/0.793 kg/L = 25.2 L
Question:
Convert the following into SI unit: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg
Answer:
(i) 28.7 pm = 2.87 x 10^{10} m
(ii) 15.15 pm = 1.515 x 10^{10} m
(iii) 25365 mg = 2.5365 x 10^{3} kg
Question:
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:
(i) When 1 mole of carbon is burnt in air, it produces 1 mole of carbon dioxide.
(ii) When 1 mole of carbon is burnt in 16 g of dioxygen, it produces 1 mole of carbon dioxide.
(iii) When 2 moles of carbon are burnt in 16 g of dioxygen, it produces 2 moles of carbon dioxide.
Question:
Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4
Answer:
(i) Molar mass of H_{2}O = 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
(ii) Molar mass of CO_{2} = 12.01 g/mol + 2(15.99 g/mol) = 44.01 g/mol
(iii) Molar mass of CH_{4} = 1.008 g/mol + 4(12.01 g/mol) = 16.04 g/mol
Question:
If 3 moles of ethane (C_{2}H_{6}) are given, 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) Number of moles of carbon atoms: 3 moles of ethane contains 6 moles of carbon atoms. Answer: 6 moles of carbon atoms.
(ii) Number of moles of hydrogen atoms: 3 moles of ethane contains 18 moles of hydrogen atoms. Answer: 18 moles of hydrogen atoms.
(iii) Number of molecules of ethane: 3 moles of ethane contains 3 molecules of ethane. Answer: 3 molecules of ethane.
Question:
Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012
Answer:
(i) 4.8 x 10^{{3}} (ii) 2.34 x 10^{{5}} (iii) 8.008 x 10^{{3}} (iv) 5.00 x 10^{{2}} (v) 6.0012 x 10^{{0}}
Question:
Calculate the atomic mass (average) of chlorine using the following data : % Natural Abundance ^{35}Cl75.77, Molar Mass 34.9689 and Abundance ^{37}Cl24.23, Molar Mass 36.9659
Answer:
Step 1: Calculate the total mass of chlorine atoms with 35Cl isotope.
Total mass of 35Cl = 75.77% x 34.9689 = 26.48 g
Step 2: Calculate the total mass of chlorine atoms with 37Cl isotope.
Total mass of 37Cl = 24.23% x 36.9659 = 9.05 g
Step 3: Calculate the total mass of chlorine atoms.
Total mass of chlorine atoms = 26.48 g + 9.05 g = 35.53 g
Step 4: Calculate the atomic mass (average) of chlorine.
Atomic mass (average) of chlorine = 35.53 g/2 = 17.765 g
Question:
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.
Answer:
Step 1: Calculate the molar mass of the oxide. Molar mass = (69.9 g/mol x 55.845 g/mol) + (30.1 g/mol x 16.00 g/mol) = 159.7 g/mol
Step 2: Calculate the empirical formula of the oxide. Empirical formula = (69.9/159.7) Fe + (30.1/159.7) O = 0.437 Fe + 0.188 O
Step 3: Determine the molecular formula of the oxide. Molecular formula = (0.437 Fe + 0.188 O) x 2 = 0.874 Fe + 0.376 O = Fe_{2}O_{3}
Question:
Calculate the mass of sodium acetate (CH_{3}COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^{1}.
Answer:
Step 1: Calculate the number of moles of sodium acetate required to make 500 ml of 0.375 molar aqueous solution.
Number of moles = 0.375 moles/L × 500 ml = 0.188 moles
Step 2: Calculate the mass of sodium acetate required to make 500 ml of 0.375 molar aqueous solution.
Mass of sodium acetate = 0.188 moles × 82.0245 g/mol = 15.35 g
Question:
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) 15 ppm (by mass) = 0.0015% (by mass)
(ii) Molality = (mass of solute (g))/(mass of solvent (kg))
Molality = (15 ppm (by mass) × 0.001 g/1 ppm) / (1 kg)
Molality = 0.0000015 kg/kg = 1.5 x 10^{5} mol/kg
Question:
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 2.00×10^{3} g dinitrogen reacts with 1.00×10^{3} 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) The mass of ammonia produced can be calculated using the following equation:
Mass of ammonia = 2 x (mass of dinitrogen) x (mass of dihydrogen)
= 2 x (2.00 x 10^{3} g) x (1.00 x 10^{3} g)
= 4.00 x 10^{6} g
(ii) No, none of the two reactants will remain unreacted as the equation is a balanced equation.
(iii) No, none of the reactants will remain unreacted.
Question:
What is the SI unit of mass? How is it defined?
Answer:
Answer: The SI unit of mass is the kilogram (kg). It is defined as the mass of the international prototype of the kilogram, which is a cylinder made of a platinumiridium alloy stored in a vault at the International Bureau of Weights and Measures in France.
Question:
Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41gmL^{1} and the mass per cent of nitric acid in it being 69%.
Answer:
Step 1: Calculate the mass of nitric acid in the sample. Mass of nitric acid = (69/100) x (1.41 g/mL)
Step 2: Calculate the moles of nitric acid in the sample. Moles of nitric acid = (Mass of nitric acid)/(Molar mass of nitric acid)
Step 3: Calculate the concentration of nitric acid in moles per litre. Concentration of nitric acid = (Moles of nitric acid)/(Volume of sample in litres)
Question:
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:
 Calculate the molar mass of the oxide of iron: Molar mass of iron = 55.85 g/mol Molar mass of dioxygen = 32.00 g/mol
Molar mass of oxide of iron = 55.85 g/mol + 32.00 g/mol = 87.85 g/mol

