Chapter No. 1
BASIC CONCEPTS
MCQs
MCQs
Q.1 Smallest particle of an element which may or may not have independent existence
(a) a molecule (b) an atom
(c) an ion (d) an electron
Q.2 Swedish chemist J. Berzelius determined the
(a) atomic no. (b) atomic volume
(c) atomic mass (d) atomic density
Q.3 The number of atoms present in a molecule determine its
(a) molecularity (b) basicity
(c) acidity (d) atomicity
Q.4 When an electron is added to a unipositive ion we get
(a) anion (b) cation
(c) neutral atom (d) molecule
Q.5 CO+ is an example of:
(a) free radical (b) cationic molecular ion
(c) an ionic molecular ion
(d) stable molecule
Q.6 Relative atomic mass is the mass of an atom of an element as compared to the mass of
(a) oxygen (b) hydrogen
(c) nitrogen (d) carbon
Q.7 Isotopes are the sister atoms of the same element with similar chemical properties and different
(a) atomic number (b) atomic mass
(c) atomic volume (d) atomic structure
Q.8 The instrument which is used to measure the exact masses of different isotopes of an element called
(a) I.R. Spectrophotometer (b) U.V. Spectrophotometer
(c) Mass Spectrometer (d) Colourimeter
Q.9 Mass spectrometer separates different positive isotopic ions on the basis of their
(a) mass value (b) m/e value
(c) e/m value (d) change value
Q.10 Simplest formula that gives us information about the simple ratio of atoms in a compound is called
(a) structural formula (b) molecular formula
(c) empirical formula (d) molar ratio
Q.11 Percentage of oxygen in H2O is
(a) 80% (b) 88.8%
(c) 8.8% (d) 9.8%
Q.12 More abundant isotope of an element is one with
(a) even atomic no. (b) odd atomic no.
(c) Even mass no. (d) odd mass no.
Q.13 Large no. of isotopes are known for the elements whose masses are multiple of
(a) two (b) four
(c) six (d) eight
Q.14 When 0.01 kg of CaCO3 is decomposed the CO2 produced occupies a volume at S.T.P.
(a) 2.2414 dm3 (b) 22.414 dm3
(c) 22414 dm3 (d) 224014 dm3
Q.15 The no. of covalent bond in 10gm of NH3 are
(a) 6.022 x 1023 (b) 1.062 x 1023
(c) 10.62 x 1024 (d) 1.062 x 1024
Q.16 No. of molecules present in 10gm of water are
(a) 3.37 x 1023 (b) 33.7 x 1023
(c) 3.37 x 1024 (d) 3.037 x 1024
Q.17 The no. of covalent bonds present in 10gm of water are
(a) 6.074 x 1023 (b) 6.74 x 1023
(c) 6.074 x 1024 (d) 6.74 x 1024
Q.18 The least no. of molecules present in 30 gm of
(a) N2O (b) NO
(c) NO2 (d) N2O3
Q.19 Which of the following has highest percentage of nitrogen
(a) (NH4)2SO4 (b) NH4H2PO4
(c) (NH4)2HPO4 (d) (NH4)3PO4
Q.20 0.1 mole of Na3PO4 completely dissociates in water to produce Na+
(a) 6.02 x 1022 (b) 6.02 x 1023
(c) 1.806 x 1023 (d) 1.806 x 1022
Q.21 Efficiency of chemical reaction can be checked by calculating
(a) amount of limiting reactant
(b) amount of the reactant in excess
(c) amount of the product formed
(d) amount of the reactant unused
Q.22 A limiting reactant is one
(a) which is present in least amount
(b) which produces minimum no. of moles of product
(c) which produces maximum no. of moles of product
(d) does not effect the amount of product
Q.23 Stoichiometry is the branch of chemistry which deals with the study of quantitative relationship among the various
(a) reactants (b) products
(c) Reactants and products (d) all of above
Q.24 500 cm3 of H2 gas at STP contradictions of hydrogen
(a) 6.02 x 1023 (b) 3.01 x 1022
(c) 2.68 x 1022 (d) 1.34 x 1022
Q.25 Largest number of H+ ions are produced by complete ionization of
(a) 0.01 mole of HCl (b) 0.0050 mole of H2SO4
(c) 0.000334 moles of H3PO4
(d) all above
Q.26 The Avogadro’s number is
(a) 6.02 x 1024 (b) 6.02 x 10–24
(c) 6.02 x 10–23 (d) 6.02 x 1023
Q.27 The largest number of H+ are produced by complete ionization of
Q.27 The largest number of H+ are produced by complete ionization of
(a) 0.100 2 moles of HCl (b) 0.051 moles of H2SO4
(c) 0.0334 moles of H3PO4 (d) All of the above
Q.28 A sample of pure matter is
(a) element (b) compound
(c) substance (d) mixture
Q.29 nm stands for
(a) Newton
(c) Newton
Q.30 One calorie is equal to
(a) 4.184 J (b) 41.84 J
(c) 0.4184 J (d) 0.04184 J
Q.31 The number of moles of CO2 which contains 8.0 gm of oxygen
(a) 0.25 (b) 0.50
(c) 1.0 (d) 1.50
Q.32 27 grams of Al will react completely with how much mass of O2 to produce Al2O3
(a) 8 gm of oxygen (b) 16 gm of oxygen
(c) 32 gm of oxygen (d) 24 gm of oxygen
Q.33 Mole of SO2 contains
(a) 6.02 x 1023 atoms of oxygen
(b) 18.1 x 1023 molecules of SO2
(c) 6.023 x 1023 atom of sulphur
(d) 4 gram of SO2
Q.34 The largest number of molecules are presenting
(a) 3.6 gram of H2O (b) 4.8 gram of C2H5OH
(c) 2.8 gm of CO (d) 5.4 gms of N2O5
Q.35 The mass of one mole of electron is
(a) 1.008 mg (b) 0.184 mg
(c) 1.673 mg (d) 0.55 mg
Q.36 Isotopes differ in
(a) properties which depend on mass
(b) arrangements of electrons in orbital
(c) chemical properties
(d) the extent to which they may be affected in electromagnetic field
(d) the extent to which they may be affected in electromagnetic field
Q.37 The volume occupied by 1.4 gm of N2 at STP is
(a) 224 dm3 (b) 22.4 dm3
(c) 1.12 dm3 (d) 112 cm3
Q.38 Many elements have fractional atomic mass. This is because
Q.38 Many elements have fractional atomic mass. This is because
(a) the mass atom is itself fractional
(b) atomic masses are average masses of isobars
(c) atomic masses are averages masses of isotopes
(d) atomic masses are average masses of isotopes proportional to relative abundance
Q.39 A limiting reactant is one which
(a) is taken in lesser quantity in grams as compared to other reactants
(b) is taken in lesser quantity in volume as compared to the other
(c) gives the maximum amount of the product which is required
(d) gives the minimum amount of the product under consideration
Q.40 Isotopes when even atomic masses are a comparatively abundant
(a) demper’s spectrograph is superior to that of Aston’s
(b) 0.1 mg of H2O has greater number of molecules then 0.1 mg of CH4
(c) the number of H+ and PO–3 ions are not equal but the number of positive and negative charges
(d) are equal when 100 molecules of H3PO4 are thrown in excess of water
Q.41 A molecule having two atoms is called
(a) monoatomic molecules (b) diatomic molecules
(c) Polyatomic molecules (d) homoatomic molecule
Q.42 An ordinary misoscope is used to measure the object of size
