I. Short answer type questions:
1. Who provides the basic theory about the nature of matter?
Ans. Dalton provides the basic theory about the nature of matter.
2. State Dalton's atomic theory.
Ans. According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms.
3. What are binary compounds?
Ans. The simplest compounds, which are made up of two different elements are called binary compounds.
Ans. The simplest compounds, which are made up of two different elements are called binary compounds.
Ex - NaCl, H2SO4, etc.
4. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans. The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction.
5. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans. The law of definite proportions were stated by Proust as “In a chemical substance the elements are always present in definite proportions by mass”.
6. What is an Atom?
Ans. All matter is made of very tiny particles called atoms and are indivisible particles, which cannot be created or destroyed in a chemical reaction.
7. What is Avogadro's number or constant?
Ans. The number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed, with a value of 6.022 × 1023. This number is called the Avogadro Constant or Avogadro Number.
8. Write the full forms of IUPAC.
Ans. International Union of Pure Applied Chemistry.
9. What is the difference between CO and Co?
Ans. CO is a compound which is carbon mono oxide consists of carbon and oxygen and Co is the symbol of cobalt which is an element.
10. Write the symbol and Latin names of elements - Iron, sodium, and potassium.
Ans. Symbol Latin name
1. Iron Ferrum
2. Sodium Natrium
3. Potassium Potash
11. Why is it not possible to see an atom with naked eyes?
Ans. It is not possible to see an atom with naked eyes because atoms are very small, they are smaller than anything that we can imagine or compare with.
Ans. It is not possible to see an atom with naked eyes because atoms are very small, they are smaller than anything that we can imagine or compare with.
12. What is a Molecule?
Ans. The smallest particle of an element or a compound that is capable of independent existence and shows all the properties of that substance. Atoms of the same element or of different elements can join together to form molecules.
13. Define valency.
Ans. The combining capacity of an element is known as its valency and can be used to find out how the atoms of an element will combine with the atom of another element to form a chemical compound.
14. How atom do exists?
Ans. Atoms of most elements are not able to exist independently. Atoms form molecules and ions. These molecules or ions aggregate in large numbers to form the matter that we can see, feel or touch.
15. What is the difference between 2H and H2.?
Ans. 2H means two atoms of hydrogen and H2 means two atoms of hydrogen or one molecule of hydrogen gas.
16. What is an ion?
Ans. Compounds composed of metals and nonmetals contain charged species. The charged species are known as ions. An ion is a charged particle and can be negatively or positively charged.
17. Name the anion and cation which constitute the molecule of Sodium chloride?
Ans. Sodium chloride's constituent particles are positively charged sodium ions (Na+) and negatively charged chloride ions (Cl– ).
18. How many atoms are there in 1 gm of hydrogen?
Ans. One atom is there in 1 gm of hydrogen.
19. Define the ionic compound?
Ans. An ionic compound is a chemical compound composed of ions held together by electrostatic forces termed ionic bonding. The compound is neutral overall but consists of positively charged ions called cations and negatively charged ions called anions.
20. Define molecular mass?
Ans. The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is, therefore, the relative mass of a molecule expressed in atomic mass units (u).
21. What is the difference between the formula unit mass and molecular mass?
Ans. The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a compound.
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of a substance whose constituent particles are ions.
22. Define mole.
Ans. A mole is the amount of substance that contains as many atoms, molecules, ions, electrons or any other elementary entities as there are carbon atoms in exactly 12 gm of.
23. Write the name of elements present in the common salt sample?
Ans. Sodium and chlorine are present in the common salt.
24. The chemical formula of a metal Sulphate is MSO4. What will be the formula of its chloride?
Ans. In the formula, MSO4 Valency of M is 2. Therefor formula of its chloride will be MCl2.
25. Write the chemical formula of ammonium chloride and sodium phosphate?
Ans. Ammonium chloride - NH4Cl
Sodium phosphate - Na3PO4
26. Define the atomic mass.
Ans. The mass of an atom of a chemical element expressed in atomic mass units which is equivalent to the number of protons and neutrons in the atom. It is also known as the mass number
II. Short answer type questions:
1. What are the Laws of Chemical Combination?
Ans. There are two laws of chemical combination.
Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction.
Law of constant proportions states that in a chemical substance the elements are always present in definite proportions by mass.
2. What is Dalton's atomic theory?
Ans. According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms.
All matter is made of very tiny particles called atoms which are indivisible particles, which cannot be created or destroyed in a chemical reaction.
3. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.
From this we have ;
1 gm of hydrogen gas required to react completely with 8 g of oxygen gas.
3 gm of hydrogen gas would be required to react completely with 8 x 3= 24 g of hydrogen gas.
4. Define chemical formula. How many atoms are present in a
(i) H2S molecule and (ii) PO4-2 ion?
(i) H2S molecule and (ii) PO4-2 ion?
Ans. The chemical formula of a compound is a symbolic representation of its composition.
(i) H2S molecule - In this molecule 2 atoms of hydrogen and 1 atoms of sulfur present. Therefore total no. of atoms = 3
(ii) PO4-2 ion - In this molecule 1 atom of phosphorous and 4 atoms of oxygen present. Therefore total no. of atoms = 5
5. Define polyatomic ion. Name two elements that exist as independent atoms?
Ans. Ions which are formed from groups of joined atoms are called compound ions or polyatomic ions. Ex - CO2-2, SO4-2,NH+4
6. Define the atomic mass unit and relative atomic mass.
Ans. Atomic mass - The atomic mass of an element is the relative mass of its atoms as compared with the mass of a carbon -12 atoms
Relative atomic mass - The relative atomic mass of an element is the number which tells how many times an atom of that element is heavier than an atom of a standard substance.
7. (a)Define atomicity.
(b)Write the atomicity of the following molecules:
(i) H2SO4 (ii) CCl4
Ans. (i) The atomicity of a substance is the number of atoms present in a single molecule of a substance. They may be monoatomic Na, Ca , diatomic O2, N2 , H2 , triatomic O3, and polyatomic S8.
(ii) Atomicity of H2SO4 - 7 ( H-2, S-1, O-4)
Atomicity of CCl4 - 5 ( C-1, Cl-4)
8. Valency of an element X is 3 while that of Y is 1.Write the formula of the compounds formed when each of them separately combines with oxygen?
Ans. Valency of an element X = 3 and O = 2
The formula of the compound formed = X2O3
Valency of Y = 1 and O = 2
The formula of the compound formed = Y2O
9. What is the formula unit mass? Calculate the formula unit masses of ZnO, Na2O, K2CO3, (NH4)2CO3 (given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u, N = 14).
Ans. The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a compound.
Formula unit mass of :
ZnO - Zn + O = 65 + 16 = 81unit.
Na2O - 2 x Na + O = 2 x 23 + 16 = 46 + 16 = 62unit
K2CO3 - 2 x K + C + 3 x O = 2 x 39 + 12 + 3 x 16 = 78 + 12 + 48 =138 unit.
(NH4)2CO3 - 2(N + 4 x H) + C + 3 x O = 2( 14 + 4 x 1)+ 12 + 3 x 16 = 36 + 12 + 48 = 96 unit
10. Calculate the molecular masses of
H2, O2, Cl2, CO2, CH4, C2H4, NH3, CH3O H.
Ans. The molecular mass of
H2 - 2 x 1 = 2 g
O2 - 2 x 16 = 42 g
CO2 - 12 + 2 x 16 = 12 + 32 = 44 g
Ans. Ions which are formed from groups of joined atoms are called compound ions or polyatomic ions. Ex - CO2-2, SO4-2,NH+4
6. Define the atomic mass unit and relative atomic mass.
Ans. Atomic mass - The atomic mass of an element is the relative mass of its atoms as compared with the mass of a carbon -12 atoms
Relative atomic mass - The relative atomic mass of an element is the number which tells how many times an atom of that element is heavier than an atom of a standard substance.
7. (a)Define atomicity.
