RBSE Class 11 Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Rajasthan Board RBSE Class 11 Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

RBSE Class 11 Chemistry Chapter 1 Important Questions Some Basic Concepts of Chemistry

Multiple Choice Questions:
 
Question 1. 
Number of moles of oxygen in 1 mol oxygen will be: 
(1) 1 
(2) 100 
(3) 6.022 × 10²³ 
(4) 0.623 × 10²³ 
Answer:
(3) 6.022 × 10²³ 

Question 2. 
Unit of equivalent weight is: 
(1) g 
(2) kg 
(3) g/L 
(4) None of these 
Answer:
(4) None of these 

Question 3. 
Volume of 1 g molecular weight of each gas at N.T.P.
(1) 224 L 
(2) 224 cc 
(3) 2.24 L 
(4) 22.4 L 
Answer:
(4) 22.4 L 

 

Question 4. 
Number of molecules present in 44 g CO₂ are: 
(1) 12 × 10²³ 
(2) 3 × 10¹⁰ 
(3) 6 × 10²³ 
(4) 1 × 10²³ 
Answer:
(3) 6 × 10²³ 

Question 5. 
Volume of 0.25 mol NH₃ gas at S.T.P: 
(1) 224 L 
(2) 5.6 L 
(3) 112 L 
(4) 22.4L 
Answer:
(2) 5.6 L 

Question 6. 
For a metal, the formula of its chloride is MC₁ and equivalent weight of metal is 9, what wi be atomic weight of element : 
(1) 9 
(2) 18
(3) 27 
(4) 3 
Answer:
(3) 27 

Question 7. 
Atomicity of any element represents: 
(1) Atomic number of element 
(2) Atomic weight of element 
(3) Avogadro number of element 
(4) Number of atoms in molecule of element 
Answer:
(4) Number of atoms in molecule of element 

Question 8. 
Equivalent weight of H₂SO₄ will be: 
(1) 98 
(2) 49 
(3) 196 
(4) 448 
Answer:
(2) 49 

Question 9. 
Total number of ions in 111g CaCO₃: 
(1) 1 mol 
(2) 2 mol 
(3) 3 mol
(4) 4 mol 
 Answer:
 (3) 3 mol
 
Question 10. 
Air is : 
(1) Compound 
(2) Mixture 
(3) Element 
(4) None of these 
Answer:
(2) Mixture 

Question 11. 
1 mol H₂O contains: 
(1) 6.022 × 10²³ oxygen atoms 
(2) 6.022 × 10²³ hydrogen atoms 
(3) 18.1 × 10²³ H₂O atoms 
(4) 3g molecules of H₂O 
Answer:
(1) 6.022 × 10²³ oxygen atoms 

Question 12. 
What is equivalent weight of oxalic acid (H₂C₂O₄2H₂O) if its molecular weight 126? 
(1) 98 
(2) 63 
(3) 196 
(4) 126 
Answer:
(2) 63 

Question 13. 
Which pair follow the law of multiple proportions? 
(1) GO and CO₂ 
(2) H₂O and D₂O
(3) NaCl and NaBr 
(4) MgO and Mg (OH)₂ 
Answer:
(1) GO and CO₂ 

Question 14. 
Equivalent weight of MnSO₄ remains half  molecular weight, when this conversion appears: 
(1) in Mn₂O₃
(2) in MnO₂  
(3) in MnO₄ 
(4) in MnO₂^- 
Answer:
(2) in MnO₂  

Question 15. 
5 mol X and 9 mol Y 'formed Z according t given reaction X + 2Y → Z, then how muc mol of Z will be formed? 
(1) 5 mol Z 
(2) 8 mol Z 
(3) 14 mol Z 
(4) 4 mol Z 
Answer:
(4) 4 mol Z 

Question 16. 
How much g of oxygen will react with 27 g aluminium? 
(1) 8 g. 
(2) 16 g 
(3) 32 g 
(4) 24 g 
Answer:
(4) 24 g 

Question 17. 
How much amount of CaCO₃ will give 56 kg of CaO? 
(1) 1000 kg 
(2) 56 kg 
(3) 2 kg 
(4) 100 kg 
Answer:
(4) 100 kg 

Question 18. 
5.3 g of Na₂CO₃ is dissolved in 500 mL of water. Molarity of solution will be 
(1) 0.1 M 
(2) 0.2 M 
(3) 0.3 M 
(4) 10 M 
Answer:
(1) 0.1 M 

Question 19. 
Which has the minimum number of molecules? 
(1) 0.1 mol CO₂ 
(2) 11 ml CO₂ 
(3) 22 g CO₂ 
(4) 22.4 × 10³ ml CO₂ 
Answer:
(4) 22.4 × 10³ ml CO₂ 

Question 20. 
Percentage of oxygen in NaOH is: 
(1) 40 
(2) 16 
(3) 8 
(4) 1 
Answer:
(1) 40 

Very Short Answer Type Questions:
 
Question 1. 
Sometimes air is considered as heterogeneous mixture, why? 
Solution: 
When dust particles present in air then it is considered as heterogeneous mixture.
 
Question 2. 
Which state of matter has indefinite shape and volume? 
Solution: 
Gaseous state. 

Question 3. 
In which state of matter, the intermolecular space is larger and smaller? 
Solution: 
In gaseous state, intermolecular space is larger whereas in solid, it is smaller. 

Question 4. 
By which force, molecules attract each other? 
Solution: 
Intermolecular force. 
 
Question 5. 
What is International System of units? 
Solution: 
International system of unit is known as S.I. It was accepted in 1960 by Conference Generale des Poids et Measures, CGPM. 

Question 6. 
What is S.I. unit of length? 
Solution: 
The S.I. unit of length is metre. 

Question 7. 
How much metre present in one micron? 
Solution:
10^-6 metre. 

Question 8. 
How much seconds are present in one solar day? 
Solution: 
86400 s. 

Question 9. 
Light year is the unit of which physical quantity? 
Solution: 
Distance. 

Question 10. 
What is difference between base units and derived units? 
 Solution: 
 Base units are those which are obtained from S other physical quantities and derived units are obtained from base units, 

Question 11. 
What is Indian standard of time? 
Solution: 
It is time obtained from caesium watch kept M in National Physical Laboratory (N.P.L). 

Question 12. 
What is Angstrom (Å) ? 
Solution: 
It is used for measurement of size of nucleus and wavelength of light. 1Å = 10^-10 m 

Question 13. 
The radius of an atom is 10^-10 m. Calculate it is in micrometre. 
Solution : 
\(10^{-10} \mathrm{~m}=\frac{10^{-10}}{10^{-6}}=10^{-4} \mu \mathrm{m}\)
∵ 1 μm = 10^-6 

Question 14. 
When vanadium is mixed with iron, it forms alloy. The density of vanadium is 5.96 g/cm³. (i What will be its S.I. unit? 
Solution: 
The density of vanadium will be 5.96 kg/m³  in S.I. or M.K.S. system. 

Question 15. 
Convert 40 calorie in joule. 
Solution:
1 cal = 42 J 
∵ 40 cal = 42 × 40 J 
= 168.0 J 

Question 16. 
How much joules are present in 1 erg? 
Solution:
1 erg = 10^-4 joule. 

