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 =6.02x10

23 molecules of H2 at STP =22.414dm

3

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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