In this post, you will get the Some basic concepts of chemistry NCERT solutions in addition to other question answer as well.
1. Calculate the molecular mass of the following: (i) H2O (ii) CO2 (iii) CH4
Answer: (i) H2O:
The molecular mass of water = (2 x Atomic mass of hydrogen) + (1 x 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 = (1 x Atomic mass of carbon) + (2 x 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 = (1 x Atomic mass of carbon) + (4 x 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.
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 = [Mass of that element in the compound/Molar mass of the compound] * 100
Now, mass percentage of sodium = [46.0 g/142.066 g]*100 = 32.379 % = 32.4 %
Mass percentage of sulphur = [32.066 g/142.066 g]*100 = 22.57 % = 22.6 %
Mass percentage of oxygen = [64.0 g/142.066 g]*100 = 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 = % Iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = % Oxygen by mass/Atomic mass of oxygen = 30.1/16 = 1.88
Now, the simplest molar ratio, Fe : O = 1.25 : 1.88 = 1 : 1.5 = 2 : 3
So, the empirical formula of 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:
C + O2 → CO2
(i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce 1 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 indicates that 1000 mL of solution contains 0.375 moles of sodium acetate.
So, Number of moles of sodium acetate in 500 mL = (0.375/1000) * 500 = 0.1875 mol
Given that the molar mass of sodium acetate is 82.0245 g/mol.
Mass of sodium acetate = No. of moles * Molar mass = 0.1875 mol * 82.0245 g/mol = 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) = 63 g/mol
So, Number of moles in 69 g of HNO3 = Given mass/Molar mass = 69 g/63 g/mol = 1.095 mol
Volume of 100 g of nitric acid solution = Mass of solution/Density of solution = 100 g/1.41 g/ml = 70.92 ml = 70.92 * 10-3 L
Concentration of nitric acid = Moles of nitric acid/Volume of solution = 1.095 mol/70.92 * 10-3 L = 15.44 M
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 = 159.5 g/mol
159.5 g of CuSO4 contains 63.5 g of copper.
100 g of CuSO4 will contain of copper = 39.81 g
Hence, the amount of copper that can be obtained from 100 g CuSO4 is 39.81 g.
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%
Mass percent of oxygen (O) = 30.1%
Number of moles of iron in iron oxide = % Iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Number of moles of oxygen in iron oxide = % Oxygen by mass/Atomic mass of oxygen = 30.1/16 = 1.88
Now, the simplest molar ratio, Fe : O = 1.25 : 1.88 = 1 : 1.5 = 2 : 3
So, the empirical formula of iron oxide is Fe2O3.
Empirical mass of Fe2O3 = 159.7 g/mol
Molar mass of Fe2O3 = 159.69 g/mol
So, n = Molar mass/Empirical mass = 159.69/159.7 = 1
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:
| Isotope of chlorine | % natural abundance | Molar mass |
|---|---|---|
| Cl-35 | 75.77 | 34.9689 |
| Cl-37 | 24.23 | 36.9659 |
Answer: The average atomic mass of chlorine is:
Hence, the average atomic mass of chlorine is 35.5 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.
Number of moles of carbon atoms in 3 moles of C2H6 = 2 × 3 = 6 moles.
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.
Number of moles of carbon atoms in 3 moles of C2H6 = 3 × 6 = 18 moles.
(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.
Number of molecules in 3 moles of C2H6 = 3 × 6.023 × 1023 = 18.069 × 1023 molecules.
11. What is the concentration of sugar (C12H22O11) in mol/L, if its 20 g are dissolved in enough water to make a final volume upto 2 L?
Answer: Mass of sugar = 20 g
Molar mass of sugar = (12*12)+(22*1)+(11*16) = 342 g/mol
Volume of solution = 2 L
Moles of sugar = Mass of sugar/Molar mass of sugar = 20 g/342 g/mol = 0.0585 mol
Molarity = Moles of sugar/Volume of solution = 0.0585 mol/2 L = 0.02925 mol/L
12. If the density of methanol is 0.793 kg/L, 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
Moles of methanol = Molarity * Volume(in L) = 0.25 M * 2.5 L = 0.625 mol
Mass of methanol = Moles * Molar mass = 0.625 mol * 32 g/mol = 20 g = 0.02 Kg
Volume required = Mass/Density = 0.02/0.793 = 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
as: 1 Pa = 1 Nm-2
If mass of air at sea level is 1034 gcm-2, calculate the pressure in Pascal.