Question 17. 
Write the 0.000213 cm in scientific notation. 
Solution: 
2.13 × 10^-4 cm  
= 2.13 × 10^-6 m 

Question 18. 
Convert the 20°C in Fahrenheit. 
Solution: 
°F = \(\frac{9}{5}\left({ }^{\circ} \mathrm{C}\right)\) (°C) + 32 
\(\circ \mathrm{F}=\frac{9}{5} \times 20+32\)
= 36 + 32 = 68 
20°C = 68°F 

Question 19. 
Convert 10°F in °C. 
Solution:

10°F = - 12 22 °C 

Question 20. 
What is scientific notation of measurement ? 
Solution: 
In scientific notation, the measurement is ritten in exponential form. It is represented as  N × 10ⁿ, where N is the number (digit) and n is ponent. 

Question 21. 
What do you mean by precision and accuracy? 
Solution:
Precision refers to closeness of various measurements for the same quantity where as accuracy the agreement of a particular value to the true value of e result. 

Question 22. 
Express the following in scientific notations? 
(i) 0.0012; 
(ii) 2130000; 
(iii) 2002; 
(iv) 600.0; 
(v) 5000; 
(vi) 7216.3 
Solution: 
(i) 0.0012 = 12 × 10^-3 
(ii) 2130000 = 2.13 × 10⁶ 
(iii) 2002 = 2.002 × 10^-3 
(iv) 6000 = 6.00 × 10² 
(v) 5000 = 5.0 × 10³ 
(vi) 7216.3 = 7.2163 × 10³ 

Question 23. 
What is the weight of C-12 atom in gram? 
Solution: 
1.992648 × 10^-23 g≈ 1.99 × 10^-23 g 

Question 24. 
When zero is not significant figure for any number? 
Solution: 
If zero is at left side of non-zero digit then it is ot significant figure. For example; in 0.014, both zero e not significant figures. 

Question 25. 
May all digit significant in any numbers? 
Solution: 
Yes, if all digits are non-zero or if zero exists etween two non-zero digits. 

Question 26. 
Which is more accurate measurement in 5.0 or 5.00 g ? 
Solution: 
5.00 is more accurate because in this measurement is taken upto two points after decima point. 

Question 27. 
In which ratio, element combines with mass.
Solution: 
According to law of constant proportion. 

Question 28. 
Who gave law of conservation of mass? 
Solution: 
Antoine Lavoisier in 1785. 

Question 29. 
Who gave principle of relativity? 
Solution: 
Albert Einstein. 

Question 30. 
What is the energy conservation law Einstein? 
Solution: 
E = mc² 

Question 31. 
What do you mean by a.m.u.? 
Solution: 
It is the Atomic Mass Unit. 

Question 32. 
Which isotope of carbon is used to expres the relative mass of element? 
Solution: 
C-12 isotope. 

Question 33. 
Why atomic masses of an element fractional? 
Solution: 
Due to presence of their isotopes. 

Question 34. 
How much atoms are present in 1g atom any element? 
Solution: 
6.022 × 10²² atoms. 

Question 35. 
What is atomic mass unit? 
Solution: 
Atomic mass unit is the 1/12th part of C-12 atom. 
So, 
1 a.m.u. = \(\frac{1.9926 \times 10^{-23}}{12}\) = 1.66 × 10^-24 g 

Question 36. 
What is the average atomic mass? 
Solution: 
The average atomic mass of an element refer to the atomic masses of isotopes of the elements takin into account the different abundance of the element' isotopes. 

Question 37. 
What is molecular mass? 
Solution:
Molecular mass shows how many times molecule of substance is heavier than that 1th of 12 th mass of an atom of C-12. 

Question 38. 
What is molar mass? 
Solution: 
When mass of any substance is expressed in terms of moles then it is called molar mass for instance molar mass of H₂O = 18.02 g. 

Question 39. 
Calculate the molecular mass of hydrogen if its atomicity is 2. 
Solution: 
Molecular mass Atomic mass × Atomicity 
= 1.008 × 2 = 2.016 u. 
∴ Molecular mass of hydrogen will be 2.016 u. 

Question 40. 
Calculate the number of oxygen atoms in 48u of ozone? 
Solution: 
Number of atoms of oxygen will be \(\frac{48 u}{16 u}\) = 3
So, there will be 3 atoms of oxygen. 

Question 41. 
Does molar volume of H₂S and CO₂ differ ? 
Solution: 
No, these will be same because at S.T.P., molar volume of all gases is 22.4 L. 

Question 42. 
Is law of mass conservation valid for nuclear reactions? 
Solution: 
No, this law is not valid for nuclear reactions. 

Question 43. 
What will be volume of 1 mol CO₂ at S.T.P. ? 
Solution: 
The volume of 1 mol CO₂ at S.T.P. will be 22.4 L. 

Question 44. 
If a person spends ₹ 1000000 per second, can he spend 1 mol in his life span? 
Solution:
∵₹1000000 spends = in 1 second 
∴ 6.022 × 10²³ rupees will spend 

= 2 × 10¹⁰ years. 
So, it is not possible to spend in life time. 

Question 45. 
Who expounded the following laws? 
(i) Law of constant proportions 
(ii) Law of conservation of mass 
(iii) Law of multiple proportions 
(iv) Law of reciprocal proportions 
(v) Law of gaseous volume 
Solution: 
(i) Joseph Proust, 
(ii) Lavoisier, 
(iii) Dalton, 
(iv) Ritcher, 
(v) Gay Lussac. 

Question 46. 
When Gay Lussac's law of gaseous volume is disobeyed? 
Solution: 
It is disobeyed when there is solid or liquid state of one of reactant or product. 

Question 47. 
What is Dulong Patit's law? 
Solution: 
According to this law, for any metal, the product of atomic mass and specific heat is 6.4.
So, Atomic mass × Specific heat = 6.4 

Question 48. 
Write the name of two other states of matter? 
Solution: 
(i) Plasma, 
(ii) Bose Einstein Condensate (BEC). 

Question 49. 
Substract 0.0016 from 3.82 and give answer in significant figures. 
Solution: 
\(\begin{array}{r} 3.82 \\ -0.0016 \\ \hline 3.8184=3.82 \\ \hline \end{array}\)

Short Answer Type Questions:
 
Question 1. 
Define the relative atomic mass of an element. 
Solution: 
Atomic mass of any element is that number which represents that how much an atom of element is heavier than 1/12th part of C-12 by mass or 1.008 part by mass of hydrogen. Average relative mass of an atom of S an element as compared to the mass of an atom of C-12.

 
Question 2. 
What is meant by formula of compound? 
Solution: 
The brief representation of molecules in compound is symbol of their elements is known as formula of compound. It is classified as : 

1. Empirical formula and 
2. Molecular formula. 

1.  Empirical Formula: It gives the simplest whole number ratio of the atoms of various elements present in one molecule of compound. 
2. Molecular Formula: It gives the actual number of atoms of various elements present in one molecule of the compound: 

Question 3. 
What is the relation between empirical formula and molecular formula ? 
Solution: 
Molecular formula = n × Empirical formula. 
(Where n = 1, 2, 3, ...) 

