1st Chapter Short Question Answers (Fsc Part 1) - Malik Xufyan

www.onlychemistrydiscussion.blogspot.com Malik Xufyan
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First Chapter Short Questions (Fsc Part 1)
Malik Xufyan
Msc Chemistry
Q4: What are ions? Under What condition are they produced?
Ans:
Ions
Ions are produced by gaining or loosing electrons .
ions are those substance which have positive or negative ion
cation are those specie which have positive charge by removal
of electron
Anion are those species which have negative charges by gaining
of electrons
Conditions:
Ions are produced by
(i) By dissolving ionic compounds in water
(ii) By X-rays
(iii) In mass spectrometry
(iv) By removing or adding electron in atom
Q8: Justify the following statements:
(a) 23 g of sodium and 238g of uranium have equal number
of atoms in them
Ans :
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.

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(b) Mg atom is twice heavier than that of carbon.
Ans:
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.
Mass of 1 atom of Mg= 24 g
Mass of 1 atom of C = 12 g
(c) 180g of glucose and 342 g of sucrose have the same
number of molecules but different number of atoms present
in them.
Ans:
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 (180 g) of glucose and I mole (342 g) 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 H22O11 contains 24 atoms.
Therefore, 6.02x1023 molecules of glucose contain different
atoms as compound to 6.02x1023
molecules of sucrose.
Hence, 180 g of glucose and 342 g of sucrose have the same
number of molecules but different number of atoms present in

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them.
(d) 4.9g of H2SO4 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.
Ans:
H2SO4 <------> 2H+ + SO4-2
When one molecules of H2SO4 completely ionizes in water
then it produces two H+ ion and one SO4 -2
ion,.
Hydrogen ion carries a unit positive charge whereas SO4-2
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) One mg of K2Cr2O7 has thrice the number of ions than
the number of formula units when ionized in water.
Ans:
When potassium dichromates ionizes in water , it dissociate in to
thrice the number of ions
as
K2Cr2O7 -------------------------------------> 2K+ + Cr2O7-2
K2 Cr O4 when ionizes in water produces two k+ ions and one
Cr2O7 -2
ion.

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Thus each formula unit of K2Cr2O7 produces three ions in
solution .Hence one mg of K2Cr 2O7 has thrice the number of ion
than the number of formula units 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.
Ans:
According to definition of Molar Volume:
we can write :
2g of H2 =1 mole of H2 =6.02x1023
molecules of H2 at STP =22.414dm3
16g of CH4 =1 mole of CH4 =6.02x1023
molecules of CH4 at STP =22.414 dm3
144g of CO2 =1 mole of CO2 =6.02x1023
molecules of CO2 at STP =22.144 dm3
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 ,16 g 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.
Q8: Define the following terms and give three examples of
each.
i. Gram atom:
Definition
The atomic mass of an element expressed in grams is called
gram atom of an element.
Formula:

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Number of gram atoms of a meter an element= mass of an element in
grams/ atomic mass of an element
Example:
1 gram atom of hydrogen = 1.008 gm
1 gram atom of carbon = 12.00 gm
1 gram at of uranium = 238 gm
ii. Gram Molecular Mass
Definition
The molecular mass of a substance expressed in grams is called
Gram Molecular mass or mole of substance.
Formula
Number of Molecular substance = Mass of molecular substance
in grams/ Molecular mass of molecular substance
Example
1 gram molecule of water = 18 g
1 gram molecule of H2SO4= 98 g
1 gram molecule of sucrose= 432 g
iii. Gram Formula:
Definition
The formula mass of an ionic compound expressed in grams is
called gram formula of the substance.
Formula
Number of gram formula= mass of ionic substance / formula
mass of ionic substance
Example
1 gram formula of NaCl = 58.50 gms
1 gram formula of Na2CO3 = 106 gm
1 gram formula of AgNO3 = 170 gm

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The atomic mass, molecular mass, formula mass or ionic mass
of the substance expressed in grams is called moles of those
substances.
iv. Gram ion:
Definition:
The ionic mass of an ionic specie expressed in grams is called
one gram ion or one mole of ions.
Formula
Number of gram ions = mass of ionic substance/Formua mass of ionic specie
Examples
1 gram ion of OH–1 = 17 grams
1 gram ion of SO = 96 grams
1 gram ion of CO = 60 grams
iv. Molar volume
Deifinition:
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
.
Example
1 mole of H2 =6.02 x 10 23 molecules of H2= 2.06 g of H2 = 22.414 dm3 at S.T.P
v. Avogadro’s number
Definition
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.
Representation
It is represented by NA. Its value is 6.02 x 1023
.
Example
Mass of sodium = 23 grams= 1mole=6.02 x 1023
atoms