(b)Write the atomicity of the following molecules:
(i) H2SO4 (ii) CCl4
Ans. (i) The atomicity of a substance is the number of atoms present in a single molecule of a substance. They may be monoatomic Na, Ca , diatomic O2, N2 , H2 , triatomic O3, and polyatomic S8.
(ii) Atomicity of H2SO4 - 7 ( H-2, S-1, O-4)
Atomicity of CCl4 - 5 ( C-1, Cl-4)
8. Valency of an element X is 3 while that of Y is 1.Write the formula of the compounds formed when each of them separately combines with oxygen?
Ans. Valency of an element X = 3 and O = 2
The formula of the compound formed = X2O3
Valency of Y = 1 and O = 2
The formula of the compound formed = Y2O
9. What is the formula unit mass? Calculate the formula unit masses of ZnO, Na2O, K2CO3, (NH4)2CO3 (given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u, N = 14).
Ans. The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a compound.
Formula unit mass of :
ZnO - Zn + O = 65 + 16 = 81unit.
Na2O - 2 x Na + O = 2 x 23 + 16 = 46 + 16 = 62unit
K2CO3 - 2 x K + C + 3 x O = 2 x 39 + 12 + 3 x 16 = 78 + 12 + 48 =138 unit.
(NH4)2CO3 - 2(N + 4 x H) + C + 3 x O = 2( 14 + 4 x 1)+ 12 + 3 x 16 = 36 + 12 + 48 = 96 unit
10. Calculate the molecular masses of
H2, O2, Cl2, CO2, CH4, C2H4, NH3, CH3O H.
Ans. The molecular mass of
H2 - 2 x 1 = 2 g
O2 - 2 x 16 = 42 g
CO2 - 12 + 2 x 16 = 12 + 32 = 44 g
C2H4 - 12 x 2 + 4 x 1 =28 g
NH3 - 14 + 3 x 1 = 17g
CH3OH - 12 + 3 x 1 + 16 + 1 = 32g
11. If 12 gm of carbon is burnt in the presence of 32 gm oxygen, how much carbon dioxide will be formed?
Ans. The chemical reaction between carbon and oxygen :
C + O2 → CO2
12g 32 12 + 32 = 44g CO2 formed.
12. Calculate the molar mass of (i) Sulphur molecule (ii) Ethyne (iii)Phosphorus molecule.
Ans. (i) The molar mass of Sulphur molecule = S8 = 32 x 8 = 256g
(ii) Ethyne C2H2 = 2 x 12 + 2 x 1 = 26 g
(iii) Phosphorus molecule P4 - 31 x 4 = 124g.
13. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, the atomic mass of Na = 23 u, Fe = 56 u)?
Ans. The molecular mass of Na = 23 g
No. of atoms = (Avogadro's no. x Given mass)/Molecular mass.
No. atoms = (6.022 x 1023) x 100
23
= 2.62 x 1026
The molecular mass of Fe = 56 g
No. atoms = (6.022 x 1023) x 100
56
= 1.08 x 1026
Na has more number of atoms.14. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Ans. The molecular mass of S8 = 32 x 8 = 256g
Given mass = 16 g
No. molecules = (Avogadro's no x Given mass) / molar mass
= 6.022 x 1023 x 16 / 256
= 3.76 x 1022
15. Calculate the formula unit mass of CaCl2.
Ans. Formula unit mass of CaCl2 - Ca + Cl x 2 = 20 + 2 x 35 = 90g
16. (a) Calculate the relative molecular mass of water (H2O).
(b) Calculate the molecular mass of HNO3
Ans. (a) Formula unit mass of H2O - 2 x 1 + 16 = 18 g
(b) Formula unit mass of HNO3 - H + N + O x 3 = 1 + 14 + 16 x 3 = 63 g
III. Long answer type questions:
1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water.
Ans. In this chemical reaction:
Na2CO3 + CH3OOH → CH3OONa + CO3 + H2O
5.3 g + 6 g → 8.2 g + 2.2 g + 0.9 g
11.3 g → 11.3 g
Mass of reactant = Mas of product
These observations are in agreement with the law of conservation of mass.
