Some Basic Concepts of Chemistry
Based on NCERT Class 11th Chemistry Chapter 1
Exercise
Q(i) Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Solution :
The balanced equation for the combustion of methane is :
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
(i)16 g of CH4 corresponds to one mole.
(ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O (g).
2 mol of water (H2O) = 2 × (2+16) = 2 × 18 = 36 g
1 mol H2O = 18 g H2O ⇒ \(\frac{ 18 g\, H_2O}{1 mol \,H_2O}\) = 1
Hence, 2 mol H2O × \(\frac{ 18 g\, H_2O}{1 mol \,H_2O}\)
= 2 × 18 g H2O = 36 g H2O
Q(ii) How many moles of methane are required to produce 22g CO2 (g) after combustion?
Solution :
According to the chemical equation,
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
44g CO2 (g) is obtained from 16 g CH4 (g).
[∴ 1 mol CO2(g) is obtained from 1 mol of CH4(g)]
Number of moles of CO2 (g) = 22 g CO2 (g) × \(\frac{ 1 mol\, CO_2(g)}{44 g\, CO_2 (g)}\)
= 0.5 mol CO2 (g)
Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g)
or 0.5 mol of CH4 (g) would be required to produce 22 g CO2 (g).
Q(iii) 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the amount of NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.
Solution :
A balanced equation for the above reaction is written as follows :
N2 (g) + 3H2 (g) ⇔ 2NH3 (g)
Calculation of moles :
Number of moles of N2 = 50.00kg N2 × \(\frac{ 1000(g)N_2}{1(kg)N_2}\) × \(\frac{ 1\,mol\,N_2}{28.0(g)N_2}\)
= 17.86×102 mol
Number of moles of H2 = 10.00kg H2 × \(\frac{ 1000(g)H_2}{1(kg)H_2}\) × \(\frac{ 1\,mol\,H_2}{2.016(g)H_2}\)
= 4.96×103 mol
According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction.
Hence, for 17.86×102 mol of N2 , the moles of H2 (g) required would be
= 17.86×102 mol of N2 × \(\frac{ 3mole\, H_2 (g)}{1 mol\, N_2(g)}\)
5.36 ×103 mol H2
But we have only 4.96×103 mol H2.
Hence, dihydrogen is the limiting reagent in this case.
So, NH3(g) would be formed only from that amount of available dihydrogen
i.e., 4.96 × 103 mol
Since 3 mol H2(g) gives 2 mol NH3(g)
4.96×103 mol H2 (g) × \(\frac{ 2mole\, NH_3 (g)}{3 mol\, H_2(g)}\)
= 3.30×103 mol NH3 (g)
3.30×103 mol NH3 (g) is obtained.
If they are to be converted to grams, it is done as follows :
1 mol NH3 (g) = 17.0 g NH3(g)
3.30 × 103 mol NH3 (g) × \(\frac{ 17.0g\, NH_3 (g)}{1 mol\, NH_3(g)}\)
3.30×103×17 g NH3 (g)
= 56.1×103 g NH3
= 56.1 kg NH3
Q(iv) A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.
Solution :
Mass per cent of A = \(\frac{ Mass \, of \, A}{Mass \, of \, solution}\) × 100
= \(\frac{ 2g }{2g \, of \, A \, + 18 \, g \, of \, water}\) × 100
= \(\frac{ 2g }{20g}\) × 100
= 10%
Q(v) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Solution :
molarity (M) = \(\frac{No.\, of\, moles \, of\, the\, solute}{Volume\, of\, the\, solution\, in \, ltr}\)
= \(\frac{Mass \, of \, NaOH/Molar \, mass \, of \ NaOH}{0.250L}\)
= \(\frac{4g/40g}{0.250L}\) = \(\frac{0.1 \, mole}{0.250L}\)
= \(0.4mol^{-1}\)
= 0.4M
Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent
Q(vi) The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate the molality of the solution.