Calculate the mass of each element in the oxide of iron: Mass of iron = 69.9% x 87.85 g/mol = 61.44 g/mol Mass of dioxygen = 30.1% x 87.85 g/mol = 26.41 g/mol

Calculate the moles of each element in the oxide of iron: Moles of iron = 61.44 g/mol ÷ 55.85 g/mol = 1.10 moles Moles of dioxygen = 26.41 g/mol ÷ 32.00 g/mol = 0.82 moles

Calculate the simplest whole number ratio of the elements in the oxide of iron: Ratio of iron to dioxygen = 1.10 moles ÷ 0.82 moles = 1.34

Determine the empirical formula of the oxide of iron: Empirical formula = Fe1.34O0.82
Question:
What is the concentration of sugar (C_{12}H_{22}O_{11}) in molL_{1} if its 20 g are dissolved in enough water to make a final volume up to 2L?
Answer:
Step 1: Calculate the molar mass of sugar (C_{12}H_{22}O_{11}). Molar mass of sugar = 12(12.01 g/mol) + 22(1.008 g/mol) + 11(15.999 g/mol) = 342.29 g/mol
Step 2: Calculate the number of moles of sugar. Number of moles of sugar = 20 g/ 342.29 g/mol = 0.0584 mol
Step 3: Calculate the concentration of sugar in molL^{1}. Concentration of sugar = 0.0584 mol/2L = 0.0292 molL^{1}
Question:
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa=1Nm^{2}
If mass of air at sea level is 1034 gcm^{2}, calculate the pressure in pascal.
Answer:
Step 1: Convert the mass of air to Newton. 1 gcm^{2} = 9806.65 Nm^{2}
Therefore, 1034 gcm^{2} = 1034 x 9806.65 Nm^{2} = 1013907.1 Nm^{2}
Step 2: Calculate the pressure in pascal.
Pressure = Force/Area
Therefore, Pressure = 1013907.1 Nm^{2}/1m^{2}
Pressure = 1013907.1 Pa
Question:
What will be the mass of one 12C atom in g ?
Answer:
Step 1: Determine the atomic mass of 12C The atomic mass of 12C is 12.
Step 2: Convert the atomic mass to grams 1 atomic mass unit (amu) is equal to 1.66 x 10^{24} g.
Therefore, 12 amu is equal to 19.92 x 10^{24} g.
Step 3: Calculate the mass of one 12C atom The mass of one 12C atom is 19.92 x 10^{24} g.
Question:
What do you mean by significant figures?
Answer:
Significant figures refer to the number of digits in a value that are known with some degree of certainty. Significant figures are used to provide an indication of the precision of a measurement or calculation. They are also used to express the accuracy of a measurement or calculation. For example, a value of 1.2345 has five significant figures, while a value of 1.2 has three significant figures.
Question:
If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