Question 4. 
Differentiate molarity and molality. 
tablee

Numerical Questions:

Question 1. 
If 2.12 g glucose and 2.925 g salt are mixed in 30.2 g of water. Calculate the mass of solution. 
Solution: 
Mass of glucose = 2.12 g 
Mass of salt = 2.925 g 
Mass of water = 302 g 
Total mass of solution = 2.12 + 2.925 + 302 
= 35.245 g 
So, by rounding off the number 35.245 ≈ 35.2 g 

Question 2. 
Mass of silver metal is 6.342g and its density is 1.28 g/cm³. Calculate its volume. 
Solution: 
Mass of silver = 6.342 g 
Density of silver = 128 g/cm 

By rounding off it will be 5.3 cm³, which has 2 significant figures. 

Question 3. 
Mass of 1 atom of hydrogen is 1.008 u. What will be mass of 10 atoms of hydrogen? 
Solution: 
Mass of 1 atom of hydrogen = 1.008 u 
Mass of 10 atoms of hydrogen = 10 × 1.008 u 
= 10.08 u 
Since 10 is accurate number so its significant number will be infinite. Hence the number of significant figures o resultant number should be equal to significant figure of other numbers. Thus total mass is 10.08 u. 

Question 4. 
Express the following into International 
System of Units (S.I.) : 
(i) 90 million miles (Distance between Earth and Sun) 
(ii) 90 miles/hour (Speed of Rajdhani Express Train) 
(iii) 5 Feet 6 inches (Average length of Indian males) 
Solution: 
(i) 90 millon miles - S.I. unit of distance is metre. 
So, 
1 mile = 1.60 km 
= 1.60 × 10³ metre 
= 1.60 × 10³ m 
Conversion coefficient 
= \(\frac{1.60 \times 10^3 \mathrm{~m}}{1 \mathrm{mile}}\)
For 90 million miles = 90 x 10⁶ miles 
= 90 x 10⁶ miles × conversion coefficient 
\(=\frac{90 \times 10^6 \text { miles } \times 1.6 \times 10^3 \mathrm{~m}}{1 \mathrm{mile}}\)
= 90 × 1.6 × 10⁶ m 
= 1.44 × 10¹¹ metre 

(ii) 90 miles per hour 
1 mile = 1.6 km = 1.6 × 10³ 
Conversion coefficient = \(\frac{1.6 \times 10^3 \mathrm{~m}}{1 \mathrm{mile}}\)
1 Hour = 60 × 60 s = 3.6 × 10³s 
Conversion coefficient = 

= 40 m/s

(iii) 5 Feet 6 inches = 5 × 12 + 6 = 66 inches 
1 inch = 2.54 × 10^-2 m 
Conversion coefficient = 2.54 × \(\frac{10^{-2} \mathrm{~m}}{\text { inches }}\)
66 inches = 66 inches × conversion coefficient 
= 66 × 2.54 × 10^-2 m 
= 167.64 × 10^-2 m 
= 1.68 m 

Question 5. 
Convert the volume of 10L H₂O in metre. 
Solution: 
1 L = 1000 cm³ 
1m = 100 cm 
1L = 10^-3 m 
Conversion coefficient = \(\frac{10^{-3} \mathrm{~m}^3}{1 \mathrm{~L}}\)
So 10L = 10L × conversion cofficient 
\(=\frac{10 . \mathrm{L} \times 10^{-3} \mathrm{~m}^3}{\mathrm{~L}}\)
= 10^-2m³ 

Question 6. 
How much mass of sodium chloride will be separated by 4.9 g H₂SO₄ if 6 g sodium hydrogen sulphate (NaHSO₄ ) and 1.825 g HCl formed during the reaction? 
Solution: 
The chemical equation will be 
NaCl + H₂SO₄ → NaHSO₄ + HCl 
Mass of reactants = Mass of NaCl + Mass of H₂SO₄ 
= x + 4.9 g 
Mass of products = Mass of NaHSO₄ + Mass of HCl 
6.0 g + 1.825g = 7.825 g 
According to law of mass conservation 
Mass of reactants = Mass of products x + 4.9 g = 7.825 g 
x = 7.825 - 4.9 g 
Hence, mass of NaCl = 2.925 g 

Question 7. 
When 10 g CaCO₃ is heated, then 5.6 g CaO and 2.24 litre CO₂ gas (at S.T.P.) are formed. Prove that data follow the law of mass conservation. 
Solution: 
Chemical reaction:
Mass of CaCO₃ = 10 g 
Mass of CaO = 5.6 g 
∵ Mass of 22.4L CO₂ at STP = 44 g 
Mass of 1L CO₂ at STP = \(\frac{44}{22.4} \mathrm{~g}\)
∴ Mass of 2.24L CO₂ at STP = \(\frac{44 \times 2.24}{22.4}\) = 4.4 g 
Total mass of products = 5.6 + 44 = 10 g 
Since mass of products is equal to the mass of reactants. So, data follow law of conservation of mass. 

Question 8. 
When 2.16 g metallic copper is heated with HNO₃, 2.7 g copper oxide is obtained. In other experiment, 1.15 g copper oxide is reduced to give 0.92 g copper. Show that data follow law of constant proportions. 
Solution: 
From first experiment: 
Weight of metallic copper = 2.16 g 
Weight of copper oxide = 2.7 g 
Weight of oxygen = 2.7 - 2.16 = 0.54 g 
Percentage of copper =
\(=\frac{2.16 \times 100}{2.7}=80 \%\)
Percentage of oxygen = 100 - 80 = 20% 
From second experiment : 
Weight of metallic copper = 0.92 g 
Weight of copper oxide = 1.15 g 
Weight of oxygen = 1.15 - 0.92 = 023 g 
Percentage of copper = \(\frac{0.92}{1.15}\) × 100 = 80% 
Percentage of oxygen = 100 - 80 = 20% 
In both experiments, ratio of mass of copper and oxygen are same. So, data obey the law of constant proportions. 

Question 9. 
Prove that following data obey law of constant proportions. Mass of carbon in the different samples of carbon monoxide are following: 
(i) 1.26 g carbon in 1.42 g of carbon monoxide. 
(ii) 1.008 g carbon in 1.136 g of carbon monoxide. 
Solution: 
In first sample: 
Mass of CO = 142 g 
Mass of C = 126 g 
Mass of O = 1.42 - 1.26 = 0.16 g 
% of carbon = \(\frac{126}{1.42}\) × 100 = 88.73% 
% of oxygen = 100 - 88.73  = 11.27% 
In second sample: 
Mass of CO = 1.136 g 
Mass of C = 1.008 g 
Mass of O = 1.136 - 1.008 
= 0.128 g 
Mass of carbon 
% of carbon =
\(=\frac{1.008 \times 100}{1.136}\)
= 88.73% 
% of oxygen = 100 - 88.73 = 1127% 
Since percentage of carbon and oxygen is same in given data, so it obeys law of constant composition proportions. 

Question 10. 
In a copper sulphate crystal, 25.45% copper and 36.07% water are present. If law of constant proportion is valid then calculate the weight of copper from 40 g copper sulphate crystal. 
Solution: 
From given data 
% of copper = 25.45% · 
% of water = 36.07% 
% of sulphate = 100 - (25.45 + 36.07) 
= 100 - 61.52 = 38.48% 
From the law of constant composition proportion, the % of copper sulphate in second crystal should be same as first one, then 
Amount of Cu in 100 g CuSO₄ = 25.45 g 
Amount of Cu in 40 g CuSO₄ = \(\frac{25.45 \times 40}{100}\)
= 10.18 g 

Question 11. 
In two sulphides of an element, the mass of sulphur are 53.33% and 36.36% respectively. Prove that data obey the law of multiple proportions. 
Solution: 
In first sulphide : 
Amount of sulphur = 53.33% 
Amount of element = 100 - 53.33 
= 46.47% 
The ratio of percentage of element and sulphur is 
46.67 : 53.33 = 1 : 1.14 
In second sulphide : 
Amount of sulphur = 36.36% 
Amount of element = 100 - 36.36 
= 63.64 
The ratio of percentage of element and sulphur is 
= 63.64 : 36.36 = 1 : 0.57 
Since amount of element in both sulphides is fixed. Then ratio in both amount is a multiple 0.57 : 1.14 or 1 : 2, which is multiple proportions. So data follow the law of multiple proportions. 

