Chemistry Ch 1
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F.S.C part1 Chap 1st Posted by chismtry zone on Sunday, 18 December 2011 at 01:33
Chapter No. 1
BASIC CONCEPTS MCQs Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
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
Swedish chemist J. Berzelius determined the (a)
atomic no.
(b)
atomic volume
(c)
atomic mass
(d)
atomic density
The number of atoms present in a molecule determine its (a)
molecularity
(b)
basicity
(c)
acidity
(d)
atomicity
When an electron is added to a unipositive ion we get (a)
anion
(c)
neutral atom
(b) (d)
cation molecule
CO+ is an example of: (a)
free radical
(b)
(c)
an ionic molecular ion
(d)
stable molecule
cationic molecular ion
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
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
Q.9
Q.10
(a)
I.R. Spectrophotometer
(c)
Mass Spectrometer
(b) (d)
U.V. Spectrophotometer
Colourimeter
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
Simplest formula that gives us information about the simple ratio of atoms in a compound is
called
Q.11
Q.12
Q.13
Q.14
Q.15
Q.16
(a)
structural formula
(b)
molecular formula
(c)
empirical formula
(d)
molar ratio
Percentage of oxygen in H2O is (a)
80%
(b)
88.8%
(c)
8.8%
(d)
9.8%
More abundant isotope of an element is one with (a)
even atomic no.
(c)
Even mass no.
(b) (d)
odd atomic no.
odd mass no.
Large no. of isotopes are known for the elements whose masses are multiple of (a)
two
(b)
four
(c)
six
(d)
eight
When 0.01 kg of CaCO3 is decomposed the CO2 produced occupies a volume at S.T.P. (a)
2.2414 dm3
(c)
22414 dm3
(b)
22.414 dm3 (d)
224014 dm3
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
No. of molecules present in 10gm of water are (a)
3.37 x 1023
(b)
33.7 x 1023
(c) Q.17
Q.18
Q.19
Q.20
Q.21
Q.22
Q.23
3.37 x 1024
3.037 x 1024
(d)
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
The least no. of molecules present in 30 gm of (a)
N2O
(c)
NO2
(b)
NO (d)
N2O3
Which of the following has highest percentage of nitrogen (a)
(NH4)2SO4
(c)
(NH4)2HPO4
NH4H2PO4
(b)
(NH4)3PO4
(d)
0.1 mole of Na3PO4 completely dissociates in water to produce Na+ (a)
6.02 x 1022
(c)
1.806 x 1023
6.02 x 1023
(b)
1.806 x 1022
(d)
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
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
Stoichiometry is the branch of chemistry which deals with the study of quantitative relationship
among the various
Q.24
(a)
reactants
(b)
(c)
Reactants and products
products (d)
all of above
500 cm3 of H2 gas at STP contradictions of hydrogen (a)
6.02 x 1023
(b)
3.01 x 1022
(c) Q.25
Q.26
Q.27
Q.28
Q.29
Q.30
Q.31
Q.32
Q.33
2.68 x 1022
1.34 x 1022
(d)
Largest number of H+ ions are produced by complete ionization of 0.0050 mole of H2SO4
(a)
0.01 mole of HCl
(b)
(c)
0.000334 moles of H3PO4
(d)
all above
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
The largest number of H+ are produced by complete ionization of (a)
0.100 2 moles of HCl (b)
(c)
0.0334 moles of H3PO4
0.051 moles of H2SO4 (d)
All of the above
A sample of pure matter is (a)
element
(c)
substance
(b)
compound
(d)
mixture
(b)
Nanometer
nm stands for (a)
Newton meter
(c)
Newton square meter (d)
none of the above
One calorie is equal to (a)
4.184 J
(b)
41.84 J
(c)
0.4184 J
(d)
0.04184 J
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
27 grams of Al will react completely with how much mass of O2 to produce Al2O3 (a)
8 gm of oxygen
(c)
32 gm of oxygen
(b) (d)
24 gm of oxygen
Mole of SO2 contains (a)
16 gm of oxygen
6.02 x 1023 atoms of oxygen
Q.34
Q.35
Q.36
Q.37
Q.38
(b)
18.1 x 1023 molecules of SO2
(c)
6.023 x 1023 atom of sulphur
(d)
4 gram of SO2
The largest number of molecules are presenting (a)
3.6 gram of H2O
(c)
2.8 gm of CO
4.8 gram of C2H5OH
(b)
5.4 gms of N2O5
(d)
The mass of one mole of electron is (a)
1.008 mg
(b)
0.184 mg
(c)
1.673 mg
(d)
0.55 mg
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
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
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) Q.41
Q.42
Q.43
Q.44
Q.45
Q.46
Q.47
Q.48
are equal when 100 molecules of H3PO4 are thrown in excess of water
A molecule having two atoms is called (a)
monoatomic molecules
(b)
diatomic molecules
(c)
Polyatomic molecules (d)
homoatomic molecule
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
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
Nickel has isotopes (a)
1
(b)
3
(c)
5
(d)
7
Cadmium has isotopes (a)
3
(b)
5
(c)
7
(d)
9
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
Number of gram atoms in 0.1 gm of Na is (a)
0.0043
(b)
0.0403
(c)
0.403
(d)
None of these
Molecule of haemoglobin contains atoms
Q.49
(a)
15,000
(b)
12,000
(c)
10,000
(d)
8,000
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 = = Q.2
= 0.01 grams atoms
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. 180 grams of glucose (C6H12O6) and 342 grams of sucrose (C12H22O11) are their molar masses indicating one mole of each (glucose and sucrose) one mole of a substance contains equal number of molecules i.e. 6.02 x 1023.