(a) upto 500 nm (b) upto 850 nm
(c) upto 1000 nm (d) upto 1200 nm
Q.43 1 atomic masses unit (amu) is equation
(a) 1.66 x 10–27 kg (b) 1.56 x 10–27 kg
(c) 1.76 x 10–21 kg (d) 1.8 x 10–27 kg
Q.44 Nickel has isotopes
(a) 1 (b) 3
(c) 5 (d) 7
Q.45 Cadmium has isotopes
(a) 3 (b) 5
(c) 7 (d) 9
Q.46 The pressure of vapours in the separating isotopes by mass spectrometry is kept at
(a) 10–6 torr (b) 10–4 torr
(c) 10–3 torr (d) 10–5 torr
Q.47 Number of gram atoms in 0.1 gm of Na is
(a) 0.0043 (b) 0.0403
(c) 0.403 (d) None of these
Q.48 Molecule of haemoglobin contains atoms
(a) 15,000 (b) 12,000
(c) 10,000 (d) 8,000
Q.49 Haemoglobin is heavier than a hydrogen atom
(a) 65,000 (b) 68,000
(c) 62,000 (d) 60,000
Answers
Questions | 1 | 2 | 3 | 4 | 5 |
Answers | b | C | d | c | b |
Questions | 6 | 7 | 8 | 9 | 10 |
Answers | d | b | c | b | c |
Questions | 11 | 12 | 13 | 14 | 15 |
Answers | b | c | b | a | d |
Questions | 16 | 17 | 18 | 19 | 20 |
Answers | a | b | d | d | c |
Questions | 21 | 22 | 23 | 24 | 25 |
Answers | c | b | d | c | d |
Questions | 26 | 27 | 28 | 29 | 30 |
Answers | d | d | a | b | a |
Questions | 31 | 32 | 33 | 34 | 35 |
Answers | a | d | c | a | d |
Questions | 36 | 37 | 38 | 39 | 40 |
Answers | a | c | d | d | c |
Questions | 41 | 42 | 43 | 44 | |
Answers | c | a | a | c | |
Questions | 45 | 46 | 47 | 48 | 49 |
Answers | d | a | a | c | b |
Short Question With Answer
Q.1 Calculate the grams atoms in 0.4 gm of potassium.
Ans.
Gram atoms of potassium =
= = 0.01 grams atoms
Q.2 23 grams of sodium and 238 gram of uranium have equal number of atoms in them.
Ans.
Mass of sodium = 23 gms= 1mole=6.02 x 1023 atoms
Mass of uranium = 238g=1 mole= 6.02x 1023 atoms
Both the substances have equal number of atoms because they have same no.of moles.
Q.3 Mg atom is twice heavier than that of carbon.
Ans.
The atomic mass of Mg is 24 which is to twice as mass as compared to the atomic mass of carbon i.e. 12. So Mg atom is twice heavier than that of carbon.
Q.4 180 grams of glucose and 342 gram of sucrose have the same number of molecules but different number of atoms present in them.
Ans.
Mass of glucose (C6H12O6)=180g=1mole=6.02x1023molecules
=24NAatoms
Mass of Sucrose (C12H22O11)=342g=1mole=6.02 x 1023 molecules
=45NAatoms
Q.5 4.9 g of H2SO4 when completely ionized in water have equal number of positive and negative ions, but the number of positively charged ions are twice the number of negatively charged ions.
Ans.
H2SO4 → 2H+ + SO4-2
When one mole of H2SO4 ionizes, it produces 2H+ and
SO4–2 ions. Hydrogen ions contains +1 charge while sulphate ions have – 2 charge. Hydrogen ions are twice in number than that of SO ion. Charges on both ions are equal (with opposite sign). Similarly ions produced by complete ionization of4.9 grams of H2SO4 in water will have equal +ve and –ve charges but the number of H+ ions are twice than number of negatively charged sulphate ions.
SO4–2 ions. Hydrogen ions contains +1 charge while sulphate ions have – 2 charge. Hydrogen ions are twice in number than that of SO ion. Charges on both ions are equal (with opposite sign). Similarly ions produced by complete ionization of
Q.6 One mg of K2CrO4 has thrice the number of ions than the number of molecules when ionized in excess of water.
Ans. K2CrO4 → 2K+ + CrO4–2
When K2CrO4 ionizes in water, its one molecule gives three ions i.e. two K+ and one CrO4–2 (chromate) ions. The ratio between the number of molecules and number of ions than the number of molecules when ionized in water.
Q.7 Two grams of H2, 16g of CH4 and 44 gram of CO2 occupy separately the volumes of 22.414 dm3 at STP, although the sizes and masses of molecules of three gases are very different from each other.
Ans.
One mole of gas at STP occupies a volume of 22.4 dm3 sizes and masses of molecules of different gas do not affect the volume. Normally it is known that in the gaseous state, the distance between the molecules is 300 times greater that their diameter. Therefore two grams of H2, 16 grams of CH4 and 44 grams of CO2 (1 mole of each gas) separately occupy a volume of 22.4 dm3. This is called molar volume.
2gH2=1mole, 16gCH4=1 mole, 44gCO2=1 mole
1mole=22.414dm3
Q.8 Define Stoichiometry ?
Ans.
Stoichiometry is the branch of chemistry which gives a quantitative relationship between reactants and products in balanced chemical equation.
Q.9 What is limiting reactant? How does it control the quantity of the product formed? Explain with three examples. /Many chemical reactions taking place in our surroundings involve limiting reactants give examples?
Ans.
The reactant which controls (limits) the amount of product formed during a chemical reaction is called limiting reactant. In our surrounding many chemical reactions take place which involve limiting reactants some of these reactions are:
(i) Burning of coal to form CO2---Coal is limiting reactatnt C + O2 ® CO2
(ii) Burning of sui gas to form CO2 and H2O
CH4 + 2O2 ® CO2 + 2H2O
(iii) Rusting of iron----iron is limiting reactant
In above reactions oxygen is always in excess, while other reactants are consumed earlier. So other reactants are limiting reactants.
Q.10 One mole of H2O has two moles of bands, three moles of atoms, ten moles of electron and twenty–eight moles the total fundamental particles present in it.
Ans.
One molecule of H–O–H has two bounds between hydrogen and oxygen. There are three atoms i.e. two
H atoms and one O atom, therefore one mole of H2O contains two moles of bonds and three moles of atoms
(2 moles of H atoms and one mole of O atoms).
H atoms and one O atom, therefore one mole of H2O contains two moles of bonds and three moles of atoms
(2 moles of H atoms and one mole of O atoms).
Similarly, there are eight elections in oxygen and one electron in each of the two, H atoms one molecule of H2O so has 10 electrons, so one mole of water contains 10 moles of electrons. There are 28 moles of all fundamental particles in one mole of water i.e.
10 moles of electrons.
10 moles of protons.
8 moles of neurons (8 neutrons in oxygen and there is no neutron in hydrogen) 28 moles of fundamental particles.
Q.11 One mole of H2SO4 should completely react with two moles of NaOH. How does Avogadro’s number help to explain it?
Ans.