Question 12. 
Different amounts of A and B mutually react and form three compounds as X, Y and Z. 
0.3 g A + 0.4 g B→ 0.7 g X 
18.0 g A + 43.0 g B → 66.0 g Y 
40.0 g A + 159.99 g B → 199.99 g Z 
On the basis of above data, prove the law of multiple proportions. 
Solution: 
In compound X, Y and Z, the amounts of B 
react with A, are 

Hence, amounts of B, reacted with A are in following ratio : 
1.33 : 2.66: 4.0 
or 
1: 2 : 3 
So law of multiple proportion is obeyed. 

Question 13. 
Reaction of H₂ and Cl₂ in presence o sunlight gives HCl gas. Calculate the volum of HCl gas obtained from 20 mL of hydrogen 
Solution: 

∵1 volume hydrogen gives = 2 volumes HCl 
∴ 20 volumes hydrogen will produce = 2 × 20 ml HCl 
= 40 ml HCI 

Question 14. 
Two oxides are formed from a metal. From these oxides, 1g of each is reduced and yiel 0.798 g and 0.888 g metal respectively. Thes results obey which law of chemica combination? 
Solution: 
In first oxide, amount of oxide = 1.000 g 
Amount of metal = 0.798 g 
Amount of oxygen = 1.000 - 0.798 
=0.202 g 
∴ 0.202 g oxygen is combined with = 0.798 g of metal 
∴ 1g oxygen will be combined with = \(\frac{0.798}{0.202}\) g of metal 
= 3.95 g 
In second oxide, 
Amount of oxide = 1.00 g 
Amount of metal = 0.888 g 
Amount of oxygen = 1.00 - 0.888 
= 0.112 g 
0.112 g oxygen is combined with  = 0.888 g of metal 
∵ 1g oxygen will be combined with = \(\frac{0.888}{0.112}\) = 7.92 g 
So, amount of metal combined to oxygen are 3.95 an 7.92 respectively which are in simple ratio of 1 : 2. S given data follow the law of multiple proportions. 

Question 15. 
If nitric oxide is formed by followin equation : 
N₂(g) + O₂(g) → 2NO (g) 
then calculate the volume of nitroge required to produce 30 L of NO. 
Solution:

∵ 2 L of NO is obtained by = 1 L of nitrogen 
∴ 30 L of NO will be obtained by = 30/2 = 15 L of Nitrogen 

Question 16. 
6.488 g of lead reacts with 1.002 g of oxygen form PbO₂. This PbO₂ is also obtained whe lead nitrate is heated, in which percentage o oxygen is 13.38%. Does above information prove the law of constant composition proportions? 
Solution: 
In first oxide, 
Mass of lead peroxide = 6.488 g + 1.002 g = 7.490 g 
∵ 7.49 g PbO₂ contains = 1.002 g oxygen 
∴ 100 g PbO₂ will contain = \(\frac{1.002 \times 100}{7.49} \mathrm{~g}\)
= 13.38% 
Since in second experiment, percentage of oxygen is 13.38% which is equal to first. So, it proves law of constant composition proportions. 

Question 17. 
Calculate the molar mass of following: 
(i) CH₃COOH, 
(ii) P₄, 
(iii) S₈, 
(iv) PCl₅, 
(v) SO₂Cl₂ 
Solution: 
(i) Molar mass of CH₃COOH 
= 2 × 12 + 4 × 1 + 2 × 16 
24 + 4 + 32 = 60 g/mol 
(ii) Molar mass of P₄ 
= 31 × 4 = 124 g/mol 
(iii) Molar mass of Sg 
= 32 × 8 = 256 g/mol 
(iv) Molar mass of PCl₅ = 31 + 35.5 × 5 
= 31+ 177.5  = 208.5 g/mol 
(v) Molar mass of SO₂Cl₂ = 32 + 16 × 2 + 2 × 35.5 
= 32 + 32 + 71.0 
= 135 g/mol. 

Question 18. 
Calculate the number of atoms and gram atoms in 120 g carbon. 
Solution: 
Gram atoms =
Number of atoms = No. of gram atoms × 6.022 × 10²³ 
= 10 × 6.022 × 10²³ 
= 6.022 × 10²⁴ atoms 

Question 19. 
Calculate the mass of 1 carbon atom. 
Solution: 
Atomic mass of carbon = 12 g 
Atomic mass of 1 mol of carbon = 12 g 
or Atomic mass of 6.022 × 10²³ atoms of carbon = 12 g 
So, atomic mass of 1 carbon atom = \(\frac{12}{6.022 \times 10^{23}} \mathrm{~g}\)
= 1.99 × 10^-23g 

Question 20. 
Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium 23 u) 
Solution: 
1 mol = Atomic mass 
∵ 23 g sodium = 1 mol sodium 
= 0.25 mol sodium 

Question 21. 
Calculate the number of sodium atoms in 2.3 g of sodium. (Atomic weight of Na = 23) 
Solution:
∵ No. of sodium atoms in 23 g sodium = 6.022 × 10²³ 
 ∴ No. of sodium atoms in 2.3 g sodium = \(\frac{6.022 \times 10^{23}}{23} \times 2.3\)
= 6.022 × 10²² sodium atoms 

Question 22. 
Calculate the number of Cl ̄ ions in 0.585 g of NaCl. 
Solution: 
Since 58.5 g NaCl contains = 6.022 × 10²³ 
NaCl molecules. 
Thus, no. of NaCl molecules in 0.585 g will be 
= \(\frac{6.022 \times 10^{23} \times 0.585}{58.5}\) NaCl molecules 
= 6.022 × 10²¹ NaCl molecules 
1 molecule of NaCl contains = Cl^- 
6.022 × 10²¹ molecule of NaCl will contain 
 = 6.022 × 10²¹ CI ions. 

Question 23. 
Find the molecular formula of ammonia formed by combination of hydrogen and nitrogen. 
Solution: 

So, molecular formula of ammonia will be NH₃ 
or mass of n × (NH₃) = 17 n 
But molecular mass of NH₃ = 17 
So \(n=\frac{17}{17}=1\)
So, molecular formula of ammonia will be NH₃. 