Answer : Pressure is defined as force acting per unit area of the surface.
P = 1.01332 × 105 kgm-1s-2
We know,
1 Pa = 1 kgm-1s-2
So, 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.
15. Match the following prefixes with their multiples:
| S No. | Prefixes | Multiples |
|---|---|---|
| (i) | micro | 106 |
| (ii) | deca | 109 |
| (iii) | mega | 10-6 |
| (iv) | giga | 10-15 |
| (v) | femto | 10 |
Answer: The correct matches is as:
| S No. | Prefixes | Multiples |
|---|---|---|
| (i) | micro | 10-6 |
| (ii) | deca | 1010 |
| (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 = 15/106 *100 = 1.5*10-3 %
(ii) 100 g of the sample contains 1.5 × 10-3g of CHCl3.
So, 1000 g or 1 Kg of the sample contains 1.5 × 10-2g of CHCl3.
Molar mass of CHCl3 = 119.5 g/mol
Mass of solvent (sample) = 1 Kg
Molality of chloroform in water = 1.5 × 10-2/119.5*1 = 1.25 * 10-4 mol/Kg
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 x 10-3
(ii) 234, 000 = 2.34 x 105
(iii) 8008 = 8.008 x 103
(iv) 500.0 = 5.000 x 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) There are 2 significant figures.
(ii) There are 3 significant figures.
(iii) There are 4 significant figures.
(iv) There are 3 significant figures.
(v) There are 4 significant figures.
(vi) 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:
| S No. | 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
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 2:4:2:5.
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.
(b)
22. If the speed of light is 3.0 x 108 ms-1, calculate the distance covered by light in 2.00 ns.
Answer: Time taken to cover the distance = 2.00 ns = 2.00 x 10-9 s
Speed of light = 3.0 x 108 ms-1
Distance travelled by light in 2.00 ns = Speed of light x Time taken = (3.0 x 108 ms-1)*(2.00 x 10-9 s) = 0.600 m = 6.00 x 10-1 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 mol of A reacts with 1 mol of B. Thus, 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) 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 mol of atom A combines with 1 mol of molecule B. Thus, 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) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 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.
24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2(g)+ H2(g) → NH3(g)
(i) Calculate the mass of ammonia produced if 2.00 x 103g dinitrogen reacts with 1.00 x 103g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Answer: Balancing the given chemical equation: N2(g)+ 3H2(g) → 2NH3(g)
(i) 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 dihydrogen = (6 g/28 g) * (2.00 × 103 g) = 428.6 g
Given amount of dihydrogen is 1.00 × 103 g, which is in excess as compared to dinitrogen.
Hence, N2 is the limiting reagent.
From the reaction, 28 g of N2 produces 34 g of NH3
Mass of ammonia produced by 2000 g of N2 = (34 g/28 g)*(2000 g) = 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 mole of Na2CO3 means 106 g of Na2CO3. So, 0.5 mol of Na2CO3 = 53 g Na2CO3
While 0.50 M of Na2CO3 = 0.50 mol/L Na2CO3. Hence, 0.50 M Na2CO3 means that 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: 2H2 + O2 → 2H2O
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:
Since, 1 pm = 10-12 m
So, 28.7 pm = 28.7 × 10-12 m = 2.87 × 10-11 m
(ii) 15.15 pm:
Since, 1 pm = 10-12 m
So, 15.15 pm = 15.15 × 10-12 m = 1.515 × 10-11 m
(iii) 25365 mg:
Since, 1 mg = 10-3 g
25365 mg = 2.5365 × 104 × 10-3 g = 2.5365 × 101 g
Also, 1 g = 10-3 Kg
So, 2.5365 × 101 g = 2.5365 × 101 × 10-3 kg = 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)
Answer: NA is Avogadro’s number whose value is 6.022 x 1023 particles (atoms/molecules/ions).