The balanced chemical equation between H2SO4 and NaOH
H2SO4 + 2NaOH ® Na2SO4 + H2O
H2SO4® 2H+ + SO4-2
2NaOH®2Na+ + 2OH-
2H+ + 2OH- ® 2H2O
2NA 2NA
This is an acid base reaction, one mole of H2SO4 releases two moles of H+ ion in solution. It needs two moles of OH ions for complete neutralization. So two moles of NaOH which releases two moles of OH are required to react with one mole of H2SO4. One mole of H2SO4 releases twice the Avogadro’s number of H+ ions and it will need the Avogadro’s number of OH ions for complete neutralization.
Q.12 N2 and CO have same number of electrons, protons and neutrons.
Ans.
Both N2 and CO have same number of electrons, protons and neutrons as it is clear from the following explanation.
For N2 No. of electrons in N2 = 7 + 7 = 14
No. of protons in N2 = 7 + 7 = 14
No. of neutrons = 7 + 7 = 14
For CO number of electrons
in C = 6
No. of electrons in O = 8
Total no. of protons = 6 + 8 = 14
No. of neutrons in C = 6
No. of neutrons in O = 8
Total no. of neutrons = 6 + 8 = 14
Q.13 How many molecule, of water are in 12 gram of ice?
Ans.
Mass of ice (water) = 12.0 gm
Molar mass of water = 18 g/mol
No. of molecules of water
=
=
No. of molecules of water = 0.66 x 6.02 x 1023
= 3.97 x 1023
Q.14 Differentiate between limiting and non–limiting reactant ?
Ans. Limiting Reactant:
A limiting reactant is a reactant and that controls the amount of the product formed in a chemical reaction.
Non–Limiting Reactant:
The reactant which produces the excess amount of the product is called non–limiting reactant.
Q.15 Distinguish between actual yield and theoretical yield ?
Ans. Actual Yield:
The amount of the products obtained in a chemical reaction is called actual yield based on experiment.
Theoretical (Experiment) Yield:
The amount of the products calculated from the balanced chemical equation is called theoretical yield.
Q.16 What do you mean by percent yield? Give its significance ?
Ans.
The yield which is obtained by dividing actual yield with theoretical yield and multiplying by 100 is called percent yield.
% yield = x 100
Significance:
(i) % yield indicates the efficiency of reaction.
(ii) More is the percent yield higher will be the efficiency of reaction.
Q.17 Why actual yield is less than the theoretical yield?
Ans.
(a) Side reaction may takes place
(b) All the reactant may not be converted into products
(c) Mecahanical loss may occur like during
e.g Filtration, evaporation, crystallization, distillation etc.
Q.18 Calculate the mass of 10–3 moles of MgSO4.
Ans.
MgSO4 is an ionic compound. We will consider its formula mass instead of molecular mass.
Number of moles of substance
=
Formula mass of MgSO4 = 120 gm/ml
Number of moles of MgSO4 = 10–3 moles
Applying formula
10–3 =
Mass of MgSO4 = 120 x 10–3 = 0.12 moles
Q.19 Define Avogadro’s number ?
Ans.
Avogadro’s number is the number of atoms, molecules and ions in one gram atom of an element, one gram molecule of a compound and one gram ion of substance, respectively. It is equal to 6.02 x 1023.
Q.20 Define mole ?
Ans.
The molecular mass of a substance expressed in grams is called molecule or gram mole or simply the mole of a substance.
Moles of substance =
1 mole of water = 18.0 g
1 mole of H2SO4 = 98.0 g
Q.21 Define isotopes ?
Ans.
Atoms of the same element which have different masses but same atomic numbers are called isotopes. For example carbon has three isotopes.
12C6 13C6 14C6 and expressed as C–12, C–13 and C–14. Similarly hydrogen has three isotopes H H H called protium, deuterium and tritium.
Q.22 Define (i) ions (ii) Positive ion (iii) Negative ion.
Ans. Ion
As specie having positive or negative charges are called ions. For example Cl–1, NO, Na+, Ca++.
Positive Ion (Cation):
A specie has +ve charge is called positive ion and attracted towards Cathode . For example Na+, K++, Ca++.
Negative Ion (Anion)
A specie which has negative charge is called negative ion and attracted towards anode . For example F–1,Cl–1,Br–1andS–2P–3,C–4,SO, Cr2O, CO.
Q.23 Define and explain the molecular ion ?
Ans.
When a molecule loses or gains an electron, molecular ion is formed. For example CH4+, CO+, N2+. Cationic molecular, ions are more abundant than anionic ions.
The molecular ions find applications of in calculation of molecular mass of a compound. The molecular ions also help in the determination of structure of macro molecules.
The break down of molecular ions obtained from the natural products can give important information about their structure.
Q.24 What do understand by the relative atomic mass ?
Ans.
Relative atomic mass is the mass of an atom of element as compared to the mass of an atom of carbon taken as 12.
The unit used to express the relative atomic mass is called atomic mass unit (amu). It is th of the mass of one carbon atom. The relative atomic mass of 12C6 is 12.00 amu. The relative atomic mass of H is 1.0078 amu.
Q.25 Define Gram atom ?
Ans.
Ans.
The atomic mass of an element expressed in grams is called gram atom of an element.
Number of gram atoms of a meter an element
=
For example 1 gram atom of hydrogen = 1.008 gm
Q.26 Define gram ion ?
Ans.
The ionic mass of an ionic specie expressed in grams is called one gram ion or one mole of ions.
Number of gram ions =
Q.27 Define gram formula and moles ?
Ans.
The formula mass of an ionic compound expressed in grams is called gram formula of the substance.
Number of gram formula or moles of a substance
=
The atomic mass, molecular mass, formula mass or ionic mass of the substance expressed in grams is called moles of those substances.
Q.28 Define molar volume ?
Ans.
The volume occupied by one mole of an ideal gas at standard temperature and pressure (STP) is called molar volume. The volume is equal to 22.414 dm3.
Q.29 Define and explain atomicity ?
Ans.
The number of atoms present in a molecule is called the atomicity. The molecule can be monoatomic, diatomic and triatomic etc. If the molecule contains one atom it is monoatomic, if it contains two atoms it is diatomic, and if it contains three atoms it is triatomic. Molecules of elements may contain one two or more same type of atoms. For example He, Cl2, O3, P4, S8. The molecules of compounds consist of different kind of atoms. For example HCl, NH3, H2SO4, C6H12O6.
Q.30 Define an atom and molecule ?
Ans. Atom:
Atom is now defined as the smallest particle of an element, which may or may not have independent existence. For example He and Ne atoms have independent existence. While atoms of hydrogen, nitrogen and oxygen do not exist independently.
Molecule:
A molecule is the smallest particle of a pure substance( element or Compound ) which can exist independentally. For example N2, O2, Cl2, HCl, NH3 and H2SO4 are examples of molecules.
Q.31 What do you mean by empirical formula and molecular formula? How they are related to each other ?
Ans. Empirical Formula:
It is the simplest formula that gives information about the simple ratio of atoms present in a compound.
In an empirical formula of a compound Ax By, there are X atoms of an element A and y atoms of an element B.
Molecular Formula:
The formula of a substance which is based on the actual molecule is called molecular formula. It gives the usual number of atoms present in the molecule. For example molecular formula of benzene is C6H6, while that of glucose is C6H12O6. The molecular formula and empirical formula are related to each other by the following relationship.
Molecular formula = n x (Empirical formula)
Where “n” is simple integer.
Q.32 Is it true many compounds have same empirical and molecular formula ?
Ans.