Question 24. 
Calculate the mass of 3.01 × 10²² molecules of NH₃. 
Solution: 
Molar mass of NH₃ = 14+ 3 = 17 g/mol 
∵ Mass of 6.022 × 10²³ molecule of NH₃ = 17 g/mol 
∴ Mass of 3.01 x 10²² molecules of NH₃
\(=\frac{17 \times 3.01 \times 10^{22}}{6.022 \times 10^{23}}\)
= 8.5 × 10^-1 
= 0.85 g/mol 

Question 25. 
Find out the number of atoms in 2.31 × 10^-6  copper. (Atomic mass of Cu 63.5 a.m.u.). 
Solution: 
63.5 g Cu = 6.022 × 10²³ Cu-atoms. 
2.31 × 10^-6 Cu = \(\frac{6.022 \times 10^{23} \times 2.31 \times 10^{-6}}{63.5}\)
= 0.219 × 10¹⁷ 
= 2.19 × 10¹⁶ Cu-atoms 

Question 26. 
The density of liquid mercury is 13.6 g/cm³ Calculate the number of moles in 1 L mercury metal. 
Solution:
Mass of Hg = Density > Volume 
= 13.6 × 1000 = 13600 g 
No. of moles of Hg =
\(=\frac{13600}{200}\)
= 68 mol 

Question 27. 
Find out the gram molecules in 4.9 g of H₂SO₄. 
Solution: 
Molecular mass of H₂SO₄ 
= 2 × 1 + 32 + 16 × 4 
= 2 + 32 + 64 = 98 g 
∵ 98 g H₂SO₄ = 1 g/mol 
∴ 4.9 g H₂SO₄ = 1/98 × 4.9 g/mol 
= 0.05 g mol 

Question 28. 
Sodium carbonate decahydrate (Na₂CO₃ 10H₂O) is an important industrial chemical. Calculate its formula mass. 
Solution:
Na₂CO₃.10H₂O
=[2 × 23 + 12 + 3 × 16 + 10(2 × 1 + 16)] g 
= [46 +12 + 48 + 180] 
= 286 g 

Question 29. 
Calculate the mass of following: 
(i) 1 mol atom of N 
(ii) 5 moles atom of Al 
(iii) 5.0 moles of Na⁺ ion 
(iv) 10 moles of NaOH 
Solution: 
(i) Atomic mass of N = 14 g 
So mass of 1 mol N = 14 g 

(ii) Mass of 5 moles atom of Al 
Atomic mass of Al = 27 g 
∵ Mass of 1 mol of Al = 27 g 
∴ Mass of 5 moles of Al = 5 x 27 = 135 g 

(iii) Mass of 5.0 moles of Na⁺ 
Atomic mass of Nation = 23 g 
∴ Mass of 1 mol of Na⁺ = 23 g 
∴ Mass of 5.0 moles of Na⁺ = 5 x 23 = 115 g 

(iv) Mass of 10 moles of NaOH 
Mass of NaOH = 40 g 
∴ Mass of 1 mol NaOH = 40 g 
∵Mass of 10 moles of NaOH = 10 × 40 = 400 g 

Question 30. 
Convert the following into mol: 
(i) 12 g SO₂ gas, 
(ii) 25 g H₂O, 
(iii) 22 g CO₂ gas 
Solution: 
(i) Atomic weight of SO₂ gas = 32 + 32 = 64g 
64 g SO₂ = 1 mol SO₂ 
12 g SO₂ = \(\frac{12}{64}\) mol SO₂ 
= 0.1875 mol 

(ii) 25 g H₂O 
Molecular weight of H₂O = 2 + 16 = 18 g 
18 g H₂O = 1 mol H₂O 
25 g H₂O = \(\frac{25}{18}\) mol H₂O
= 1.39 mol H₂O 

(iii) 22g CO₂ 
CO₂ = 12 + 32 = 44 g 
44 g CO₂ = 1 mol CO₂ 
22 g CO₂  = \(\frac{22}{44}\)
= 1/2 mol of CO₂ 
= 0.5 mol CO₂ 

Question 31. 
Calculate the total number of electrons in 1.4 nitrogen gas. 
Solution: 
Molar mass of N₂ = 14 x 2 = 28 g 
No. of electrons present in 28 g N₂ = 14 mol electrons 
No. of electrons present in 1.4 g nitrogen 
= \(\frac{14 \times 1.4}{28}\) mol electrons 
= 0.7 mol electrons 
= 0.7 × 6.022 × 10²³ electrons 
= 4. 215 × 10²³ electrons 

Question 32. 
Find out the number of each atom present in 3.42g sucrose (C₁₂H₂₂O₁₁). 
Solution: 
Molecular weight of sucrose C₁₂H₂₂O₁₁ 
= 12 × 12 + 22 × 1 + 11 × 16 
= 144 + 22 + 176 = 342 g 
∵ 342 g sucrose contains = 6.022 × 10²³ 
 ∴ 3.42 g sucrose will contain 
\(=\frac{6.022 \times 10^{23} \times 3.42}{342}\)
= 6.022 × 10²¹ molecules 
∵1 molecule of sucrose = 12 carbon atoms 

 ∴ 6.022 × 10²¹ molecules of sucrose will contain 
= 12 × 6.022 × 10²¹ carbon atoms 
= 72.24 × 10²¹ carbon atoms 
= 7.224 × 10²² C atoms 

∴ 1 molecule of sucrose = 22 hydrogen atoms 
∵ 6.022 × 10²¹ molecules of sucrose will contain 
= 22 × 6.022 × 10²¹ 
= 132.44 × 10²¹ H atoms 
= 1.32 × 10²³ H atoms 

∵ 1 molecule of sucrose = 11 oxygen atoms 
6.022 × 10²¹ molecules of sucrose will contain 
= 11 × 6.022 × 10²¹ oxygen atoms 
 = 6.622 × 10²¹ O-atoms 
= 6.622 × 10²² O-atoms 

Question 33. 
Calculate the number of aluminium ions present in 1.21 g of aluminium oxide (Al₂O₃). 
Solution: 
Molecular weight of Al₂O₃ = 2 × 27 + 3 × 16 
= 54 + 48 = 102 g/mol 
No. of aluminium ions present in 102 g Al₂O₃ 
\(=\frac{2 \times 1.21}{102}\) mol aluminium ions 
= 0.024 mol aluminium ion 
= 0.024 × 6.022 × 10²³ aluminium ions 
= 0.144 × 10²³ Al ions 
= 1.44 × 10²² Al ions 

Question 34. 
Calculate the number of atoms present in 3.2 g of sulphur. 
Solution: 
Atomic weight of S = 32 
No. of atoms in 32 g S = 6.022 × 10²³ 
No. of atoms in 3.2 g S = \(\frac{6.022 \times 10^{23} \times 32}{32}\)
= 6.022 × 10²² 

Question 35. 
What will be the number of molecules in SO₂ in which 8.0 g oxygen is present? 
Solution: 
Molecular weight of SO₂ = 32 + 2 × 16 = 64 g
n 32 g oxygen will present in = 64 g SO₂ 
8 g oxygen will present in = \(\frac{64 \times 8.0}{32}\)
= 16 g SO₂ 
64g SO₂ contains = 6.022 × 10²³ molecules 
16 g SO₂ contains = \(\frac{6.022 \times 10^{23} \times 16}{64}\)
= 1.505 × 10²³ molecules 

Question 36. 
If 1021 molecules are removed from 200 mg of H₂S. Calculate the number of moles present in H₂S. 
Solution: 
Molecular weight of H₂S = 2 + 32 = 34 g 
Number of molecules in 34 g H₂S = 6.022 × 10²³ 
Number of molecules in 200 × 10^-3 g H₂S 
\(=\frac{6.022 \times 10^{23} \times 200 \times 10^{-3}}{34}\)
= 35.41 × 10²⁰ 
= 3.54 × 10²¹ molecules. 
Number of molecules removed = 10²¹ 
So, no. of remaining molecules 
molecules 
= 3.54 × 10²¹ - 1.00 × 10²¹ 
= 2.54 × 10²¹ molecules. 
∵6.022 × 10²³ H₂S molecules are present = 1 mol H₂S 
∵ For 2.54 × 10²¹ H₂S molecules = imm mol H₂S 
= 4.2 × 10^-3 mol H₂S 