2. Write the symbols of (i) Silver (ii) Mercury (iii) Magnesium (iv) Chromium?
Ans. Symbol of
(i) Silver - Ag
(ii) Mercury - Hg
(iii) Magnesium - Mg
(iii) Chromium - Cr
3. (a) Write a chemical formula of a compound using zinc ion, phosphate ion, Sodium carbonate, lead carbonate, and Ammonium chloride?
(b) Calculate the ratio by mass of atoms present in a molecule of carbon dioxide (CO2).
Ans. (a)
4. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3.
(i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3.
Ans. (i)Al2(SO4)3 - Aluminum Sulphate
(ii) CaCl2 - Calcium Chloride
(iii) K2SO4 - Potassium Sulphate
(iV) KNO3 - Potassium Nitrate
(V) CaCO3 - Calcium Carbonate.
5. Define
(i) Diatomic
(ii) Triatomic
(iii)Tetra atomic
(iv) Polyatomic molecule of elements.
(i) Diatomic
(ii) Triatomic
(iii)Tetra atomic
(iv) Polyatomic molecule of elements.
Ans. (i) Diatomic - Diatomic elements are elements that form molecules consisting of two atoms. Ex - O2 , N2, H2
(ii) Triatomic - Triatomic elements are elements that form molecules consisting of three atoms. Ex - O3
(iv) Tetra atomic - Tetra atomic elements are elements that form molecules consisting of four atoms. Ex - P4
(iV) Polyatomic molecules of elements - Polyatomic atomic elements are elements that form molecules consisting of more than 4 atoms. Ex - S8 .
6. What is the modern-day symbol of atoms of different elements Hydrogen, Carbon, Oxygen, Phosphorous, Sulphur, Iron, Copper, Lead, Silver, Gold, Platinum, and Mercury?
Ans.
7. Calculate the no. of moles in :
(i) 17 gm of H2O (ii) 5.75 gm of sodium (iii) 52 gm of He (iv) 34 gm of NH3 (v) 60 gm. of Ca.
Ans. (i) Molecular mass of H2O - 2 x 1 + 16 = 18 g
18 g of water = 1 mole.
Therefore 17 g of water = 1/18 x 17 = 17/18 = 0.94 mole.
(ii) Molecular mass of Sodium Na = 23 g
23 g of sodium = 1 mole.
5.75 g of sodium = 1/23 x 5.75 = 5.75/23 = 0.25 mole.
(iii) Molecular mass of He = 4 g
4 g 0f He = 1 mole
52 g of He = 1/4 x 52 = 52/4 g = 13 mole.
(iv) Molecular mass of ammonium = 14 + 3 x 1 =17 g.
17 g of ammonium = 1 mole
34 g of ammoniu, = 1/17 x 34= 34/17 = 2 mole.
(v) Molecular mass of calcium = 40 g
40 g of molecular mass = 1 mole.
60 g of molecular mass = 1/40 x 60 = 60/40 = 3/2 = 1.5 mole.
8. (a) Calculate the no. of molecules in 8 gm of oxygen?
(b) 5 gm of calcium combines with 2gm of oxygen to form a compound. Find the molecular formula of the compound. (A.M of Ca=40u O=16u).
Ans. (a) The molecular mass of oxygen is 16 g.
Given mass = 8 g
No. of molecules = Avogadro's no. x Given mass /molecular mass
= 6.022 x 1023 x 8 / 16
= 3.011 x 1023
(b) Number of moles in 5g of calcium = mass / molar mass
= (5 / 40) = 0.125 moles
Number of moles in 2g of oxygen = mass / molar mass
= (2 / 16) = 0.125 moles
Since the number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.
= (2 / 16) = 0.125 moles
Since the number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.
