1st year chemistry 1st chapter short question answers - 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 K2Cr 2O7 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 M ass 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 & N 2

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 year chemistry 1st chapter short question answers - Malik Xufyan

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