Question 37. 
What will the number of molecules (density of water = 1g/mL) in 1L of water? 
Solution: 
Mass of 1L water 1 x 1 = 1g
So, 
no. of molecules = \(\frac{1}{18}\) × 6.022 × 10²³ 
= 3.34 × 10²² molecules 

Question 38. 
The molecular weight of haemoglobin is 89600. It has 0.25% (by mass) iron. Calculate the number of iron atoms present in one molecule of haemoglobin. 
Solution: 
According to question, 
100 g haemoglobin contains Fe = 0.25 g 
So, 89600 g haemoglobin will contain Fe 
= \(\frac{0.25}{100}\) × 89600 = 224 g Fe 
So, 1 mol haemoglobin contains Fe 
or 
= \(\frac{224}{56}\)
4 g Fe atoms 
Hence, the number of Fe atoms in 1 mol haemoglobin will be 4 g. 

Question 39. 
If 1 g of any gas at N.T.P. occupies 1400 mL volume, what is molecular weight of gas? How much molecules are present in gas at N.T.P.? 
Solution: 
Weight of 1400 mL gas = 1 g 
At N.T.P., weight of 22400 mL gas = \(\frac{22400}{1400}\)
Molecular weight of gas = 16 g 
Number of molecules, at N.T.P. in 22400 mL gas 
= 6.022 × 10²³ molecules 
Hence, number of molecules in 1 mL gas 
\(=\frac{6.022 \times 10^{23}}{22400}\)
= 2.688 × 10¹⁹ molecules 

Question 40. 
What will be volume of 61 g oxygen at N.T.P.? 
Solution: 
Molecular weight of O₂ = 16 x 2 = 32 
Volume of 32 g O₂ at N.T.P. = 22.4 L 
Volume of 16 g O₂ at N.T.P. =  \(\frac{22.4 \times 16}{32}\)
= 11.2 L 

Question 41. 
What will be the number of molecules in  10 mL water vapour at S.T.P. ? 
Solution: 
At S.T.P., volume of 6.022 × 10²³ molecules = 22.4 L 
At S.T.P., 22400 mL water vapour contains  = 6.022 × 10²³ molecules 
So, at S.T.P., 10 mL of water vapour will contain 
\(=\frac{6.022 \times 10^{23} \times 10}{22400}\)
= 2.69 × 10²⁰ molecules 

Question 42. 
Boron occurs in nature in its two isotopes whose atomic masses are 10 and respectively. What will be the % abundance of that sample which has average atomic mass as 10.8? 
Solution: 
Suppose, percentage of B10  = x 
B11 = 100 - x
Average atomic mass of boron 
\(\begin{aligned} & =\left(\frac{10 x}{100}\right)+\left[\frac{11(100-x)}{100}\right] \\ 10.8 & =\left(\frac{10 x}{100}\right)+\left[\frac{1100-11 x}{100}\right] \end{aligned}\)
10.8 × 100 = 1100 - x 
1080 = 1100x 
So, percentage of B10 and percentage of B11 = 100 - 20 = 80% 

Question 43. 
Find out the percentage of K, Cl and oxyge in KClO₃. 
Solution: 
Molar mass of KClO₃ 
= 39 + 35.5 + 3 × 16 = 122.5 
% of K = \(\frac{39 \times 100}{122.5}\) = 31.83% 

% of Cl = \(\frac{35.5}{122.5} \times 100\) = 28.97% 
% of O = \(\frac{48}{122.5} \times 100\) = 39.20% 

Question 44. 
Find out the percentage of crystallized wate in pure sample of washing soda (Na₂CO₃·10H₂O). 
Solution: 
Molecular weight of Na₂CO₃·10H₂O 
= 2 × atomic weight of Na + atomic weight of C + 3 atomic weight O + 10 × atomic weight of H₂O. 
= 2 × 23 + 12 + 3 × 16 + 10 × 18 
= 46 + 12 + 48 + 180
= 286 
Amount of H₂O in 286 g of Na₂CO₃·10H₂O = 180 g 
100 g Na₂CO₃·10H ₂O = \(\frac{180 \times 100}{286}\)
= 62.94% 

Question 45. 
In a compound, Na, S, H and O are presen The percentage of Na, S and H are 14.28% 9.92%, 6.20% respectively. Determine th molecular formula of compound. If a H-atoms are present in H₂O form  compound then give its crystallized formula when its molar mass is 322 g. 
Solution:
% of O = 100 - (% of Na + % of S + % of H) 
= 100 - (14.28 + 9.92 + 6.20) = 100 - 30.40 = 69.60 

So, empirical formula of compound = Na₂H₂OO₁₄ 
Empirical formula mass = 23 × 2 + 32 + 20 + 14 × 16 
46 + 32 + 20 + 224 
= 322 g 
Molar mass of compound = 322 g 

Molecular formula of compound is Na₂SH₂OO₁₄. The formula of crystallized compound will be Na₂SO₄10H₂O. 

Question 46. 
When a compound is analysed, following data were obtained. C = 54.5%, H= 9.9%, O = 36.41%, if vapour density of compound is 44.1 then determine its molecular formula. 
Solution: 

Empirical formula = C₂H₄O 
Vapour density = 44.1 
Molecular Weight = 2 × Vapour density 
= 2 × 44.1882 

So, molecular formula = 2 × C₂H₄O = C₄H₈O₂ 

Question 47. 
5.325 g carbon compound was analysed and it was found that it contains 3.758 g carbon, 0.316 g hydrogen and 1.251 g oxygen. Determine its empirical formula and molecular formula if its vapour density is 68. 
Solution: 
Mass of compound = 5.325 g 
Mass of carbon = 3.758 g 
Mass of hydrogen = 0.316 g 
Mass of oxygen = 1251 g 
% of carbon = \(\frac{3.758}{5.325} \) × 100 = 70.57% 
% of hydrogen  \(=\frac{0.316}{5.325}\) × 100 = 5.93% 
% of oxygen = \(\frac{1.251}{5.325}\) × 100 = 23.50% 

Empirical formula of compound = C₄H₄O 
Empirical formula mass = (4 × 12) + (4 × 1) + 16 = 68 
Molecular mass of compound = 2 × Vapour density 
= 2 × 68 = 136 

So, molecular formula = 2 × C₄H₄O 
= C₈H₈O₂ 

Question 48. 
50 g nitrogen gas and 8 g hydrogen gas are combined to produce NH₃ gas. Which is the limiting reagent of reaction? Calculate the amount of NH₃ gas produced. 
Solution: 
Balanced chemical equation is 

Number of moles of nitrogen = 50 g x \(\frac{1 \mathrm{~mol}}{28 \mathrm{~g}}\) = 1.786 mol 
Number of moles of hydrogen = 8 g × \(\frac{1 \mathrm{~mol}}{2 \mathrm{~g}}\) = 4 mol 
According to equation, 
Number of moles of H₂ reacts with 1.786 mol N₂ 
= 3/1 × 1.786 = 5.358 moles 
But moles present in H₂ = 4 
So complete H₂(g) will take part in reaction. So, H₂ iş limiting reagent. 
Amount of NH₃ (g) formed by 3 mol of H₂ = 2 mol 
Ammonia formed by 4 mol of H₂ = \(\frac{2}{3} \times 4 \mathrm{~mol}\)
= 2.66 mol 
2.66 mol = 2.66 × 17 =  45.22 g 