There are many compounds, whose empirical formulas and molecular formulas are the same. For example H2O, CO2, NH3 and C12H22O11 have the same empirical and molecular formulas. Their simple multiple n is unity. Actually value of “n” is the ratio of molecular mass and empirical formula mass.
n =
Q.33 Ethylene glycol is used in automobile antifreeze. It has 38.7% carbon, 9.7% hydrogen and 51.6% oxygen. Its molar mass is 62 gms mole–1. Determine its empirical and molecular formula ?
Ans.
C = 38.7%, H = 9.7%, O = 51.6%
Dividing above %ages by atomic mass.
We get molar ratios
C = = 3.225
H = = 9.7
O = = 3.225
Dividing above molar ratio by least ratio we get atomic ratio.
C = = 1
H = = 3
O = = 1
Empirical formula is CH3O
Molar mass = 62
Empirical formula mass = 12 + 3 + 16 = 31
Now
n =
= = 2
Molar formula = n x Empirical formula
= 2 x CH3O
Molecular formula = C2H6O2
Hence molecular formula of Ethylene glycol = C2H6O2
Q.34 The combustion analysis of an organic compound shows it to contain 65.44% carbon 5.5% hydrogen and 29.06% of oxygen. What is the empirical formula of the compound? If the molecular mass of the compound is 110.15. Calculate the molecular formula of the compound.
Ans.
First of all divide the percentage of each element by its atomic mass to get the number of from atoms or moles.
No. of gram atoms of carbon =
= 5.45 gram atoms of C
No. of gram atoms of hydrogen =
= 5.45 gram atoms of H
No. of gram atoms of oxygen =
= 1.82 gram atoms of 0
Mole ratio C : H : O
4.45 5.45 1.82
Divide number of grams atoms by the smallest number
C : H : O
: :
3 : 3 : 1
Carbon, hydrogen and oxygen are present in the given organic compound in ratio of 3 : 3 : 1. So the empirical formula is C3H3O.
In order to calculate the molecular formula first calculate the empirical formula mass.
Empirical formula mass = 3 x 12 + 3 x 1 + 16
= 36 + 3 + 16 = 55.05
Molar mass of the compound = 110.15
h = = = 2
Molecular formula = n x empirical formula
= 2 x C3H3O
= C6H6O2
Q.35 Give relationships, between the amounts of substances and number of particles. There are three useful relationships ?
Ans.
1. Number of atoms of an element = x NA
2. Number of molecules of a compound
= x NA
3. Number of ions of ionic species = x NA
NA is the Avogadro’s number. The value is 6.02 x 1023.
Q.36 What are the types of relationships of stoichiometric calculations ?
Ans.
There are three types of relationships of stoichiometric calculations.
1. Mass–Mass Relationship
The relationship in which the mass of one substance is given and the mass of other substance is calculated.
2. Mass–mole or mole–mass relationship
The relationship in which mass of one substance is given and moles of other substance is to be calculated or vice versa.
3. Mass–volume or volume mass relationship
The relationship in which the mass of one substance is given and the volume of other substance is to be calculated or vice versa.
Q.37 Law of conservation of mass has to be obeyed during the stoichiometric calculations ?
Ans.
Stoichiometric calculations are based on balanced chemical equation and equation is balanced on the basis of Law of conservation of mass e.g
C+O2→ CO
In this equation stoichiometric calculations are not possible because it is not a balanced equation and it is not obeying Law of coseravtion.
TEXT BOOK EXERCISE
Q1. Select the most suitable answer from the given ones in each question.
(i) The mass of one mole of electrons is
(a) Properties which depend upon mass
(b) Arrangement of electrons in orbital
(c) Chemical properties
(d) The extent to which they may be affected in electromagnetic field
(ii) Which of the following statements is not true?
(a) isotopes with even atomic masses are comparatively abundant
(b) isotopes with odd atomic masses and even atomic number are comparatively abundant
(c) atomic masses are average masses of isotopes.
(d) Atomic masses are average masses of isotopes proportional to their relative abundance
(iii) Many elements have fractional atomic masses, this is because
(a) The mass of the atom is itself fractional
(b) Atomic masses are average masses of isobars
(c) Atomic masses are average masses of isotopes.
(d) Atomic masses are average masses of isotopes proportional to their relative abundance
(iv) The mass of one mole of electrons is
(a) 008mg (b) 0.55mg (c) 0.184mg (d) 1.673mg
(v) 27g of Al will react completely with how much mass of O2 to produce Al2O3
(a) 8g of oxygen (b) 16g of oxygen
(c) 32g of oxygen (d) 24g of oxygen
(vi) The number of moles of CO2 which contain 8.0 g of oxygen.
(a) 0.25 (b) 0.50 (c) 1.0 (d) 1.50
(vii) The largest number of molecules are present in
(a) 3.6g of H2 O (b) 4.8g of C2H5 OH
(c) 2.8 g of CO (d) 5.4g of N2O5
(viii) One mole of SO2 contains
(a) 6.02x1023 atoms of oxygen
(b) 18.1x1023 Molecules of SO2
(c) 6.02x1023 atoms of sulphur
(d) 4 gram atoms of SO2
(ix) The volume occupied by 1.4 g of N2at STP is
(a) 2.24 dm3 (b) 22.4dm3
(c) 1.12 dm3 (d) 112 cm3
(x) A limiting reactant is the one which
(a) is taken in lesser quantity in grams as compared to other reactants
(b) is taken in lesser quantity in volume as compared to the other reactants
(c) give the maximum amount of the product which is required
(d) give the minimum amount of the product under consideration
Ans: (i)a (ii)d (iii)d (iv)b (v)d (vi)a (vii)a (viii)c (ix)c (x)d
Q2: Fill in the blanks:
(i) The unit of relative atomic mass is-----------
(ii) The exact masses of isotopes can be determined by ------------spectrograph.
(iii) The phenomenon of isotopes was first discovered by --------------
(iv) Empirical formula can be determined by combustion analysis for those compounds which have-----------and -----------in them.
(v) A limiting reagent is that which controls the quantities of -------------
(vi) I mole of glucose has-----------atoms of carbon ---------------of oxygen and ----------of hydrogen.
(vii) 4g of CH4 at OoC and I atm pressure has ---------molecules of CH4.
(viii) Stoichiometry calculations can by performed only when -------------law is obeyed.
Ans: (i) amu (ii) mass (iii) Soddy (iv) carbon, hydrogen
(v) Products (vi) 6NA,6NA,12NA
(vii) 1.505x1023 (viii) conservation and multiple proportion
Q3: Indicate true or false as the case may be:
(i) Neon has three isotopes and the fourth one with atomic mass 20.18 amu.
(ii) Empirical formula gives the information about he total number of atoms present in the molecule
(iii) During combustion analysis Mg(CIO4)2 is employed to absorb water vapors.
(iv) Molecular formula is the integral multiple of empirical formula and the integral multiple can never be unity.
(v) The number of atoms in 1.79 g of gold and 0.023g of sodium are equal.
(vi) The number of electrons in the molecules of CO an dN2 are 14 each, so 1 mg go each gas will have same number of electrons.
(vii) Avogadro’s hypothesis is applicable to all types of gases, i.e., ideal and non-ideal .
(viii) Actual yield of a chemical reaction may by greater than the theoretical yield.
Ans. (i) False (ii) False (iii) True (iv) False
(v) False (vi) True (vii) False (viii) False
Q4: What are ions? Under What condition are they produced?