Solution:
M = 3 mol L–1
Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
( The formula, such as NaCl, is used to calculate the formula mass instead of molecular mass as in the solid state sodium chloride does not exist as a single entity. Thus, the formula mass of sodium chloride is atomic mass of sodium + atomic mass of chlorine = 23.0 u + 35.5 u = 58.5 u)
Mass of 1L solution = 1000 × 1.25 = 1250 g
(since density = 1.25 g mL–1)
Mass of water in solution = 1250 –175.5 = 1074.5 g
Molality = No. of moles of solute / Mass of solvent in kg
= 3 mol / .0745kg = 2.79 m
Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration.
The solution of higher concentration is also known as stock solution.
Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.
Q 1.1: Calculate the molecular mass of the following:
(i) H2O (ii) CO2 (iii) CH4
Answer: (i) H2O: (ii) CO2: (iii) CH4:
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
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
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
Q1.2: 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 = \(\frac{ mass \, of \, that \, element \, in \, the \, compund}{Molar \, mass \, of \, the \, compund}\) × 100 Mass percent of sodium: = \(\frac{46.0g}{142.066 g}\) × 100 = 32.379 = 32,4 % Mass percent of sulphur: = \(\frac{32.066g}{142.066 g}\) × 100 = 22.57 = 22.6% Mass percent of oxygen: = \(\frac{64.00g}{142.066 g}\) × 100 = 45.049 = 45.05%
Q1.3: 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] Relative moles of iron in iron oxide: = \(\frac{Percent \, of \, iron \, by \, mass}{Atomic \, mass \, of \, iron}\) = \(\frac{69.9}{55.85}\) = 1.25 Relative moles of oxygen in iron oxide: = \(\frac{Percent \, of \, oxygen \, by \, mass}{Atomic \, mass \, of \, oxygen}\) = \(\frac{30.1}{16}\) = 1.88 Simplest molar ratio of iron to oxygen:
% of oxygen by mass = 30.1 % [Given]
= 1.25: 1.88
= 1: 1.5
= 2: 3
∴ The empirical formula of the iron oxide is Fe2O3.
Q 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.
Answer: The balanced 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 air, combustion is complete. Therefore,C02 produced from the combustion of 1 mole of carbon = 44 g. (ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant. Hence,C02 produced = 22 g. (iii) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole of carbon.C02 produced again is equal to 22 g.
Q1.5: 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 means that 1000 mL of the solution contain sodium acetate = 0.375 mole ∴ Number of moles of sodium acetate in 500 mL = \(\frac{0.375}{1000}\) × 500 = 0.1875 mole Molar mass of sodium acetate = 82.0245 g mole–1 (Given) ∴ Required mass of sodium acetate = (82.0245 g mol–1)×(0.1875 mole)
= 15.38 g
Q1.6: 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) Number of moles in 69 g of HNO3 = \(\frac{69 g}{63 g mol^{-1}}\) = 1.095 mol Volume of 100g of nitric acid solution = \(\frac{Mass \, of \, solution}{Density \, of \, solution}\) = \(\frac{100 g}{1.41 g mol^{-1}}\) = 70.92 mL = 70.92 x 10-3 L Concentration of nitric acid = \(\frac{1.095 \, mole}{70.92 × 10^{-3} L}\) = 15.44 mol/L ∴ Concentration of nitric acid = 15.44 mol/L
= {1 + 14 + 3(16)} g mol–1
= 1 + 14 + 48
= 63 g mol–1
Q 1.7: 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) 159.5 g of CuSO4 contains 63.5 g of copper. ⇒ 100 g of CuSO4 will contain = \(\frac{63.5 × 100 \, g}{159.5}\) of copper ∴ Amount of copper that can be obtained from 100 g CuSO4 = \(\frac{63.5 × 100 \, g}{159.5}\) = 39.81 g
= 63.5 + 32.00 + 64.00
= 159.5 g
Q1.8: 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.
(Atomic mass: Fe = 55.85, O = 16.00 amu)Calculation of Empirical Formula. See Q3.