Question 49. 
3.0 g H₂ reacts with 30.0 g O₂ to form H₂O. What is the limiting reagent of reaction? Calculate the amount of H₂O formed. Also calculate the amount of excess reactant. 
Solution: 
Following will be chemical reaction: 

For limiting reagent, 
4 g H, reacts with = 32 g O₂
3 g H2 will react with = \(\frac{32 \times 3}{4}\) = 24 g O₂
But according to problem, we have 30 g 02. viz O₂ is in excess, So, the limiting reagent for this reaction is ydrogen. 
4 g H₂ form = 36 g H₂O 
3 g H₂ will produce = \(\frac{36 \times 3}{4}\)
= 27 g H₂O 

Question 50. 
Combustion of 1g Mg is furnished in a closed vessel which contains 0.5 g O₂. What is limiting reagent for reaction? Calculate the amount of MgO produced in reaction. 
Solution: 
Balanced equation is: 

∵ 48 g Mg react with = 32 g O₂ 
∴ 1g Mg will react with = \(\frac{32}{48}\)
But, as per question, amount of O, is 0.5 g which is less 
O2 is limiting reagent. 
Determination of amount of MgO 
32 g O₂ react with Mg and gives 80 g MgO. 
0.5 g O₂ will react with Mg to give = \(\frac{80 \times 0.5}{32}\)
= 1.25 g MgO

Question 51. 
The molecular weight of a salt is 40 g mol-1, if 4 g of salt is dissolved in 250 g H₂O. Calculate the molality of solution. 
Solution:
Weight of solute 'W' = 4g
Weight of solvent 'W' 250 g 

Now , molality
\(\begin{aligned} & =\frac{W_B \times 1000}{M_B \times W_A} \\ & =\frac{4 \times 1000}{40 \times 250} \end{aligned}\)
= 0.4 mol/Kg

Question 52. 
If 10 g oxalic acid is dissolved in 500 n solution (density = 1.1g mL^-1) then calculate the percentage mass of oxalic acid. 
solution. 
Solution: Weight of solute 'WR' = 10 g 
Density of solvent 'd' = 1.1 g/mL 
Mass of solution = Volume x Density 
= 500 × 1.1 
= 550 g 
Mass percentage 

= 1.82% 

Question 53. 
Density of 49% (weight percent) solution H₂SO₄ is 1.109 g/mL. Calculate the molality a molarity? 
Solution: 
49% (weight percent) means 49 g solu present in 100 g solution. 
Weight of solute 'WB' = 49 g 
Weight of Solvent 'WA' =100 - 49 = 51 g 
Molecular weight of solute 'MB' = 98 g mol^-1 (H₂SO₄)
Density of solution d = 1.109 g/mL 
So molality (m) 
\(\begin{aligned} & =\frac{W_B \times 100}{M_B \times W_A} \\ & =\frac{49 \times 1000}{98 \times 51} \end{aligned}\)
= 9.8 mol/kg 

Molarity (M) 

= 5.54 mol/L 

Question 54. 
In the mixture of 80 g water and 90 g methyl alcohol, calculate the mole fraction of bo compounds. 
Solution: 
Weight of H₂O 'WA' = 80 g 
Molecular weight of H₂O 'MA'= 18 g/mol 
Weight of CH₃OH 'WB' = 90 ġ 
Molecular Weight of CH₃OH 'MB' = 32 g/mol 
Mole fraction of CH₃OH =

= 0.387 
Mole fraction of H₂O, xA =  1 - XB 
= 1 - 0.387 
= 0.613 

Question 55. 
If 5.85 g NaCl is dissolved in 500 g H₂O. Calulate the molality of solution. 
Solution: 
Weight of solute 'WB' = 5.85 g 
Molecular weight of solute 'M' = 23 + 35.5 
= 58.5 g/mol 
Weight of solvent 'WA'  = 500 g
Molality (m) 
\(\begin{aligned} & =\frac{W_B \times 1000}{M_B \times W_A} \\ & =\frac{5.85 \times 1000}{58.5 \times 500} \end{aligned}\)
= 0.2 mol/kg 

Question 56. 
Density of aqueous solution of 2 M oxalic acid is 1.02 g/mL, if molecular weight of oxalic acid is 126g/mol then calculate the molality of solution. 
Solution: 
Molarity 'M' = 2 M 
Density 'd' = 1.02 g/mL 
Molecular weight of solute 'MB' = 126 g/mol 
Molality (m) = ? 
2 M' means that 2 mol of solute is dissolved in 1 L solution. 
Volume of solution = 1000 mL 
Density of solution = 1.02 g/mL 
Weight of solution = ? 
So Weight of solution = Density × Volume 
= 1.02 × 1000 = 1020 g
Moles of solute =
'WB' = Moles of solute × Molecular weight 
= 2 × 126 = 252 g 
Weight of solvent 'WA'  = Weight of solution - Weight of solute 

= 2.604 mol/kg.
 
Question 57. 
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) if solution contains 20% mass of C₂H₆O₂. 
Solution: 
20% C₂H₆O₂ means that 20 g C₂H₆O₂ is So present in 100 g solution. 
So, 
Weight of solute WB = 20 g 
Weight of solute WA = 100 - 20 = 80 g 
Weight of solvent W1 Molar mass of solute MB 
= 2 × 12 + 6 × 1 + 2 × 16 = 62 g/mol 
Molar mass of solvent MA = 18 g 

= 0.068 
Mole fraction of H₂O = 1 - X(glycol) 
= 1 - 0.068 
= 0.932 
So, mole fraction of glycol = 0.068 
mole fraction of H₂O = 0.932

 

Question 58. 
Write down the chemical equation for a reaction' and balance it by using fit and lapse method. 
Magnesium nitrite + Water → Magnesium hydroxide + Ammonia 
Solution: 
Chemical reaction will be: 
Mg₃N₂ + H₂O → Mg(OH)₂ + NH₃ 
Balancing of Equation 
(i) To balance the atoms of Mg 
Mg₃N₂ + H₂O → 3Mg(OH)₂ + NH₃ 
(ii) To balance atoms of O 
Mg₃N₂ + 6H₂O₃ → Mg(OH)₂ + NH₃ 
(iii) To balance atoms of N 
Mg₃N₂ + 6 H₂O → 3Mg(OH)₂ + 2NH₃ 
(iv) H atoms are already balanced. 
So
complete balanced equation will be 
Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃↑ 

Question 59. 
Find out the mass of CaO obtained when 100 kg of 90% pure CaCO₃ is heated. 
Solution :
CaCO₃ → CaO+ CO₂ 
Molar mass of CaCO₃ = 100 g 
90% pure CaCO₃ means 90 kg CaCO₃ is present 100 kg CaCO₃. 
So amount of pure CaCO₃ in 100 kg = 90 kg 
According to reaction, 

amount of CaO formed from 100g CaCO₃ = 56 g 
or from 100 kg CaCO₃, amount of CaO formed = 56 kg 
Therefore, CaO formed from 90 kg CaCO₃ 
= \(\frac{56}{100}\) = 50.4 kg 