9. Write the chemical formula of bicarbonates of Na+, K+1, Al+3, Mg+2, Ca+2 and Zn+2.
Ans. The chemical formula of bicarbonates of
(i) Na+ - NaHCO3
(ii) K+1 - KHCO3
Ans. The chemical formula of bicarbonates of
(i) Na+ - NaHCO3
(ii) K+1 - KHCO3
(iii) Al+3 - Al(HCO3)3
(iv) Mg+2 - Mg(HCO3)2
(v) Ca+2 - Ca(HCO3)2
(vi) Zn+2 - Zn(HCO3)2
10. The percentage of three elements Ca, C and O2 in a sample of calcium carbonate is given as Ca=40%, C=12.0%, O=48%. If the law of proportion is true, what weights of these elements will be present in 1.5g of another sample of Ca Carbonate? (A.m. of Ca=20, C=12 O=16).
Ans. The percentage of three elements Ca, C and O2 in a sample of calcium carbonate is given as Ca=40%, C=12.0%, O=48%.
Wt of Ca in 1.5 g fo sample of Calcium Carbonate = 40/100 x 1.5 = 0.6 g
Wt of C in 1.5 g fo sample of Calcium Carbonate = 12/100 x 1.5 = 0.18g
Wt of O in 1.5 g fo sample of Calcium Carbonate = 48/100 x 1.5 = 0.72g
11. (a) An element X has valency 3. Write the chemical formula of its chloride.
(b) If an element Y has valency 2 then write the chemical formula of its potassium salt?
Ans. (a) If X has valency 3 then the chemical formula of its chloride will be =XCl3
(b) If Y has valency 2 then the chemical formula of its potassium salt will be =K2Y
12. Calculate the number of moles for the following:
(i) 52 g of He (finding mole from mass)
(ii) 12.044 ×1023 number of He atoms (finding mole from a number of particles).
(iii) 3.011 x 1023 no. of oxygen atoms.
Ans. (i) No. moles in 52 g of He = Given mass/Molecular mass = 52/4
= 13 moles.
(ii) No. moles of 12.044 ×1023 number of He atoms = Given no. of atoms /Avogadro's no. = 12.044 ×1023/6.022 ×1023 = 2 moles.
(iii) No. moles in 3.011 × 1023 number of O atoms
= Given no. of atoms /Avogadro's no.
= 3.011 ×1023/6.022 ×1023 = 0.5 moles.
13. Calculate the mass for the following:
(i) 0.5 mole of N2 gas
(ii) 0.5 mole of N atoms
(iii) 3.011 × 1023 number of N atoms.
(iv) 6.022 × 1023 number of N2 molecules.
Ans. (i) The molecular mass of N2 gas = 2 x 14 = 28g
mass = molecular mass x no. of moles = 28 x 0.5 = 14g
(ii) The molecular mass of N atoms = 14g
mass = molecular mass x no. of moles =14 x 0.5 = 7g
(iii) The molecular mass of N atoms = 14g
No. of moles = Given no. of atoms /Avogadro's no. = 3.011 × 1023/ 6.022× 1023 = 0.5 moles
mass = molecular mass x no. of moles
mass = 14 x 0.5 = 7g
(iv)The molecular mass of N2 molecules. = 28g
No. of moles = Given no. of atoms /Avogadro's no. = 6.022 × 1023/ 6.022× 1023 = 1 moles
mass = molecular mass x no. of moles
mass = 28 x 1 = 28g
14. When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Ans. The chemical reaction between carbon and oxygen:
C + O2 → CO2
12g 32g 44g (mass of the reactants = mass of the products.)
The ratio between C and O = 12:32 = 3:8
When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8 gm oxygen
will use and the same 11g of CO2 will be produced.
The law of conservation of mass will govern the answer.
15. A 0.24 gm of a sample of a compound of oxygen and boron was found by analysis to contain 0.096 gm of boron and 0.144 gm of oxygen.
(i) Calculate the % composition of the compound by weight.
(ii) Find the no. of moles in 0.144g of oxygen gas.