(i) 1 g Au = 1/197 mol = NA/197 atoms = 3.06 × 1021 atoms
(ii) 1 g Na = 1/23 mol = NA/23 atoms = 26.2 × 1021 atoms
(iii) 1 g Li = 1/7 mol = NA/7 atoms = 86.0 × 1021 atoms
(iv) 1 g Cl2 = 1/71 mol = NA/71 molecules = NA/35.5 atoms = 16.96 × 1021 atoms
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 is to be 1 g/cm3.
Answer: Assuming that the volume of water is 1 L.
Since, the density of water is 1 g/cm3, so, mass of water = 1000 g
Molar mass of water = 18 g/mol
Number of moles of water = Mass of water/Molar mass of water = 1000/18 = 55.55 mol
Molarity of ethanol solution = Moles of ethanol/Volume of solution = 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 = 12/6.023 x 1023 = 1.993 × 10-23 g
31. How many significant figures should be present in the answer of the following calculations?
(i) (0.02856 x 298.15 x 0.112)/0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Answer: (i) Least precise number of calculation = 0.112
Number of significant figures in the answer = Number of significant figures in the least precise number = 3
(ii) Least precise number of calculation = 5.364
Number of significant figures in the answer = Number of significant figures in 5.364 = 4
(iii) 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 | Molar mass (g/mol) | % Abundance |
|---|---|---|
| Ar-36 | 35.96755 | 0.337 |
| Ar-38 | 37.96272 | 0.063 |
| Ar-40 | 39.9624 | 99.600 |
Answer: The average molar mass of argon is:
Hence, the average molar mass of Argon is 40 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
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
So, 52u of He = (52/4) atom of He = 13 atoms of He
(iii) 4 g of He = 6.022 × 1023 atoms of He
52 g of He = [(6.022 × 1023 x 52)/4] atoms of He = 7.8286 × 1024 atoms of He.
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.
3.38 g of CO2 will contain carbon = (12/44)*3.38 = 0.9217 g
18 g of water contains 2 g of hydrogen.
0.690 g of water will contain hydrogen = (2/18)*0.690 = 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, so total mass is of carbon and hydrogen only.
Mass of the compound = 0.9217 g + 0.0767 g = 0.9984 g
Percentage of carbon in the compound = (0.9217/0.9984)*100 = 92.32 %
Percentage of hydrogen in the compound = (0.0767/0.9984)*100 = 7.68 %
Now, the empirical formula of the compound is:
| Element | %age | Atomic mass | Mole | Mole ratio | Simplest whole no. ratio |
|---|---|---|---|---|---|
| Carbon | 92.32 | 12 | 92.32/12 = 7.69 | 7.69/7.68 = 1.001 | 1 |
| Hydrogen | 7.68 | 1 | 7.68/1 = 7.68 | 7.68/7.68 = 1 | 1 |
So, the empirical formula of the compound is CH.
(ii) Weight of 10.0 L of the gas (at STP) = 11.6 g
We know that, 1 mole of gas = Molar mass of gas = 22.4 L at STP
So, weight of 22.4 L of gas = (11.6*22.4)/10 = 25.98 g = 26 g
Hence, the molar mass of the gas is 26 g/mol.
(iii) Empirical formula mass = 12+1 = 13 g
Molecular mass = 26 g
n = Molecular mass/Empirical formula mass = 26/13 = 2
Molecular formula = (Empirical formula)n = (CH)2 = C2H2
Hence, the molecular formula of welding fuel gas is 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?
Answer: Moles of HCl = Molarity x Volume(L) = 0.75 M x 0.025 L = 0.01875 mol
From the balanced chemical equation, 2 mol HCl reacts with 1 mol CaCO3
So, 0.01875 mol HCl reacts with CaCO3 = 0.01875/2 = 0.009375 mol
Molar mass of CaCO3 = 40+12+3(16) = 100 g/mol
Mass of CaCO3 = Mole*Molar mass = 0.009375 mol * 100 g/mol = 0.9375 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?