Question 60.
On complete combustion of gaseous mixture of propane and butane, 10L CO₂ is formed. Calculate the composition of gaseous mixture. 
Solution: 
Suppose, volume of propane in mixture = xL 
Volume of butane in mixture = (3 - x) L 

1L C₃Hg gives CO₂ = 3 L. 
x L C₃H₈ gives CO₂ = 3x L 

1L C₄H₁₀ give CO₂ = 4L 
(3 - x) LC₄H₁₀ gives CO₂ = 4 × (3 - x) L 
Volume of free CO₂ = 10 L 
3x + 4(3 - x) = 10 
3x + 12 4x = 40 
- x = 10 -12 
x = 2 
Volume of propane = 2 L 
Volume of butane = (3 - 2) = 1L 

Competitive Exam Questions:
 
Question 1. 
The number of oxygen atoms in 4.4 g CO₂ : 
(a) 1.2 × 10²³
(b) 6 × 10²³ 
 (c) 6 × 10²³  
 (d) 12 × 10²³ 
Answer:
(a) 1.2 × 10²³

Question 2. 
When 2.46 g hydrated salt (MgSO₄.x H₂O) completely dehydrated, it gives 1.2 g dehydrated salt. If molecular mass of this salt is 120 g mol^-1, then value of x will be : 
(a) 2  
(b) 4 
(c) 5
(d) 6 
(e) 7 
Answer:
(e) 7 

Question 3. 
In a compound, the weight of sulphur is 8%. Least melecular weight is : 
(a) 200 
(b) 400 
(c) 155 
(d) 355 
Answer:
(b) 400 

Question 4. 
In which amount (g), of dioxygen there will  be 1.8 × 10²² molecules? 
(a) 9.60 
(b) 0.096 
(c) 96.0 
(d) 0.960 
Answer:
(d) 0.960 

Question 5. 
When 50 mL of 16.9% AgNO₃ is mixed with 50 mL of 5.8% NaCl. The weight of precipitate will be : 
(a) 28 g 
(b) 3.5 g
(c) 7 g 
(d) 14 g 
Answer:
(c) 7 g 

Question 6. 
What is the volume of H₂O used in decompostion of 1.368 kg sucrose? 
(a) 0.072 dm 
(b) 0.720 dm³
(c) 0.18 dm³  
(d) 0.018 dm 
Answer:
(a) 0.072 dm 

Question 7. 
The maximum number of water molecules is present in: 
(a) 18 molecules of H₂O 
(b) 1.8 g H₂O 
(c) 18 g H₂O 
(d) 18 mol H₂O 
Answer:
(d) 18 mol H₂O 

Question 8. 
If Avogadro number NA, 6.022 × 10²³ mol^-1 converts into 6.022 × 10²⁰ for carbon, then change in mass of 1 mol of carbon will be:
(a) 1.2 × 10^-3 g
(b) 12 × 10^-3 g
(c) 12 x 10^-3 g
(d) 12 × 10^-3 g
Answer:
(b) 12 × 10^-3 g

Question 9. 
When 20.0 g magnesium carbonte decomposes to give CO₂ and 8.0 g MgO, then what will be purity of Mg in sample? 
(a) 75 
(b) 96 
(c) 60 
(d) 84 
Answer:
(d) 84 

Question 10. 
When Mohr's salt is dissolved in excess of water, then how much ions molecules formed? 
(a) 6 
(b) 4 
(c) 10 
(d) 5 
Answer:
(d) 5 

Question 11. 
Which contains highest number of atoms? 
(a) 24 g of 12 C 
(b) 56 g of 56 Fe 
(c) 27 g of 27 Al 
(d) 108 g of Ag 
Answer:
(a) 24 g of 12 C 

Question 12. 
1.0 g Mg is burnt with 0.56 g O₂ in closed vessel, which reectant is left in excess and how much? 
(a) Mg, 0.44g 
(b) 02, 0.28 g 
(c) Mg, 0.16 g 
(d) 02 0.16 g 
Answer:
(c) Mg, 0.16 g 

Question 13. 
When 22.4L H₂ (g) is mixed with 11.2L Cl₂ (g) each at S.T.P., then how many moles of HCl(g) will be formed? 
(a) 0.5 mol HCl (g) 
(b) 1.5 mol HCl (g) 
(c) 1 mol HCl (g) 
(d) 2 mol HCl (g) 
Answer:
(c) 1 mol HCl (g) 

Question 14. 
How much silver (At. wt. 108) will be displaced by such amount of electricity which is used to displace 5600 mL O₂ at S.T.P? 
(a) 54.0 g 
(b) 108.0 g 
(c) 5.4 g 
(d) 10.8 g 
Answer:
(b) 108.0 g 

Question 15. 
What will be number of H⁺ ions present in 10 million parts of pure water in 1.33 cm' 3 at 25°C? 
(a) 6.023 million 
(b) 60 million 
(c) 8.01 million 
(d) 80.23 million 
Answer:
(c) 8.01 million 

Question  16. 
What will be volume of 0.1 M oxalic acid which is completely oxidised by 20 mL of 0.025 M solution of KMnO₄? 
(a) 12.5 mL 
(b) 25 mL 
(c) 125 mL 
(d) 37.5 mL 
Answer:
(a) 12.5 mL 

Question 17. 
At S.T.P, 2 g of 100% pure divalent carbonate produces 448 cc CO₂ on its complete thermal decomposition. What will be equivalent weight of metal? 
(a) 40
(b) 20  
(c) 28
(d) 12 
(e) 56 
Answer:
(b) 20  

Question 18. 
Which one of the following is the lightest? 
(a) 0.2 mol of H₂ gas 
(b) 6.022 × 10²² molecules of N₂ 
(c) 0.1 g of silver 
(d) 0.1 mol of O ₂(g) 
(e) 1 g H₂O 
Answer:
(c) 0.1 g of silver 

Question 19. 
The total number of electrons in 18 mL of H₂O will be : 
(a) 6.02 × 10²³ 
(b) 6.02 × 10²⁵ 
(c) 6.02 x 10²⁴ 
(d) 6.02 × 18 × 10²³ 
Answer:
(c) 6.02 x 10²⁴ 

Question 20. 
If the molecular weight of H₃PO, is M, its equivalent weight will be: 
(a) M 
(b) M/2 
(c) M/3 
(d) 2M 
Answer:
(b) M/2 

Question 21. 
Which has maximum number of molecules among the following? 
(a) 8g H₂ 
(b) 64g SO₂ 
(c) 44g CO₂ 
(d) 48g O₃ 
Answer:
(a) 8g H₂ 

Question 22. 
A  hydrocarbon contains 80% carbon, empirical formula of this will be : 
(a) CH₂
(b) CH₃  
(c) CH₄ 
(d) C₂H₃ 
Answer:
(b) CH₃  

Question 23. 
Arrange the following masses in increasing order: 
(I) 1 atom of oxygen 
(II) 1 atom of nitrogen 
(III) 1 x 10^-10 mol of oxygen 
(IV) 1 x 10^-10 mol of copper 
(a) II < I < III < IV 
(b) 1 < II < III < IV
(c) III < II < IV < I
(d) IV < II < III < I  
(e) II < IV < I< III 
Answer:
(a) II < I < III < IV