Ans. (i) % of the boron of the compound by weight
= 0.096 gm of boron x 100 /0.24 = 40%
% of the oxygen of the compound by weight
= 0.144 gm of oxygen x 100 /0.24 = 60%
(ii) The no. of moles in 0.144 g of oxygen gas .= Given mass/Molecular mass
= 0.144/16 = 0.009moles
16. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Ans. The chemical formulae of the following.
(a) Magnesium chloride - MgCl2
(b) Calcium oxide - CaO
(c) Copper nitrate - CuNO3
(d) Aluminium chloride - AlCl3
(e) Calcium carbonate - CaCO3
17. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium Sulphate.
Ans. (a) Quick lime (CaO) - Calcium and oxygen.
(b) Hydrogen bromide (HBr) - Hydrogen and bromine.
(c) Baking powder (NaHCO3) - Sodium, hydrogen, carbon and oxygen.
(d) Potassium Sulphate (K2SO4) - Potassium, sulphate, and oxygen.
18. Calculate the molar mass of the following substances.
(a) Ethyne - C2H2
(b) Sulphur molecule - S8
(c) Phosphorus molecule- P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid - HCl
(e) Nitric acid - HNO3
Ans. The molar mass of
(a) Ethyne - C2H2 - 2 x 12 + 2 x 1 = 26g
(b) Sulphur molecule - S8 - 8 x 32 = 256 g
(c) Phosphorus molecule- P4 - 4 x 31 = 124g
(d) Hydrochloric acid HCl - 1 x 1 + 35 = 36g
(e) Nitric acid - HNO3 - 1 x 1 + 14 + 3 x 16 = 63g
19. What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of Aluminium atoms (Atomic mass of Aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)
Ans.(a) Mass of 1 mole of nitrogen atom - 14 g
(b) Mass of 4 moles of Aluminium atoms - 4 x 27 = 108g
(c) 10 moles of sodium sulphite (Na2SO3) - 10 (2 x 23 + 32 + 3 x 16) =1260g
20. Calculate the number of Aluminium ions present in 0.051 g of Aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. The atomic mass of Al = 27 u).
Ans. The atomic mass of Al = 27u
Given mass = 0.051g
No. of aluminium ions = (Avogadros no. x Given mass )/Molecular mass
No. of aluminium ions = (6.022 x 1023 x 0.051)/27 = 1.1375 x 1021 ions.
21. Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide.
Ans. The formula of
(a) Sodium oxide - Na2O
(b) Aliminium chloride - AlCl3
(c) Sodium sulphide - Na2S
(iv) Magnesium hydroxide Mg(OH)2
22. Convert into a mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Ans. (a) 12 g of oxygen gas - molecula mass of oxygen gas = 2 x 16 = 32 g
No.of mole = Given mass/molecular mass = 12/32= 0.38 mole.
(b) 20 g of water - molecular mass of water = 2 x 1 + 16 = 18 g
No.of mole = Given mass/molecular mass = 20/18= 1.11 mole.
(c) 22 g of carbon dioxide - molecular mass of carbon dioxide
= 12 + 2 x 16=44g
No.of mole = Given mass/molecular mass = 22/44= 0.5 mole.
23. Calculate the no of particles in each of the following:
(i) 7g of nitrogen molecules
(ii) 0.5 mole of carbon atoms.
(iii) 8 gm of oxygen molecule.
(iv) 0.1 mole of carbon.
(v) 42 g of N atoms.
(Given N=14, C=12, N0=6.022 × 1023 )
Ans. No. of particles in
(i) 7g of nitrogen molecules - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 7)/ 14 = 6.022 × 1023/2 = 3.011 × 1023no. of particles.
(ii) 0.5 mole of carbon atoms - Avogadro's no x given mole
= 6.022 × 1023 x 0.5 = 3.011 × 1023no. of particles.
(iii) 8g of oxygen molecules - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 8)/ 32 = 6.022 × 1023/4 = 1.51 × 1023no. of particles.
(iv) 0.1mole of carbon atoms - Avogadro's no x given mole
= 6.022 × 1023 x 0.1 = 6.022 × 1022no. of particles.
(v) 42 g of N atoms - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 42)/ 7 = 6.022 × 1023 x 6 = 36.13 × 1023no. of particles.