Answer: From the balanced chemical equation,
1 mol MnO2 reacts completely with 4 mol of HCl, i.e., 87 g of MnO2 reacts with 146 g of HCl.
So, 5.0 g of MnO2 will react with HCl = (146/87)*5 = 8.4 g
Hence, 8.4 g of HCl will react completely with 5.0 g of MnO2.
37. What is the SI unit of molarity?
Answer: The SI unit of molarity is mol dm-3 or mol L-1, and it is represented as M.
38. What do you mean by stoichiometric coefficients in a chemical equation?
Answer: The coefficients of reactant and product involved in a chemical equation represented by the balanced form, are known as stoichiometric coefficients, e.g., 2H2 + O2 → 2H2O
The stoichiometric coefficients are 2, 1, 2 respectively.
39. Give an example of a molecule in which the ratio of the molecular formula is six times the empirical formula.
Answer: The compound is glucose. Its molecular formula is C6H12O6, while its empirical formula is CH2O.
40. What is an atom according to Dalton’s atomic theory?
Answer: According to Dalton’s atomic theory, an atom is the ultimate particle of matter which cannot be further divided.
41. Why air is not always regarded as homogeneous mixture?
Answer: This is due to the presence of dust particles.
42. Define the term unit of measurement.
Answer: It is defined is the standard of reference chosen to measure a physical quantity.
43. Define the law of conservation of mass.
Answer: It states that matter can neither be created nor be destroyed.
44. How is empirical formula of a compound is related to its molecular formula?
Answer: Molecular formula = (Empirical formula)n, where n is positive integer.
n is the ratio of molecular mass to the empirical formula mass of that compound.
45. Name two factors that introduce uncertainty into measured figures.
Answer: (i) Reliability of measuring instrument.
(ii) Skill of the person making the measurement.
46. State Avogadro’s law.
Answer: Equal volumes of all gases under the conditions of same temperature and pressure contain the same number of molecules.
47. How are 0.5 m of NaOH different from 0.5 M of NaOH?
Answer: 0.5 m of NaOH means 0.5 mol (20 g) NaOH in 1 Kg of solvent, while 0.5 M means 0.5 mol (20 g) NaOH in 1 L of solution.
48. What is the number of significant figures in 1.050 x 107?
Answer: There are four significant figures.
49. What is ‘1 amu’ or ‘1 u’?
Answer: 1 amu or 1 u is 1/12th mass of an atom of carbon-12.
50. Define molality. Is molality depend on temperature?
Answer: Molality is defined as the moles of solute per kilogram of solvent.
Molality = Number of moles of solute/Mass of solvent (in Kg)
Mass does not change with temperature, so molality is independent of temperature.
51. Why molarity changes with temperature?
Answer: Molarity is defined as number of moles of solute per litre of solution.
Molarity = Number of moles of solute/Volume of solution (in L)
Volume changes with temperature. The volume increases as the temperature increases, and decreases as the temperature decreases. That’s why molarity changes with temperature.
52. Calculate the number of moles in:
(i) 400 g of sulphuric acid
(ii) 100 L of sulphur dioxide at STP
(iii) 12.044*1024 molecules of oxygen
(iv) 16 g of carbon
Answer: (i) Molar mass of H2SO4 = 98 g/mol
Number of moles of sulphuric acid = Given mass/Molar mass = 400/98 mol = 4.082 mol