Answer: Empirical formula mass of Fe203 =( 2 x 55.85 + 3 x 16.00 )= 159.7 g mol–1 n =\( \frac{Molecular\, mass}{ Emperical \, formula \, mass}\) = \(\frac{159.8}{159.7}\) = 0.999 = 1 Hence, molecular formula is same as empirical formula, viz.,Fe203.
Q1.9.Calculate the atomic mass (average) of chlorine using the following data:
Answer: Fractional abundance of 35Cl = 0.7577 , Molar mass = 34.9689 Fractional abundance of 37Cl = 0.2423 , Molar mass = 36.9659 Average atomic mass = (0.7577)(34.9689) + (0.2423)(36.9659) = 26.4959 + 8.9568 = 35.4527
Q 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
Answer: (i) 1 mole of C2H6 contains 2 moles of carbon atoms. (ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms. (iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.
Number of moles of carbon atoms in 3 moles of C2H6
= 2 × 3 = 6
Number of moles of carbon atoms in 3 moles of C2H6
= 3 × 6 = 18
Number of molecules in 3 moles of C2H6
= 3 × 6.023 × 1023 = 18.069 × 1023
Q1.11. 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: Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol -1 No of moles in 20g of sugar = \(\frac{20\, g}{342\, g\,mol_{-1}}\) = 0.0585 mol Molar concentration = \(\frac{No\, of\, moles \, of\, the\, solute}{Volume\, of\, the\, solution\, in \, ltr}\) = \(\frac{0.0585 \, mol}{2L}\) = 0.02925 mol L–1 ∴ Molar concentration of sugar = 0.02925 mol L–1
Q1.12. 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) Molarity of methanol solution = \(\frac{0.032\, kg\, mol^{-1}}{0.793\,kg\,L^{-1}}\) = 24.78 mol L–1 (Solution to be prepared) (24.78 mol L–1) V1 = (2.5 L) (0.25 mol L–1)
= 32 g mol–1
= 0.032 kg mol–1
(Since density is mass per unit volume)
Applying,
M1V1 = M2V2 (Given solution)
V1= 0.0252 L
V1 = 25.22 mL
Q1.13: Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below: 1Pa = 1N m–2If 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= \(\frac{F}{A}\) = \(\frac{1034g × 9.8 ms^{-2}}{cm^2}\) = \(\frac{1034g × 9.8 \,ms^{-2}}{cm^2}\) × \(\frac{1\,kg}{1000\,gm}\) × \(\frac{(100)^2\,cm^2}{1\,m^3}\) = 1.01332 × 105 kg m–1 s–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
Pressure = 1.01332 × 105 Pa
Q1.14. What is the S.I. unit of mass?
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.
Q1.15. Match the following prefixes with their multiples:
| Sr | Prefixes | Multiples |
| 1 | micro | 106 |
| 2 | deca | 109 |
| 3 | mega | 10-6 |
| 4 | giga | 10-15 |
| 5 | femto | 10 |
Answer: micro = 10-6 , deca = 10, mega = 106 , giga = 109 femto = 10-15
Q1.16. What do you mean by significant figures?
Answer: The digits in a properly recorded measurement are known as significant figures. It is also defined as. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.
Q1.17. A sample of drinking water was found to be severely contaminated with chloroform, CHCly 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 = \(\frac{15}{10^6}\) × 100 =1.5 x 10-3 % (ii) 100 g of the sample contains 1.5 × 10–3 g of CHCl3. ∴ Molality of chloroform in water = \(\frac{1.5 × 10^{-2} }{Molar\,mass\,of\, CHCL_3}\) Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5) ∴ Molality of chloroform in water = \(\frac{1.5 × 10^{-2} }{119.5}\) = 0.0125 × 10–2 m = 1.25 × 10–4 m
⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.
= 119.5 g mol–1
Q1.18. Express the following in scientific notation:
(i) 0.0048 (v) 6.0012 (ii) 234,000 (iii) 8008 (iv) 500.0
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 × 100
Q1.19. 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.