(iv) Mg+2 - Mg(HCO3)2
(v) Ca+2 - Ca(HCO3)2
(vi) Zn+2 - Zn(HCO3)2
10. The percentage of three elements Ca, C and O2 in a sample of calcium carbonate is given as Ca=40%, C=12.0%, O=48%. If the law of proportion is true, what weights of these elements will be present in 1.5g of another sample of Ca Carbonate? (A.m. of Ca=20, C=12 O=16).
Ans. The percentage of three elements Ca, C and O2 in a sample of calcium carbonate is given as Ca=40%, C=12.0%, O=48%.
Wt of Ca in 1.5 g fo sample of Calcium Carbonate = 40/100 x 1.5 = 0.6 g
Wt of C in 1.5 g fo sample of Calcium Carbonate = 12/100 x 1.5 = 0.18g
Wt of O in 1.5 g fo sample of Calcium Carbonate = 48/100 x 1.5 = 0.72g
11. (a) An element X has valency 3. Write the chemical formula of its chloride.
(b) If an element Y has valency 2 then write the chemical formula of its potassium salt?
Ans. (a) If X has valency 3 then the chemical formula of its chloride will be =XCl3
(b) If Y has valency 2 then the chemical formula of its potassium salt will be =K2Y
12. Calculate the number of moles for the following:
(i) 52 g of He (finding mole from mass)
(ii) 12.044 ×1023 number of He atoms (finding mole from a number of particles).
(iii) 3.011 x 1023 no. of oxygen atoms.
Ans. (i) No. moles in 52 g of He = Given mass/Molecular mass = 52/4
= 13 moles.
(ii) No. moles of 12.044 ×1023 number of He atoms = Given no. of atoms /Avogadro's no. = 12.044 ×1023/6.022 ×1023 = 2 moles.
(iii) No. moles in 3.011 × 1023 number of O atoms
= Given no. of atoms /Avogadro's no.
= 3.011 ×1023/6.022 ×1023 = 0.5 moles.
13. Calculate the mass for the following:
(i) 0.5 mole of N2 gas
(ii) 0.5 mole of N atoms
(iii) 3.011 × 1023 number of N atoms.
(iv) 6.022 × 1023 number of N2 molecules.
Ans. (i) The molecular mass of N2 gas = 2 x 14 = 28g
mass = molecular mass x no. of moles = 28 x 0.5 = 14g
(ii) The molecular mass of N atoms = 14g
mass = molecular mass x no. of moles =14 x 0.5 = 7g
(iii) The molecular mass of N atoms = 14g
No. of moles = Given no. of atoms /Avogadro's no. = 3.011 × 1023/ 6.022× 1023 = 0.5 moles
mass = molecular mass x no. of moles
mass = 14 x 0.5 = 7g
(iv)The molecular mass of N2 molecules. = 28g
No. of moles = Given no. of atoms /Avogadro's no. = 6.022 × 1023/ 6.022× 1023 = 1 moles
mass = molecular mass x no. of moles
mass = 28 x 1 = 28g
14. When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Ans. The chemical reaction between carbon and oxygen:
C + O2 → CO2
12g 32g 44g (mass of the reactants = mass of the products.)
The ratio between C and O = 12:32 = 3:8
When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8 gm oxygen
will use and the same 11g of CO2 will be produced.
The law of conservation of mass will govern the answer.
15. A 0.24 gm of a sample of a compound of oxygen and boron was found by analysis to contain 0.096 gm of boron and 0.144 gm of oxygen.
(i) Calculate the % composition of the compound by weight.
(ii) Find the no. of moles in 0.144g of oxygen gas.
Ans. (i) % of the boron of the compound by weight
= 0.096 gm of boron x 100 /0.24 = 40%
% of the oxygen of the compound by weight
= 0.144 gm of oxygen x 100 /0.24 = 60%
(ii) The no. of moles in 0.144 g of oxygen gas .= Given mass/Molecular mass
= 0.144/16 = 0.009moles
16. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Ans. The chemical formulae of the following.