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Mass of uranium = 238 g =1 mole = 6.02x 1023
atoms
vi. Stoichiometry
Definition
Stoichiometry is the branch of chemistry which gives a
quantitative relationship between reactants and products in
balanced chemical equation.
Assumptions of Stoichiometry:
a) reactants are completely converted into products
b) no side reactions occurs
c) while doing calculations, the law of conversion of mass and
the law of definite proportion are obeyed.
vii. Percentage yield
Defininition
The yield which is obtained by dividing actual yield with
theoretical yield and multiplying by 100 is called percent yield.
Formula
Percentage yield is efficiency of reaction which is determined by
Q25: Explain the following with reasons.
(j) Law of conservation of mass has to be obeyed during
Stoichiometric calculations.
Ans
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 la

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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) Many chemical reactions taking place in our surrounding
involves the limit reactants.
Ans:
According to the defininion of limiting reactant:
A limiting reactant is one which has limited quantity and
consumed first in a chemical reaction.
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.
Example:
1. petrol burns in excess of oxygen present in air
2. Rusting of iron in the excess of oxygen present in air.
(iii) No individual neon atom in the sample of the element
has a mass of 20.18amu.
Ans:
According to the average atomic mass:
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.

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Calculation:
Average atmic mass = 20 x 90.92 + 21 x 0.26 + 22 x 8.82/ 100
=20.18 amu.
(iv) One mole of H2 SO4 should completely react with two
moles of NaOH. How does Avogadro, s number help to
explain it.
Ans:
H2 SO4 + 2NaOH --------------------> Na2 SO4 + H2 O
1 mole 2moles
2 moles of H+
ions 2 moles of OH ions
2x6.02x1023
H+
ions 2x6.02x1023
OH ions
Hence one 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 NaOH.
(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.
Ans:
Since one molecule of H2O has two covalent bonds between H
and O atoms.
There are 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

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atoms, ten mole
of electrons and twenty eight moles of total fundamental particle
present in it in detail u can write that:
Bonds:
1 molecule of H20 contains bonds = 2
6.02 X 10 23
molecules contain bonds = 2 X 6.02 X 10 23
Thus 1 mole of H20 contains bonds = 2 moles
Atoms:
1 molecule of H20 contains atoms= 3
6.02 X 10 23
molecules contains atoms = 3 X 6.02 X 10 23
Thus 1 mole of H20 contains atoms = 3 moles
Electrons:
One molecue of H2O contain 2 H atoms and 1 O atom
Since
One O atom contains electrons = 8
One H atom contains electrons = 1
two H atom contains electrons = 2
Hence 1 molecule of H2O contains = 2+8 = 10 e
6.02 X 10 23
molecules conatains = 10 X 6.02 X 10 23
Thus 1 mole of H2O contains electrons =3 moles
Total Fundamental particles :
1 oxygen atom contains = 8 electrons , 8 protons , 8 necutrons
1 oxygen atom contains total fundamental particles = 8+8+8=24
1 hydrogen atom contains = 1 electron, 1 proton , 0 neutron
1 Hydrogen atom contain total fundamental particles = 1+1+0=
2
2 Hydrogens atoms contains total fundamental particles= 4
Hence

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1 molecule of water contains fundamental particles = 24+4= 28
6.02 X 10 23
molecules contain fundamental particles = 28 X
6.02 X 10 23
Thus 1 mole of water contains bonds = 28 moles
(vi) N2 and CO have the same number of electrons, protons
and neutrons.
Ans:
For N2
In N2 there are 2 N atoms which contain 14 electrons (2x7),
14 protons (2x7) and 14 neutrons (2x7) .
For CO
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).
For CO & N2
Hence , N2 and CO have the same number of electrons,
protons and neutrons.

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Remember that electrons, protons and neutrons of atoms remain
conserved during the formation of molecules in a chemical
reaction.
Q. Why actual yield is less than theoretical yield?
Ans:
There are three reasons for that
1. A practically inexperienced worker has many shortcomings
and cannot get the expected yield.
2. The processes like filtration, separation by distillation,
separation by separating funnel , washing , drying and
crystallization , if not properly carried out , decreased the
actual yield.
3. Some of the reactants might take part in competing side
reaction andreduce the amount of desired product.
Q. Differentiate between empirical and molecular formula
Sr.# Emperical Formula Molecular Formula
1. The empirical formula is the
simplest formula for a
compound
A molecular formula is the same
as or a multiple of the empirical
formula
2. Ionic and covalents
compounds have empirical
formula
Ionic compounds do not
have molecular formula
3. Example:
Glucose and benzene have
CH20 and CH respectively
Example :
Molecular formula of
benzene and glucose are
C6H6 and C6H1206
4. Formula: Formula

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Empirical formula =
molecular formula / n
Moleularformula=
n(empirical formula)

1st Chapter Short Question Answers (Fsc Part 1) - Malik Xufyan

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1st Chapter Short Question Answers (Fsc Part 1) - Malik Xufyan