Ans: Ions can be produced by the following processes:
(i) By dissolving ionic compounds in water
(ii) By X-rays
(iii) In mass spectrometry
(iv) By removing or adding electron in atom
Q4:
(a) What are isotopes? How do you deduce the fractional atomic masses of elements form the relative isotopes abundance? Give two examples in support of your answer. ( See detail in Sublime subjective)
(b) How does a mass spectrograph show the relative aboundace of isotopes of an element? . ( See detail in Sublime subjective)
(c) What is the justification of two strong peaks in the mass spectrum for bromine; while for iodine only one peak at 127 amu , is indicated?
Ans The two strong peak in the mass spectrum for bromine represent two different isotopes of bromine having nearly equal natural abundances. Only one peak at 127 amu in the mass spectrum for iodine indicates that it has only one isotope of atomic mass 127 amu.
Remember that!
Height of the peaks ∞ Relative abundance of isotopes
No. of peaks = No. of isotopes
Q5: Silver has atomic number 47 and has 16 known isotopes but two occur naturally I,e, Ag _____107 . and Ag _____109 . Given the following mass spectrometric data, calculated the average atomic mass of silver,
Isotopes mass (amu) percentage abundance
107Ag 106.90509 51.84 109 Ag 108.90476 48.16
Solution: The mass contribution for silver are:
Isotopes Fractional abundance isotopic mass mass contribution
107Ag 
107 0.5184x107=55.4688
107 0.5184x107=55.4688109Ag 
107 0.4816x109=52.4944
107 0.4816x109=52.4944 Fractional atomic mass of silver =107.9632
Hence the fractional atomic mass of silver is =107.9632 Ans.
Q6: Boron with atomic number 5 has two naturally occurring isotopes. Calculate the percentage abundance of 10B and 11B from the following information.
Average atomic mass of boron =10.81 amu
Isotopic mass of 10B =10.0129 amu
Isotopic mass of 11B =11.0093
Solution: Let, the fractional abundance of 10B =x
The fractional abundance of 11B =1-x
Remember that the sum of the fractional abundances of isotopes must be equal to one, now, The equation to determine the atomic mass of element is
(fractional abundance) (isotopic mass) (fractional abundance of 10B)(isotopic mass of 10B )+(fractional abundance of 11B) (isotopic mass of 11B) =Average atomic mass of Boron
(x)(10.0129)+(1-x)(11.0093) =10.81
10.0129x+11.00093x =10.81
10.0129x-11.00093x =10.81-11.0093
-0.9964x =-0.1993
x =
Fractional abundance of 10B =0.2000
Fractional abundance of 11B =(1-0.2000)=0.8000
By percentage the fractional abundance of isotope is
%of 10B =0.2000x100 =20% Answer
% of 11B =0.8000x100 =80%Answer
Q7: Define the following terms and give three examples of each.
(i) Gram atom (ii) Gram molecular mass
(iii) Gram molecular mass (iv) Gram ion (v) molar volume (vi) Avogadro’s number (vii) Stoichiometry (viii) Percentage yield
Q8: Justify the following statements:
(a) 23 g of sodium and 238g of uranium have equal number of atoms in the (b) Mg atom is twice heavier than that of carbon
(c) 180g of glucose and 342 g of sucrose have the same number of molecules but different number of atoms present in them.
(d) 4.9g of H2 SO4 when completely ionized in water , have equal number of positive and negative charges but the number of positively charged ions are twice the number of negatively charged ions.
(e) One mg of K2 Cr O4 has thrice the number of ions than the number of formula units when ionized in water.
(f) Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the volumes of 22.414 dm3 , although the sizes and masses of molecules of three gases are very different from each other.
Solution:
(a) 23g of Na =1 mole of Na =6.02x1023 atoms of Na
238g of U =1 mole of U =6.02x1023 atoms of U.
Since equal number of gram atoms(moles) of different elements contain equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of uranium contain equal number of atoms , i , e ,6.02x1023 atoms.
(b) Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12) therefore, Mg atom is twice heavier than that of carbon. Or
Mass of 1 atom of Mg=
Mass of 1 atom of C =
Since the mass of one atom of Mg is twice the mass of one atom of C, therefore, Mg atom is twice heavier than that of carbon.
(c) 180 g of glucose = 1 mole of glucose =6.02x1023 molecules of glucose 342 g of sucrose=1mole of sucrose =6.02x1023 molecules of sucrose
Since one mole of different compounds has the same number of molecules.
Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the same number (6.02x1023)of molecules. Because one molecule of glucose, C6H12O6 contains 45 atoms whereas one molecules of glucose, C12 H22 O11 contains 24 atoms. Therefore, 6.02x1023 molecules of glucose contain different atoms as compound to6.02x1023 molecules of sucrose. Hence, 180 g of glucose and 342g og sucrose have the same number of molecules but different number of atoms present in them.
(d) H2 SO4 
2H+ + SO
2H+ + SO When one molecules of H2 SO4 completely ionizes in water it produces two H+ ion and one SOion,. Hydrogen ion carries a unit positive charge whereas SOion carries a double negative charge. To keep the neutrality, the number of hydrogen are twice than the number of soleplate ions. Similarly the ions produced by complete ionization of 4.8g of H2 SO4 in water will have equal number of positive and negative but the number of positively charged ions are twice the number of negatively charged ions.
ion,. Hydrogen ion carries a unit positive charge whereas SOion carries a double negative charge. To keep the neutrality, the number of hydrogen are twice than the number of soleplate ions. Similarly the ions produced by complete ionization of 4.8g of H2 SO4 in water will have equal number of positive and negative but the number of positively charged ions are twice the number of negatively charged ions. (e) H2 SO4 2H+ + SO
2H+ + SO K2 Cr O4 when ionizes in water produces two k+ ions one C O ion. Thus each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2 Cr O4 has thrice the number of ion than the number of formula units ionized in water.
ion. Thus each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2 Cr O4 has thrice the number of ion than the number of formula units ionized in water.(f) 2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP =22.414dm3 16g of CH4 =1mole of CH4 =6.02x1023 molecules of CH4 at STP =22.414dm3 144g of CO2 =1mole of CO2 =6.02x1023 molecules of CO2 at STP =22.144dm3
Although H2 , CH4 and CO2 have different masses but they have the same number of moles and molecules . Hence the same number of moles or the same number of molecules of different gases occupy the same volume at STP. Hence 2 g of H2 ,16g of CH4 and 44 g of CO2 occupy the same volume 22.414 dm3 at STP. The masses and the sizes of the molecules do not affect the volumes.
Q10: Calculate each of the following quantities
(a) Mass in grams of 2.74 moles of KMnO4 .
(b) Moles of O atoms in 9.0g of Mg (NO3)2 .
(c) Number of O atoms in 10.037g of Cu SO4 .5H2 O.
(d) Mass in kilograms of 2.6x 1020 molecules of SO2 .
(e) Moles of C1 atoms in 0.822g C2H4C12 .
(f) Mass in grams of 5.136 moles of silver carbonate .
(g) Mass in grams of 2.78x1021 molecules of CrO2 C12 .
(h) Number of moles and formula units in 100g of KC1O3 .
(i) Number of K+ ions C1O
ions, C1 atoms, and O atoms in (h)
ions, C1 atoms, and O atoms in (h)Solution:
(a) No of moles of KMnO4 =2.74moles
formula mass of KMnO4 =39+55+64=158g mol -1
Mass of KMnO4 =?