(a) Magnesium chloride - MgCl2
(b) Calcium oxide - CaO
(c) Copper nitrate - CuNO3
(d) Aluminium chloride - AlCl3
(e) Calcium carbonate - CaCO3
17. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium Sulphate.
Ans. (a) Quick lime (CaO) - Calcium and oxygen.
(b) Hydrogen bromide (HBr) - Hydrogen and bromine.
(c) Baking powder (NaHCO3) - Sodium, hydrogen, carbon and oxygen.
(d) Potassium Sulphate (K2SO4) - Potassium, sulphate, and oxygen.
18. Calculate the molar mass of the following substances.
(a) Ethyne - C2H2
(b) Sulphur molecule - S8
(c) Phosphorus molecule- P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid - HCl
(e) Nitric acid - HNO3
Ans. The molar mass of
(a) Ethyne - C2H2 - 2 x 12 + 2 x 1 = 26g
(b) Sulphur molecule - S8 - 8 x 32 = 256 g
(c) Phosphorus molecule- P4 - 4 x 31 = 124g
(d) Hydrochloric acid HCl - 1 x 1 + 35 = 36g
(e) Nitric acid - HNO3 - 1 x 1 + 14 + 3 x 16 = 63g
19. What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of Aluminium atoms (Atomic mass of Aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)
Ans.(a) Mass of 1 mole of nitrogen atom - 14 g
(b) Mass of 4 moles of Aluminium atoms - 4 x 27 = 108g
(c) 10 moles of sodium sulphite (Na2SO3) - 10 (2 x 23 + 32 + 3 x 16) =1260g
20. Calculate the number of Aluminium ions present in 0.051 g of Aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. The atomic mass of Al = 27 u).
Ans. The atomic mass of Al = 27u
Given mass = 0.051g
No. of aluminium ions = (Avogadros no. x Given mass )/Molecular mass
No. of aluminium ions = (6.022 x 1023 x 0.051)/27 = 1.1375 x 1021 ions.
21. Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide.
Ans. The formula of
(a) Sodium oxide - Na2O
(b) Aliminium chloride - AlCl3
(c) Sodium sulphide - Na2S
(iv) Magnesium hydroxide Mg(OH)2
22. Convert into a mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Ans. (a) 12 g of oxygen gas - molecula mass of oxygen gas = 2 x 16 = 32 g
No.of mole = Given mass/molecular mass = 12/32= 0.38 mole.
(b) 20 g of water - molecular mass of water = 2 x 1 + 16 = 18 g
No.of mole = Given mass/molecular mass = 20/18= 1.11 mole.
(c) 22 g of carbon dioxide - molecular mass of carbon dioxide
= 12 + 2 x 16=44g
No.of mole = Given mass/molecular mass = 22/44= 0.5 mole.
23. Calculate the no of particles in each of the following:
(i) 7g of nitrogen molecules
(ii) 0.5 mole of carbon atoms.
(iii) 8 gm of oxygen molecule.
(iv) 0.1 mole of carbon.
(v) 42 g of N atoms.
(Given N=14, C=12, N0=6.022 × 1023 )
Ans. No. of particles in
(i) 7g of nitrogen molecules - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 7)/ 14 = 6.022 × 1023/2 = 3.011 × 1023no. of particles.
(ii) 0.5 mole of carbon atoms - Avogadro's no x given mole
= 6.022 × 1023 x 0.5 = 3.011 × 1023no. of particles.
(iii) 8g of oxygen molecules - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 8)/ 32 = 6.022 × 1023/4 = 1.51 × 1023no. of particles.
(iv) 0.1mole of carbon atoms - Avogadro's no x given mole
= 6.022 × 1023 x 0.1 = 6.022 × 1022no. of particles.
(v) 42 g of N atoms - (Avogadro's no x given mass) /molecular mass
= 6.022 × 1023 x 42)/ 7 = 6.022 × 1023 x 6 = 36.13 × 1023no. of particles.
No comments:
Post a Comment