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 of 4.9 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. 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:
is limiting reactatnt C + O2 CO2
(i)
Burning of coal to form CO2---Coal
(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). 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 2NaOH2Na+ + 2OH2H+ + 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
Ans.
Ion
(ii) Positive ion
(iii) Negative 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 Ans.
Define and explain the molecular ion ?
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 Q.25
12C6 is 12.00 amu. The relative atomic mass of
H is 1.0078 amu.
Define Gram atom ?
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 1 gram atom of carbon = 12.00 gm 1 gram at of uranium = 238 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 = 1 gram ion of OH–1 = 17 grams 1 gram ion of SO = 96 gram 1 gram ion of CO = 60 gram 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 = 1 gram formula of NaCl = 58.50 gms 1 gram formula of Na2CO3 = 106 gm 1 gram formula of AgNO3 = 170 gm 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:
4.45
5.45
O 1.82
Divide number of grams atoms by the smallest number C: :
H:
O
3:
1
: 3:
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 = = = 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. (i)
Select the most suitable answer from the given ones in each question.
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)
(v)
008mg (b)
0.55mg
(c)
0.184mg
(d)
1.673mg
27g of Al will react completely with how much mass of O2 to produce Al2O3
(a)
8g of oxygen
(c)
32g of oxygen (vi)
(viii)
16g of oxygen (d)
24g of oxygen
The number of moles of CO2 which contain 8.0 g of oxygen. (a)
(vii)
(b)
0.25
(b)
0.50
(c)
1.0
(d)
1.50
The largest number of molecules are present in (a)
3.6g of H2 O
(c)
2.8 g of CO
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
(b) (d)
4.8g of C2H5 OH 5.4g of N2O5
(ix)
The volume occupied by 1.4 g of N2at STP is
(x)
(a)
2.24 dm3
(c)
1.12 dm3
22.4dm3
(b) (d)
112 cm3
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
Q2:
Fill in the blanks:
(iv)b
(v)d
(vi)a
(vii)a
(viii)c (ix)c
(x)d
(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)
(v)
Products
mass
(iii)
Soddy (iv) carbon, hydrogen
(vi) 6NA,6NA,12NA
(vii)
1.505x1023
Q3:
Indicate true or false as the case may be:
(viii) conservation and multiple proportion
(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.
Q4:
(i)
False
(ii)
False
(v)
False
(vi)
True
(iii)
True
(iv)
(vii)
False
(viii)
False False
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 107
Ag
109
Ag
106.90509
108.90476
51.84
48.16
Solution:
The mass contribution for silver are:
Isotopes
Fractional abundance isotopic mass mass contribution
107
Ag
107
0.5184x107=55.4688
109Ag
107
0.4816x109=52.4944
Fractional atomic mass of silver
=107.9632
Hence the fractional atomic mass of silver is =107.9632 Q6:
Ans.
Boron with atomic number 5 has two naturally occurring isotopes. Calculate the percentage abundance of 10B and 11B from the following information.
Solution:
Average atomic mass of boron
=10.81 amu
Isotopic mass of 10B
=10.0129 amu
Isotopic mass of 11B
=11.0093
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 )+(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
10
B)(isotopic mass of10B
x
=
Fractional abundance of 10B
=0.2000
Fractional abundance of 11B
=(1-0.2000)=0.8000
By percentage the fractional abundance of isotope is
Q7:
%of 10B
=0.2000x100 =20% Answer
% of 11B
=0.8000x100 =80%Answer
Define the following terms and give three examples of each. (i)
Gram atom
(iii)
Gram molecular mass
molar
(ii)
volume
(iv) (vi)
Stoichiometry
(viii)
Gram molecular mass Gram ion
Avogadro’s
number
(v) (vii)
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,
6.02x1023molecules 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
When one molecules of H2 SO4 completely ionizes in water it produces two H+ ion and one SO
ion,. Hydrogen ion carries a unit positive charge whereas SO
ion 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 H2SO4 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
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. (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)
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 =32+32=64 g mol-1
Molecular mass of SO2 . 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 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
= 0.822g
=
=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)
=52+32+71=155g mol-1
Molecular mass of CrO2C12 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
No of formula units
No of moles x Avogadro,s No =0.816mole x 6.02x1023 formula units =4.91x1023 formula units
No of K+ ions
(i)
No of CIO
ions
=4.91x1023 Answer =4.91x1023 Answer
No of CIO ions
=4.91x1023 Answer
No of O atoms
= 4.91x1023 x3 =14.73x1023 =1.473x1024 Answer
Q 11
(a)
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?