Formula used:
Mass of KMnO4 = no of mole of KMnO4 x formula mass of KMnO4
=2.74 mol x 158 g mol-1
=432.92g Answer
(b) Mass of Mg (NO3)2 =9g
Formula mass of Mg (NO3)2 =24+28+96=148g mol -1
No of moles of O atoms =?
Formula used:
No of mole of Mg (NO3)2 =
Now, I mole of Mg (NO3)2 contains = 6moles of O atoms
0..06 moles of Mg (NO3)2contains =6x0.6
=0.36 moles of O atoms
Alternatively,
148g of Mg (NO3)2 contains =6moles of O atoms
g of Mg (NO3)2contains =
=0.36 mole Answer
(c) Mass of CuSO4. 5H2O=10.037g
Formula mass of CuSO4. 5H2O=63.54+32+64+90
=249.546g mol -1
No of moles of CuSO4. 5H2O =?
No of moles of CuSO4. 5H2O =
=
Now, 1 mole of CuSO4 .5H2O contains 9moles of O atoms
0.04 mole of CuSO4 .5H2O contains=9x0.04
=0.36 moles of O atoms
Now, I mole of O atoms contains =6.02x1023 O atoms
0.36 mole of O atoms contains =6.02x1023 x0.36 oxygen atoms
=2.17x1023 oxygen atoms
=2.17x1023 atoms Answer
(d) No of molecules of SO2 . =2.6x1020 molecules
Molecular mass of SO2 . =32+32=64 g mol-1
Now, Avogadro’s number , NA =6.02x1023 molecules of SO2
Mass of SO2 molecules 
=27.64x10-3 g
=
=27.64x10-6 kg
=2.764x10-3 kg Answer
(e) Mass of C2 H4C1 = 0.822g
Molecular mass of C2 H4C1 =24+4+71=99 g mol-1
No of moles of C2 H4C1 =
Now, 1 mole of C2 H4C1 contains =2moles of C1 atoms
8.3x10-3mole of C2 H4C1 contains =2x8.3x10-3 mole of atom
=16.6x10-3
=0.0166mole of C1 atom
=0.017 mole Answer
(f) No of mole of Ag2 CO3 =5.136moles
Formula mass of Ag2 CO3 =215.736+12+48=275.736 g mol-1
Mass of Ag2 CO3=No of moles of Ag2 CO3xformula mass of Ag2 CO3
=5.136molx275.736 g mol-1
=416.18g
=1416.2 g Answer
(g) Molecular mass of CrO2C12 =52+32+71=155g mol-1
NA =6.02x1023 molecules mol-1
Molecules of CrO2C12==2.78x1021 molecules
Now, mass of CrO2C12
= =
=71.578x10-2 g
=0.71578
=0.716 g Answer
(h) Mass of KCIO3 =100g
Formula mass of KCIO3 =39x35.5+48=122g mol-1
No of moles of KCIO3 =?
No of moles of KCIO3 = 
=
=0.816mole Answer
=0.816mole Answer No of formula units No of moles x Avogadro,s No
=0.816mole x 6.02x1023 formula units
=4.91x1023 formula units
(i) No of K+ ions =4.91x1023 Answer
No of CIO 
ions =4.91x1023 Answer
ions =4.91x1023 Answer No of CIO ions =4.91x1023 Answer
No of O atoms = 4.91x1023 x3
=14.73x1023 =1.473x1024 Answer
Q 11 Aspartame he artificial sweetener, has a molecular formula of C14 H18 N2O5 .
(a) What is the mass of one mole of aspartame?
(b) How many moles are present in 52g of aspartame?
(c) What is the mass in grams of 10.122 moles of aspartame?
(d) How many hydrogen atoms are present in 2.34g of aspartame?
(a) Molecular mass of aspartame =168+18+28+80=295g mol-1
Mass of 1 mole of aspartame =294g mol-1 Answer
(b) Mass of aspartame =52g
Molecular mass of aspartame =294g mol-1
No of moles of aspartame =
=
=0.1768 mol
=0.177 mol Answer
(c) No moles of aspartame = 10.122 moles
Molecular mass of aspartame =294g mol-1
Mass of aspartame =No of moles x Molar mass
=10.122mol x 294g mol-1
=2975.87 g Answer
(d) Mass of aspartame =243g
Molar mass of aspartame =294g mol -1
No of molecules of aspartame=?
No of molecules of aspartame=
xNA
xNA =
=
=4.98x1021 molecules.
Now,1 molecule of aspartame contains =18 H atoms
4.98x 1022 molecules =18x4.98x1021 H atoms
=89.64x1021H atoms
=8.964x1022 H atoms Answer
Q 12: A sample of 0.600 mole of a metal M reacts completely with excess of fluorine to from 46.8g MF2 .
(a) How many moles of F are present in the sample of MF2 that forms.
(b) which elements is represented by the symbol M ?
Solution:
(a) Formula of compound =MF2
No of moles of M =0.6 mol
Mass of MF2 =46.8g
The molar of M:F in the compounds;
No of moles of F =0.6x2=1.2mol Answer
Mass of F =No of moles of Fx At . mass of F
=1.2x19=22.8g
Mass of compound =46.8g
Mass of metal, M =46.8-22.8
=24
At mass of M =
=
(b) The atomic mass of the elements, M =40
The metal is calcium, Ca Answer
Q 13 : In each pair , choose the larger of the indicated quantity ,or state if the samples are equal.
(a) Individual particles: 0.4 mole of oxygen molecules or0.4mole of oxygen atom.
(b) Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms
(c) Mass: 0.6 mole of C2 H4 or 0.6mole of 12
(d) Individual particles: 4.0g N2O4 or 3.3g SO2
(e) Total ions: 2.3 moles of NaC1O3 or 2.0mole of MgC12
(f) Molecules: 11.0g of H2Oor 11.0g H2O2
(g) Na+ ion: 0.500 moles of NaBr or 0.0145kg NaC1
(h) Mass: 6.02x1023 atoms of 235U or 6.02x1023 atoms of 238U
Ans:
(a) Number of molecules =moles x NA
Number of O2 molecules =0.4x6.02x1023 =2.408x1023 molecules
No of O atoms=0.4x6.02x1023=2.108x1023 atoms
There are equal number of individual particles in 0.4 mole of oxygen molecules and 0.4 mole of oxygen atom. In general, equal number of moles of different substances contains equal number of particles.
Both are equal Answer
(b) Mass of substance = moles x molar mass
Mass of oxygen atoms =0.4x16=64g
Mass of ozone, O3 molecules =0.4x48=19.2g
0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms.
Ozone Answer
(c) Mass of C2H4 =0.6x28=1.68g
Mass of 12 =0.6x127=254g
0.6mole of 12 have larger mass than 0.6 mole of C2H4
12 Answers
(d) No of molecules =
No of molecules in N2 O4 =
x6.02x1023 =2.62 x1023 molecules
x6.02x1023 =2.62 x1023 moleculesNo of molecules in SO2 =x6.02x1023 =3.1x1022 molecules
3.3g of SO2 have larger number of individual particles than 4.0 g of N2 O4 .