Q1.20. 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
Q1.21. The following data were obtained when dinitrogen and dioxygen react together to form 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 1:2:1:5. 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. (i) 1 km = 1 km × \(\frac{1000\,m}{1\,km}\) × \(\frac{100\,cm}{1\,m}\) × \(\frac{10\,mm}{1\,cm}\) ∴ 1 km = 106 mm 1 km = 1 km × \(\frac{1000\,m}{1\,km}\) × \(\frac{1\,pm}{10^{-12}\,m}\) ∴ 1 km = 1015 pm (ii) 1 mg = 1 mg × \(\frac{1\,g}{1000\,mg}\) × \(\frac{1\,kg}{1000\,g}\) ∴ 1 mg = 10-6 kg 1 mg = 1 mg × \(\frac{1\,g}{1000\,mg}\) × \(\frac{1\,ng}{10^{-9}\,g}\) ∴ 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 × \(\frac{1 \,dm × 1\,dm ×1 \,dm }{10\,cm × 10\,cm × 10\,cm }\) cm3 1 ml = 10-3 dm3 ∴ 1 ml = 10-3 L = 10-3 dm3
Hence, 1 km = 106 mm = 1015 pm
Q1.22 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: Distance travelled by light in 2.00 ns = Speed of light × Time taken = (3.0 × 108 ms–1) (2.00 × 10–9 s)
Time taken to cover the distance = 2.00 ns = 2.00 × 10–9 s
Speed of light = 3.0 × 108 ms–1
= 6.00 × 10–1 m
= 0.600 m
Q1.23: 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.
Q1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + 3H2(g) —–> 2NH3 (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, 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. 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. 28 g of N2 produces 34 g of NH3. Hence, mass of ammonia produced by 2000 g of N2 = \(\frac{34g}{28g} × 2000\) = 2428.57 g (ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain un-reacted (iii) Mass of dihydrogen left unreacted = 1.00 ×103 g – 428.6 g = 571.4 g
N2(g) + 3H2(g) → 2NH3(g)
⇒ 2.00 × 103 g of dinitrogen will react with \(\frac{6g}{28g} × 2.00 × 10^3\) dihydrogen
Q1.25. How are 0.50 mol Na2C03 and 0.50 M Na2C03 different?
Answer: Molar mass of Na2C03= (2 x 23) +( 12) +( 3 x 16) = 106g mol-1 0.50 mol Na2C03 means 0.50 x 106 g = 53 g 0. 50 M Na2C03 means 0.50 mol, i.e., 53 g Na2C03 are present in 1 litre of the solution.
Q1.26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
Answer: H2 and 02 react according to the equation 2H2(g) + 02 (g) ——>2H2O (g) Thus, 2 volumes of H2 react with 1 volume of 02 to produce 2 volumes of water vapour. Hence, 10 volumes of H2 will react completely with 5 volumes of 02 to produce 10 volumes of water vapour.
Q1.27. Convert the following into basic units:
(i) 28.7 pm (ii) 15.15 µs (iii) 25365 mg
Answer: (i) 28.7 pm: (ii) 15.15 µs: (iii) 25365 mg:
1 pm = 10-12 m
∴ 28.7 pm = 28.7 × 10-12 m
= 2.87 × 10-11 m
1 µs = 10-6 m
∴ 15.15 µs = 15.15 × 10-6 s
= 1.515 × 10-5 s
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 × 101 × 10-3 kg
∴25365 mg = 2.5365 × 10-2 kg
Q1.28. 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) (Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)
Answer: (i) 1 g Au = \(\frac{1}{197}\) mol × 6.022 × 1023 atoms (ii) 1 g Na = \(\frac{1}{23}\) mol × 6.022 × 1023 atoms (iii) 1 g Li = \(\frac{1}{7}\) mol × 6.022 × 1023 atoms (iv) 1 g of Cl2(g) = \(\frac{1}{71}\) mol × 6.022 × 1023 atoms Hence, 1 g of Li (s) will have the largest number of atoms.