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 Molar mass of aspartame
=243g =294g mol -1
No of molecules of aspartame=? No of molecules of aspartame=
xNA
= = =4.98x1021 molecules. Now,1 molecule of aspartame contains
=18 H atoms =18x4.98x1021 H atoms
4.98x 1022 molecules
=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 No of moles of M Mass of MF2
=MF2
=0.6 mol =46.8g
The molar of M:F in the compounds;
No of moles of F Mass of F
=0.6x2=1.2mol Answer =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 (b)
Mass of substance
Answer
= 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
= x6.02x1023
No of molecules in N2 O4 =
No of molecules in SO2 =x6.02x1023
=2.62 x1023 molecules =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 =2.0x6.02x1023 x3=3.6x1024 ions
No of formula units of MgC12
No .of ions in one formula unit of MgC12 =3 =1.20x1024 x3=3.6x1024 ions
Total no of ions in MgC12
2.0moles of MgC12 contain lager number of total ions than 2.3 moles of NaC1o3MgC1 Answer (f)
No of molecules
=
No of molecules in H2 O2 No of molecules in H2 O2
= =
NA x6.02x1023=3.68x1023 molecules x6.02x1023=1.95x1023 molecules
11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2 H2 O2Answer (g)
No of formula units No of formula units NaBr
=moles xNA =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 No of formula units of NaC1
=
=3.01x1023 Na+ ions
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 238
U Answer
Q 13: (a)
Calculate the percentage of nitrogen in the four important fertilizer i.e., (i)NH3 (ii)NH2CONH2(Urea) (iii)(NH4)2SO4
(b)
(iv)NH4 NO3
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 %of N
=28g =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
(II)
Mol-mass of ((NH2 )2 SO4
(III)
Q 14:
==26.96% Answer =28+9+31+64=132g
Mass of N
=28g
Mass of P
=
%of N
=
=21.21% Answer
%of P
=
=23.48% Answer
Mol-mass of (NH2 )2 SO4
=42+12+31+64=149g
Mass of N
=42g
Mass of P
=31g
%of N
=
%of P
=
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
=180g mol-1
Mol-mass of 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 =6 O –atoms
1 molecule of glucose contains 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 At. Mass of C=12g mol-1
% of H =9.7g
% of O=51.6g
At. Mass of H=1.008g mol-1
No of moles of C
=
No of moles of H
=
No of moles of O
=
At. Mass of O =16g mol-1
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
10
:
12
:
2
:
1
Atomic Ratio
Empirical formula
=C10 H12 N2 O
Empirical formula mass
=120+12+28+16=176g mol-1 =176g mol-1
Molecular mass n= Q17:
An unknown metal M reacts with S to from a compound with a formula M2S3 .If 3.12 gof
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 Mass of M
= M2 S3 =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 Q19:
=52
The octane present in gasoline burns according to the following equation. 2C8 H18 (i)
+ 2502(g)
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)
(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)
2 moles
Now,
2 moles of C8 H18 1 mole of C8 H18
=16 moles of CO2 = =8 moles of CO2 Answer
(c)
6 moles
2C8 H18 (i)
+ 2502(g)
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)
2 moles
Now,
1800moles
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 8 moles of CO2
=2moles of C8 H18 = =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 =1 molx 114 g mol-1
ofC8H18
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 =32g mol-1
Molar mass of S No of moles of S
=
0.74 mole 0.94 mole
2A1 2 mole
Now,
+
3S 3 mole
A12 S3 1 mole
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 0.94 moles of S
=1 moles of A12 S3 = =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
2 mole
Now,
A12 S3
3 mole
3 moles of S 0.94 moles of S
=2 moles of A1 =
= 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 Q20:
=20-17=3g
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
Solution: 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 2 moles
Now , 2moles of N2H4 3.125moles of N2H4
3N2 +4 H2O
1mole
=3moles of N2 = =4.69 mole of N2
3moles
Q22:
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 Molar mass of N2
=4.69 moles =28g
Mass of N2
mol-1
=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
When 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
60g
Now,
SiC +
2CO
40g
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% (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) (b)
Define yield. How do we calculate the percentage yield of a chemical reaction? 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 1 mole
+2NaOH
Na2 SO4
2moles
2 moles of H+ ions
2 moles of OH ions
2x6.02x1023 H+ ions
2x6.02x1023 OH ions
+ 2H2 O
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 (6 C proton +8 O proton ) and 14 neutrons (6 neutrons +8 O neutrons).Hence , N2 and CO have the same number of electrons, protons and neutrons. Remember that electrons, protons and neutrons of atoms remain conserved during the formation of molecules in a chemical reaction.
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13 COMMENTS:
Farooq ali said... very nice .................................................... 7 APRIL 2013 21:14
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Muhammad Mesum said... Thanks man............ classy, great effort 13 JUNE 2013 04:28
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israr khan said... nice buddy 6 SEPTEMBER 2013 13: 02
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sania saeed said... there are some little mistakes 5 OCTOBER 2013 11:20
erum khan said...
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khurram ch said... good 9 DECEMBER 2013 03:2 3
Rose Jack said... Thorough prep .... :-) 5 JANUARY 2014 04:30
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