SO2 Answer
(e) No of formula units =Moles x NA
No of formula units of NaC1O3 =2.3x6.02x1023=1.38x1024 formula units
No of ions in 1 formula units of NaC1O3=2
Total no of ions in MgC12 =2x1.38x1023=2.76x1024 ions
No of formula units of MgC12 =2.0x6.02x1023 x3=3.6x1024 ions
No .of ions in one formula unit of MgC12 =3
Total no of ions in MgC12 =1.20x1024 x3=3.6x1024 ions
2.0moles of MgC12 contain lager number of total ions than 2.3 moles of NaC1o3-
MgC1 Answer
(f) No of molecules =
NA
NANo of molecules in H2 O2 =
x6.02x1023=3.68x1023 molecules
x6.02x1023=3.68x1023 moleculesNo of molecules in H2 O2 =
x6.02x1023=1.95x1023 molecules
x6.02x1023=1.95x1023 molecules11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2
H2 O2Answer
(g) No of formula units =moles xNA
No of formula units NaBr =0.5x6.02x1023=3.01x1023 formula units
One formula units o NaBr contain Na+ ions =1
3.01 x1023 formula unit of NaBr contains Na +ions =3.01x1023 Na+ ions
No of formula units of NaC1 =
x6.02x1023 =1.49x1023formula units
x6.02x1023 =1.49x1023formula units One formula unit of NaC1 contains Na+ ions =1
1.49x1023 formula units of NaC1 contains =1.49x1023 Na+ ions
0.500 moles of NaBr contains lager number of Na+ ions than 0.0145kg ofNaC1.
NaBr Answer
(h) Mass of atoms of an element =
Mass of 235Uatoms =x6.02x1023 =235g
Mass of 238U atoms =x6.02x1023=238g
238U Answer
Q 13:
(a) Calculate the percentage of nitrogen in the four important fertilizer i.e.,
(i)NH3 (ii)NH2CONH2(Urea) (iii)(NH4)2SO4 (iv)NH4 NO3
(b) Calculate the percentage of nitrogen and phosphorus in each of the following:
(i) NH4H2PO4 (ii) (NH4)) PO4 (iii) (NH4)4 PO4
Solution:
(a) Mol-mass of NH3 =14+4=17g
Mass of N =14g
% of N =x100
=82.35% Answer
(b) Mol-mass of NH2 CONH2 =28+4+12+16=60g
Mass of N =28g
%of N =x100
=46.35% Answer
(c) Mol-mass of (NH2 )2 SO4 =28+8+32+64=132g
Mass of N =28g
% of N =x100
=21.21% Answer
(d) Mol-mass of (NH2 )2 SO4 =28+4+48=80g
Mass of N =28g
%of N =x100
=35% Answer
(I) Mol-mass of (NH2 )2 SO4 =14+6+31+64=115g
Mass of N =14g
Mass of P =31g
%of N =x100=12.17% Answer
%of P ==26.96% Answer
(II) Mol-mass of ((NH2 )2 SO4 =28+9+31+64=132g
Mass of N =28g
Mass of P =
%of N = =21.21% Answer
%of P = =23.48% Answer
(III) Mol-mass of (NH2 )2 SO4 =42+12+31+64=149g
Mass of N =42g
Mass of P =31g
%of N =
%of P =
Q 14: Glucose C6 H12 O6 is the most important nutrient in the cell for generating chemical potential energy. Calculate the mass% of each element in glucose and determine the number of C,H and O atoms in 10.5g go the sample.
Solution:
Mol-mass of glucose C6 H12 O6 =72+12+96=180g
Mass of C =72
Mass of H =12
Mass of O =96
% of C = =40% Answer
% of H = =6.66% Answer
% of O = =53.33% Answer
Mass of C6 H12 O6 =10.5g
Mol-mass of C6 H12 O6 =180g
Mol-mass of =180g mol-1
No of moles of C6 H12 O6 =
No of molecules of glucose =No of moles x NA
=0.058 molx 6.02x1023 molecules mol-1
=0.35x1023 molecules
=3.5x1022 molecules
Now, 1 molecule of glucose contains =6C-atoms
3.4x1022 molecules of glucose contains =6x3.5x1022 C-atoms
=21x1022 =2.1x1023 C atoms Answer
1 molecules of glucose contains =12H-atoms
3.5x1022 molecules glucose contains =12x3.5x1022
=4.2x1023 H- atoms Answer
1 molecule of glucose contains =6 O –atoms
3.5 x 1022 molecules of glucose contains =6x3.5x1022
=2.1x1023 O-atoms Answer
Q 16: Ethylene glycol is used as automobile antifreeze .It has 38.7% carbon, 9.7% hydrogen and 51.6% oxygen. Its molar mass is 62.1 grams mol-1 .Determine its molecular formula.
Solution:
% of C=38.37 g % of H =9.7g % of O=51.6g
At. Mass of C=12g mol-1 At. Mass of H=1.008g mol-1 At. Mass of O =16g mol-1
No of moles of C =
No of moles of H =
No of moles of O =
Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest ratio.
C :H :O
1 :3 :1
Empirical formula =CH3 O
Empirical formula mass =31
n= 
Molecular formula =2x CH3 O
=C2 H6 O2 Answer
Q 16: Serotonin (Molecular mass= 176g mol-1 ) is a compound that conducts nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08% O. What is its molecular formula?
Solution:
No of moles of C =
No of moles of H =
No of moles of N =
No of moles of O =
C | : | H | : | N | : | O | |
Atomic Ratio | : | : | : | ||||
10 | : | 12 | : | 2 | : | 1 |
Empirical formula =C10 H12 N2 O
Empirical formula mass =120+12+28+16=176g mol-1
Molecular mass =176g mol-1
n=
Q17: An unknown metal M reacts with S to from a compound with a formula M2S3 .If 3.12 g of M reacts with exactly 2.88 g of sulphur, what are the names of metal M and the compound M2 S3 .
Solution:
Formula of compound = M2 S3
Mass of M =3.12g
Mass of S =2.88g
Atomic mass of S =32g mol-1
No of moles of S =
No of moles of S =
The molar ratio of M: S in the compound is :
No of moles of M =
=0.06 mole
Now, No of moles of M =
At. Mass M =
The mass of M used in the formation of M2S3 is 3.12g. The product M2S3 therefore also contains 3.12g of M, because mass is conserved. The amount of M before and after reaction must be the same. Since we know both the number of moles of M and the mass of M , we can cal calculate the atomic mass of M as follows:
At. Mass of M =
=52
Atomic number, Z =52
Q19: The octane present in gasoline burns according to the following equation.
2C8 H18 (i) + 2502(g) 
16CO 2(g) + 18H2O (i)
16CO 2(g) + 18H2O (i)(a) How many moles of O2 are needed to react fully with 4 moles of actane?
(b) How many moles of CO2 can be produced from one mole of actane?
(c) How many moles of water are produced by the combustion of 6 moles of octane?
(d) If this reaction is to be used to synthesize 8 moles of CO2 how many grams of oxygen are needed? How many grams of octane will be used?
Solution:
4 moles
2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)
16CO 2(g) + 18H2O (i)(a) 2 moles 25 moles
2 moles of C8 H18 =25 moles of O2
4 moles of C8 H18 =
=50moles of O2 Answer
(b) 1 moles
2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)
16CO 2(g) + 18H2O (i)2 moles
Now, 2 moles of C8 H18 =16 moles of CO2
1 mole of C8 H18 =
=8 moles of CO2 Answer
(c) 6 moles
2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)
16CO 2(g) + 18H2O (i)2 moles
Now, 2 moles of C8 H18 =18 moles of H2 O(i)
6 moles of C8 H18 =
=54 moles of H2 O
(d) 6 moles
2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)
16CO 2(g) + 18H2O (i)2 moles 1800moles
Now, 16 moles of CO2 =25 moles of O2
8 moles of CO2 =
=12.5 moles of CO2
Mol-mass of O2 =32g mol-1
=12.5 molx 32g mol-1
=400g of O2
Now, 16moles of CO2 =2moles of C8 H18
8 moles of CO2 =
=1 mole of C8 H18
Mol-mass of C8 H18 =96+18=114g mol-1
Mass of C8 H18 =No of moles of C8 H18xMol.mass ofC8 H18 =1 molx 114 g mol-1
114g Answer
Q19: Calculate the number of grams of A12 S3 which can be prepared by the reaction of 20 g of A1 and 30 g of sulphur. How much the non-limiting reaction is in excess?