Q1.29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.
Answer: Mole fraction of C2H5OH = \(\frac{n(c_2h_5OH)}{Number \, of \, moles \, of \, solution}\) 0.040 = \(\frac{n(c_2h_5OH)}{n(c_2h_5OH)+n(H_2O)}\) ……..(i) Number of moles present in 1 L water: \(nH_2O\) = \(\frac{1000}{18\,g\,mol_{-1}}\) \(nH_2O\) = 55.55 mol Substituting the value of \(nH_2O\) in equation (i), \(\frac{n(C_2H_5OH)}{n(c_2H_5OH)+55.55}\) = 0.040 \(n(C_2H_5OH)\) = 0.040 \(n(C_2H_5OH)\) + 0.040(55.55) 0.96 \(n(C_2H_5OH)\) = 2.222 mol \(n(C_2H_5OH)\) = 2.314 mol ∴ Molarity of solution = 2.314 mol L-1 = 2.314 M
Q1.30: 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 = \(\frac{12\,g}{6.022× 10^{23}}\) = 1.993 × 10-23 g
Q1.31. How many significant figures should be present in the answer of the following?
(i) \(\frac{0.02856 × 298.15 × 0.112 }{0.5785}
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Answer: (i) The least precise term has 3 significant figures (i.e., 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.
Q1.32: Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Answer: Molar mass of Ar = 35.96755 x 0.00337 + 37.96272 x 0.00063 + 39.96924 x 0.99600 = 39.948 g mol-1
Q1.33: 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 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 4 u of He = 1 atom of He 1 u of He =\(\frac{1}{4}\) atom of He 52u of He = \(\frac{52}{4}\) atom of He = 13 atoms of He (iii) 4 g of He = 6.022 × 1023 atoms of He 52 g of He = \(\frac{6.022 × 10^{23} }{4}\) atoms of He = 7.8286 × 1024 atoms of He
Q1.34. 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 S.T.P.) 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. ∴ 3.38 g of CO2 will contain carbon = \(\frac{12\,g}{44\,g}\) × 3.38 g = 0.9217 g 18 g of water contains 2 g of hydrogen. ∴ 0.690 g of water will contain hydrogen = \(\frac{2\,g}{18\,g}\) × 0.690 g = 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 ∴ Percent of C in the compound = \(\frac{0.9217\,g}{0.9984\,g}\) × 100 = 92.32 % Percent of H in the compound = \(\frac{0.0767\,g}{0.9984\,g}\) × 100 = 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 ∴ 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 Weight of 22.4 L of gas at STP = \(\frac{11.6 \,g}{10.0\,L}\) × 22.4 L = 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= \(\frac{Molar\, mass \,of \,gas}{Empirical\, formula \, mass\, of \, gas}\) = \(\frac{26\,g}{13\,g}\) n = 2 Molecular formula of gas = (CH)n = C2H2
Q1.35. Calcium carbonate reacts with aqueous HCl according to the reaction
CaC03 (s) + 2HCl (aq) ———->CaCl2 (aq) +C02(g) +H2O(l).
What mass of CaC03 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. Amount of HCl present in 25 mL of solution. = \(\frac{27.375\,g}{1000\,mL}\) × 25 mL = 0.6844 g From the given chemical equation, CaC03 (s) + 2HCl (aq) ———->CaCl2 (aq) +C02(g) +H2O(l). 2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g). Amount of CaCO3 that will react with 0.6844 g = \(\frac{100}{71}\) × 0.6844 g = 0.9639 g
Q1.36. Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction.
4 HCl (aq) + Mn02 (s) ———–> 2 H2O (l) + MnCl2(aq) +Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u)
Answer: 1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl. \(\frac{146\,g}{87\,g}\) × 5.0 g HCl = 8.4 g of HCl Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.
5.0 g of MnO2 will react with