(ii) 22.4 L of a gas at STP = 1 mol
100 L of SO2 at STP = 100/22.4 mol = 4.464 mol
(iii) 6.022*1023 molecules of oxygen = 1 mol
12.044*1024 molecules of oxygen = (12.044*1024)/(6.022*1023) = 20 mol
(iv) Atomic mass of carbon = 12 g
Number of mole of carbon = Given mass/Atomic mass = 16/12 = 1.333 mol
53. A compound on analysis was found to contain C = 34.6%, H = 3.85% and O = 61.55%. Calculate the empirical formula.
Answer: Calculation of simplest whole number ratio is as follows:
| Element | Percentage | Atomic mas | No. of mole | Mole ratio | Simplest whole no.. ratio |
|---|---|---|---|---|---|
| C | 34.6 | 12 | 34.6/12 = 2.88 | 2.88/2.88 = 1 | 1*3 = 3 |
| H | 3.85 | 1 | 3.85/1 = 3.85 | 3.85/2.88 = 4/3 | (4/3)*3 = 4 |
| O | 61.55 | 16 | 61.55/16 = 3.85 | 3.85/2.88 = 4/3 | (4/3)*3 = 4 |
The simplest whole number ratio of elements is C:H:O = 3:4:4
So, the empirical formula of the compound is C3H4O4.
54. Calculate: (i) mass of 4 g atom of magnesium (ii) gram atom in 32 g of nitrogen (N).
Answer: (i) 1 g atom of magnesium = 24 g
4 g atom of magnesium = 4*24 = 96 g
(ii) 14 g of N = 1 g atom
32 g of N = 32/14 = 2.29 g atom
55. The density of water at room temperature is 1.0 g/ml. How many molecules are there in a drop of water if its volume is 0.05 ml?
Answer: Volume of drop of water = 0.05 ml
Density of water = 1 g/ml
Mass of water = Density x Volume = 1 g/ml x 0.05 ml = 0.05 g
Molar mass of water = 18 g/mol
Now, 18 g of water = 1 mol
0.05 g of water = 0.05/18 = 0.0028 mol
1 mol of water contain molecules = 6.022 x 1023
0.0028 mol of water contain molecules = 6.022 x 1023 x 0.0028 = 1.68 x 1021 molecules.
56. What is the molecular mass of a substance, each molecule of which contains 9 atoms of carbon, 13 atoms of hydrogen and 2.33 x 10-23 g other component?
Answer: Mass of 9 atoms of carbon = 9 x 12 u = 108 u
Mass of 13 atoms of hydrogen = 13 x 1 u = 13 u
Mass of 2.33 x 10-23 g of other component = (2.33 x 10-23)/(1.66 x 10-24) = 14.04 u
Molecular mass of the substance = 108 + 13 + 14.04 = 135.04 u
57. Calculate the number of carbon and oxygen atoms present in 11.2 L of CO2 at NTP.
Answer: Step 1: Find the number of CO2 molecules
22.4 L CO2 at NTP = 1 mol
11.2 L CO2 at NTP = 0.5 mol
Now, 1 mol CO2 = 6.022 x 1023 molecules
0.5 mol CO2 = 3.011 x 1023 molecules
Step 2: Find the carbon and oxygen atoms
1 molecule of CO2 contain 1 atom of carbon and 2 atoms of oxygen.
Number of carbon atoms in 3.011 x 1023 molecules of CO2 = 3.011 x 1023 atoms
Number of oxygen atoms in 3.011 x 1023 molecule of CO2 = 2 x 3.011 x 1023 = 6.022 x 1023 atoms.
58. KClO3 on heating decomposes to give KCl and O2. What is the volume of O2 at NTP liberated by 0.1 mole of KClO3?
Answer: The balanced chemical equation for the decomposition of KClO3 is: 2KClO3 → 2KCl + 3O2
From the chemical equation,
2 mole KClO3 on decomposition gives 3 mole of O2
0.1 mole KClO3 on decomposition gives 0.15 mole of O2
Volume of 1 mole of a gas at NTP = 22.4 L
Volume of 0.15 mole of O2 at NTP = 0.15 x 22.4 L =3.36 L
59. 10 ml of a solution of NaCl containing KCl gave on evaporation 0.93 g of the mixed salt which gave 1.865 g of AgCl by reacting with AgNO3 solution. Calculate the quantity of NaCl in 10 ml of the solution.
Answer: The balanced chemical equation for the reaction occurs is: NaCl + AgNO3 → NaNO3 + AgCl
Let the mass of NaCl and KCl be p g and q g respectively.
So, p + q = 0.93 g
Let us find AgCl formed on reacting NaCl and KCl with AgNO3 solution.