Solution:
Mass of A1 =20g
Molar mass of A1 =27g mol-1
No of moles of A1 =
Mass of S = 30g
Molar mass of S =32g mol-1
No of moles of S =
0.74 mole 0.94 mole
2A1 + 3S A12 S3
A12 S3 2 mole 3 mole 1 mole
Now, 2 moles of A1 =1 mole of A12 S3
0.74 moles of A1 =
=0.37 mole of A12 S3
Now, 3 moles of S =1 moles of A12 S3
0.94 moles of S =
=0.313 mole of A12 S3
Since S give the least number of moles of A12 S3 therefore, it is the limiting reactant.
No of moles of A12 S3 =0.313 mole
Molar mass of A12 S3 =150g mol-1
Mass of A12 S3=No of moles of A12 S3xMolar mass of A12 S3
=0.313molx 150 g mol-1
=46.95 g of A12 S3 Answer
The non-limiting reactant is A1 which is in excess. Now mass of A1 required reacting completely with 0.94 moles of S can be calculated as:
0.94 mole
2A1 + 3S A12 S3
A12 S3 2 mole 3 mole
Now, 3 moles of S =2 moles of A1
0.94 moles of S =
=
Mass of A1 =No of moles of A1 x molar mass of A1
=0.63x 27
=17g of A1
Mass of A1available =20g
Mass of A1 which reacts completely =17g with available S
Excess of A1 =20-17=3g
Q20: A mixture of two liquids, hydrazine N2H4 and N2 O4 are used as a fuel in rockets. They produce N2 and water vapors. How many grams of N2 gas will be formed by reacting 100g of N2 O4 and 200g g of N2 O4.
2N2H4 + N2O2 3N2 +4 H2O
3N2 +4 H2OSolution:
Mass of2N2H4 =100g
Mass of N2O2 =200g
Molar mass of 2N2H4 =28+4=32g mol-1
Molar mass of N2O2 =28+64=92g mol-1
No of moles of N2H4 =
No of moles of N2O2 =
3.125moles 2.174 moles
2N2H4 + N2O2 3N2 +4 H2O
3N2 +4 H2O2 moles 1mole 3moles
Now , 2moles of N2H4 =3moles of N2
3.125moles of N2H4 =
=4.69 mole of N2
Now , 1 mole of N2O2 =3moles of N2
2.174 moles of N2O4 =
=6.52 mole of N2O2
Since N2H4gives the least number of moles of N2, hence it is the limiting reactant.
Amount of N2 produced =4.69 moles
Molar mass of N2 =28g mol-1
Mass of N2 =4.69g molx 28g mol-1
=131032 g Answer
Q21: Silicon carbide (SiC) is an important ceramic material . It is produced by allowing sand (SiO2 )to react with carbon at high temperature.
SiO2 + 3C SiC + 2CO
SiC + 2COWhen 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is produced.
Solution:
Mass of SiO2 =100 kg=100000g
Mass of SiC produced =5.14 kg =51400g
100000g
SiO2 + 3C SiC + 2CO
SiC + 2CO 60g 40g
Now, 60g of SiO2 =40g of SiC
100000g of SiO2 =
=66666.67 g
Actual yield of Sic =51400 g
Theoretical yield of SiC =66666.67g
% yield =
=
=77.1%
Q22: (a) What is Stoichiometry? Give its assumptions? Mention two important law, which help to perform the Stoichiometry calculations.
(b) What is a limiting reactant? How does it control the quantity of the product formed? Explain with three examples
Q 23: (a) Define yield. How do we calculate the percentage yield of a chemical reaction?
(b) What are the factors which are mostly responsible for the low yield of the products in chemical reactions.
Q24: Explain the following with reasons.
(j) Law of conservation of mass has to be obeyed during Stoichiometric calculations.
(ii) Many chemical reactions taking place in our surrounding involves the limit reactants.
(iii) No individual neon atom in the sample of the element has a mass of 20.18amu.
(iv) One mole of H2 SO4 should completely react with two moles of NaOH. How does Avogadro, s number help to explain it.
(v) One mole H2 O has two moles of bonds , three moles of atoms , ten moles of electrons and twenty eight moles of the total fundamental particles present in it.
(vi) N2 and CO have the same number of electrons, protons and neutrons.
Ans. (i) According to law of conservation of mass, the amount of each element is conserved in a chemical reaction. Chemical equations are written and balanced on the basis of law of conversation of mass. Stoichiometry calculations are related with the amounts of reactants and products in a balanced chemical equation. Hence, law of conservation of mass has to be obeyed during stoichiometric calculations.
(ii) In our surrounding many chemical reactions are taking place which involve oxygen. In these reactions oxygen in always in excess quantity while other reactant are in lesser amount. Thus other reactants act as limiting reactants.
(iii) Since the overall atomic mass of neon in the average of the determined atomic masses of individual isotopes present in the sample of isotopic mixture .Hence, no individual neon atom in the sample has a mass of 20.18amu.
(iv) H2 SO4 +2NaOH Na2 SO4 + 2H2 O
Na2 SO4 + 2H2 O1 mole 2moles
2 moles of H+ ions 2 moles of OH ions
2x6.02x1023 H+ ions 2x6.02x1023 OH ions
Once mole of H2SO4 consists of 2 moles of H+ ions that contains twice the Avogadro’s number of H+ ions. For complete neutralization it needs 2 moles of one mole of H2 SO4 should completely react with two moles of NA OH.
(v) Since one molecule of H2O has two covalent bonds between H and O atoms. Three atoms, ten electrons and twenty eight total fundamental particles present in it. Hence, one mole of H2 O has two moles of bond, three moles of atoms, ten moles of electrons and twenty eight moles of total fundamental particle present in it.
In N2 there are 2 N atoms which contain 14 electrons (2x7),14 protons (2x7) and 14 neutrons (2x7) . In CO, there are one carbon and one oxygen atoms. It contains 14 electrons (6carbon e +8 oxygen e), 14 protons (
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Salam sir, MCQ # 16. Correct answer is 3.34*10^23 not 3.37*10^23.
Written in 2013, still helpful in 2015!!!
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(a) Mol-mass of NH3 =14+3=17g
Mass of N =14g
% of N =x100
=82.35% Answer
please help me in ans how it came after process........??
(a) Mol-mass of NH3 =14+3=17g
Mass of N =14g
% of N =x100
=82.35% Answer
please help me in ans how it came after process........??
% of N= mass of N/molar mass×100
=14/17 × 100
= 82.35 %
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Very nice helpul fir students. Great work sir. May Allah give the bless of this work. Continue this in future...Best of luck.
Very nice helpul fir students. Great work sir. May Allah give the bless of this work. Continue this in future...Best of luck.
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Where are the answers of Q no 4a and 4b
Where are the answers of Q no 4a and 4b
Question 22&23 Ka answers ni ha iss me
thanks yaar
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