Dihydrogen gas (H_{2}) + Dioxygen gas (O_{2}) → Water vapour (H_{2}O)

10 volumes of H_{2} + 5 volumes of O_{2} → x volumes of H_{2}O

10 volumes of H_{2} + 5 volumes of O_{2} → 10 volumes of H_{2}O
Question:
How are 0.50 m Na_{2}CO_{3} and 0.50 M Na_{2}CO_{3} different?
Answer:

0.50 m Na_{2}CO_{3} is a concentration expressed in molarity, which is a measure of the number of moles of solute per liter of solution.

0.50 M Na_{2}CO_{3} is a concentration expressed in molality, which is a measure of the number of moles of solute per kilogram of solvent.
Question:
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) 52 moles of Ar Answer: 3.2 x 10^{24} atoms
(ii) 52 u of He Answer: 3.2 x 10^{22} atoms
(iii) 52 g of He Answer: 8.4 x 10^{23} atoms
Question:
Calculate the molar mass of CH4.
Answer:
Step 1: Identify the elements present in CH4. Answer: Carbon (C) and Hydrogen (H)
Step 2: Find the atomic mass of each element. Answer: Carbon (C): 12.0107 g/mol; Hydrogen (H): 1.00794 g/mol
Step 3: Multiply the atomic mass of each element by the number of atoms present in CH4. Answer: Carbon (C): 12.0107 g/mol x 1 atom = 12.0107 g/mol; Hydrogen (H): 1.00794 g/mol x 4 atoms = 4.03176 g/mol
Step 4: Add the two results together to calculate the molar mass of CH4. Answer: 12.0107 g/mol + 4.03176 g/mol = 16.04246 g/mol
Question:
If the speed of light is 3.0×108ms^{1}, calculate the distance covered by light in 2.00 ns.
Answer:
Step 1: Convert 2.00 ns to seconds.
1 ns = 10^{9} s
2.00 ns = 2.00 x 10^{9} s
Step 2: Calculate the distance covered by light.
Distance = Speed x Time
Distance = 3.0 x 10^{8} m/s x 2.00 x 10^{9} s
Distance = 6.00 x 10^{1} m
Question:
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction: 4HCl(aq)+MnO_{2}(s)→2H_{2}O(I)+MnCl_{2}(aq)+Cl_{2}(g). How many grams of HCl react with 5.0 g of manganese dioxide ?
Answer:
 Calculate the moles of manganese dioxide (MnO2): MnO_{2} = 5.0 g Molar Mass of MnO_{2} = 86.94 g/mol
Moles of MnO_{2} = 5.0 g / 86.94 g/mol = 0.0575 mol
 Calculate the moles of hydrochloric acid (HCl): The equation shows that 4 moles of HCl are needed for every 1 mole of MnO2.
Moles of HCl = 4 x 0.0575 mol = 0.23 mol
 Calculate the grams of hydrochloric acid (HCl): Molar Mass of HCl = 36.5 g/mol
Grams of HCl = 0.23 mol x 36.5 g/mol = 8.395 g
Question:
Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction,
CaCO_{3}(s)+2HCl(aq)→CaCl_{2}(aq)+CO_{2}(g)+H_{2}O(l).
What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl ?
Answer:

Calculate the moles of HCl: Moles of HCl = 0.75 M * 0.025 L = 0.01875 moles

Calculate the moles of CaCO3 required: Moles of CaCO_{3} = 2 x 0.01875 moles = 0.0375 moles

Calculate the mass of CaCO_{3} required: Mass of CaCO_{3} = 0.0375 moles x 100.09 g/mol = 3.75 g
01 Some Basic Concepts of Chemistry
02 Structure of Atom
03 Classification of Elements and Periodicity in Properties
04 Chemical Bonding and Molecular Structure
05 States of Matter Gases and Liquids
06 Thermodynamics
07 Equilibrium
08 Redox Reactions
09 Hydrogen
10 The s block elements
11 Some p block elements
12 Organic Chemistry some basic principles and techniques
13 Hydrocarbons
14 Environmental