From the chemical equation, 58.5 g NaCl give AgCl = 143.5 g
p g of NaCl will give AgCl = (143.5/58.5) *(p) g
Similarly, 74.5 g KCl give AgCl = 143.5 g
q g of KCl will give AgCl = (143.5/74.5) *(q) g
Given that the mass of AgCl actually formed is 1.865 g.
Mass of NaCl in the mixture = p = 0.14 g
Mass of KCl in the mixture = 0.93 – q = 0.93 – 0.14 = 0.79 g
60. The cost of table salt (NaCl) and table sugar (C12H22O11) are Rs. 1 per Kg and Rs. 6 per Kg respectively. Calculate their cost per mole.
Answer: (i) Molar mass of NaCl = 23 + 35.5 + 58.5 g/mol
1000 g of NaCl cost = Rs. 1
So, 58.5 g NaCl cost = (1/1000) x 58.5 = Rs. 0.0585
(ii) Molar mass of C12H22O11 = (12×12)+(22×1)+(11×16) = 342 g/mol
1000 g of C12H22O11 cost = Rs. 6
So, 342 g of C12H22O11 cost = (6/1000) x 342 = Rs. 2.052
61. A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms?
Answer: We know that, 1 molecule of oxygen gas (O2) contain 2 oxygen atoms while 1 molecule ozone gas (O3) contain 3 oxygen atoms.
1 mole of a gas contains 6.022 x 1023 molecules
In flask P, 0.5 mole of oxygen gas = 0.5 x 6.022 x 1023 molecules = 0.5 x 6.022 x 1023 x 2 atoms = 6.022 x 1023 atoms
In flask Q, 0.4 mole of ozone gas = 0.4 x 6.022 x 1023 molecules = 0.4 x 6.022 x 1023 x 3 atoms = 7.23 x 1023 atoms
Hence, it is clear that flask Q contains greater number of oxygen atoms as compared to flask P.
62. Calculate the total number of electrons present in 1.6 g of methane.
Answer: Mass of methane = 1.6 g
Molar mass of methane = 12 + 4(1) = 16 g/mol
Number of mole of methane = Mass/Molar mass = 1.6/16 = 0.1 mol
1 mole of methane contains 6.022 x 1023 molecules
So, 0.1 mole of methane contains 6.022 x 1022 molecules.
1 molecule of CH4 contains 10 electrons, i.e., 6 electrons of carbon atom and 4 electrons of four hydrogen atoms.
Number of electrons in 0.1 mole methane = 6.022 x 1022 x 10 = 6.022 x 1023 electrons
Hence, the total number of electrons present in 1.6 g of methane are 6.022 x 1023.
63. The vapour density of a mixture of NO2 and N2O4 is 38.3 at 27°C. Calculate the number of moles of NO2 in 100 g of the mixture.
Answer: Vapour density of the mixture = 38.3
Molecular mass of mixture = 2 x vapour density = 2 x 38.3 = 76.6 g
Mass of the mixture = 100 g
No. of mole in the mixture = 100/76.6
Let the mass of NO2 in the mixture = x g
Mass of N2O4 in the mixture = (100-x) g
Molar mass of NO2 = 14 + 32 = 46 g/mol
Molar mass of N2O4 = 28 + 64 = 92 g/mol
No. of mole of NO2 = x/46 mol
No. of mole of N2O4 = (100-x)/92 mol
On solving x = 20.104
Mass of NO2 = 20.104 g
Mole of NO2 in the mixture = 20.104/46 = 0.437 mol = 0.44 mol
64. The vapour density of a gaseous element is 5 times of oxygen under similar conditions. If the molecule is triatomic, what will be its atomic mass.
Answer: Molecular mass of oxygen = 32 u
Vapour density of oxygen = Moleculas mass/2 = 32/2 = 16
Vapour density of gaseous molecule = 5 x Vapour density of oxygen = 5 x 16 = 80
Molecular mass of gaseous molecule = 2 x Vapour density = 2 x 80 = 160
The gaseous molecule is triatomic, i.e., its atomicity is 3.
Atomic mass of element = 160/3 = 53.33 u
So, the atomic mass of the element is 53.33 u.
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