CHEMISTRY
Time Allowed : 3 Hours ] [ Maximum Marks : 150
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Note : Draw diagrams and write equations wherever necessary.
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PART – I
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Note : i) Answer all
the questions.
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ii) Choose and write the correct
answer. 30 x 1 = 30
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1.
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The hybridisation in SF6
molecule is
a)sp3 b)sp3d2 c)sp3d d)sp3d3
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2.
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En = − 313.6/n2, If the value of En
= -34.84 to which value ‘n’ corresponds
a)4 b)3 c)2 d)1
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3.
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On moving down the group, the radius of an ion
a) Decreases b) Increases c) No change d) None of these
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4.
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Halogens belong to the group number
a) 14 b) 15 c) 17 d) 18
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5.
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Which of the following compounds will not give positive chromyl
chloride test?
a) CuCl2
b) HgCl2 c) ZnCl2 d) C6H5Cl
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6.
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The transition element with the lowest atomic number is
a) Scandium b) Titanium c) Zinc d) Lanthanum
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7.
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The electronic configuration
of Lanthanides is
a)[Xe]4f0 5d0 6s0 b)[Xe]
4f1-7 5d1 6s1
c)[Xe]4f1-14 5d1 6s2 d)[Xe]4f1-14 5d1-10
6s2
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8.
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_______form oxocations
a) Lanthanides b)
Actinides c) Noble gases d) Alkali metals
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9.
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The geometry of [Ni(CN)4]2- is
a)Tetrahedral
b)Square planar c)Triangular d) Octahedral
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10.
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92U235 nucleus
absorbs a neutron and disintegrates into 54Xe139, 38Sr94
and x. What will be
the product x?
a) 3 neutrons b) 2 neutrons c) α particle d) β particle
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11.
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A regular three dimensional arrangement of identical points in
space is called
a) Unit cell b) Space lattice c) Primitive d) Crystallography
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12.
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If ΔG for a reaction is negative, the change is
a) Spontaneous b)
Non-spontaneous c) Reversible d) Equilibrium
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13.
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The amount of heat exchanged with the surrounding at
constant temperature and pressure is
called
a) ΔE b) ΔH c) ΔS d) ΔG
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14.
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For the homogeneous gas reaction at
600 K 4NH3(g) + 5O2
(g) ⇌ 4NO(g) + 6H2O(g) the
equilibrium constant Kc has
the unit
a) (mol dm-3)-1 b) (mol dm-3) c) (mol dm-3)10
d) (mol dm-3)-9
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15.
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In the equilibrium N2 + 3H2 ⇌ 2NH3, the maximum yield of ammonia will be obtained
with the
process having
a)low pressure and high temperature b)low pressure and low
temperature
c)high
temperature and high pressure
d)high pressure and low
temperature
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16.
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Arrhenius equation is
a) k = Ae–1/RT
b) k = Ae–RT/Ea c) k = Ae–Ea/RT d) k = AeEa/RT
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17.
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Which one is the correct factor that
explains the increase of rate of reaction by a catalyst
a) shape selectivity
b) particle size
c) increase of free energy d)
lowering of activation energy
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18.
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In case of physical adsorption, there
is desorption when
a) temperature increases b) temperature
decreases
c) pressure increases d)
concentration increases
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19.
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Oil soluble dye is mixed with emulsion and emulsion remains
colourless then, the emulsion is
a) O/W b) W/O c) O/O d) W/W
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20.
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The laws of electrolysis were enunciated first by _______
a) Dalton b) Faraday c) Kekule d) Avogadro
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21.
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The ionization constant of phenol is higher than that of ethanol
because
a) phenoxide ion is bulkier than ethoxide b) phenoxide ion is stronger base
than ethoxide
c) phenoxide
ion is stablized through delocalization
d) phenoxide ion is less stable than ethoxide ion
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22.
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According to Lewis concept of acids and bases, ethers are
a) Neutral b) Acidic c) Basic d) Amphoteric
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23.
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Ether is formed when alkylhalide is treated with sodium
alkoxide. This method is known as
a) Hoffmann reaction b) Williamson's synthesis c) Wurtz synthesis d) Kolbe's reaction
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24.
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Tollen’s reagent is
a) ammoniacal cuprous chloride b) ammoniacal cuprous oxide
c) ammoniacal silver nitrate d) ammoniacal silver
chloride
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25.
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The compound which undergoes intramolecular dehydration with P2O5
is
a) acetic acid b) formic acid c) propionic acid d) Butyric acid
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26.
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Which of the following compounds has the smell of bitter almonds?
a) aniline b) nitro methane c) benzene sulphonic acid d)
nitrobenzene
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27.
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Nitromethane condenses with
acetaldehyde to give
a) nitro propane b) 1-nitro-2-propanol c) 2-nitro-1-propanol d) 3-nitro propanol
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28.
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The basic character of amines is due to the
a) tetrahedral structure b)
presence of nitrogen atom
c) lone pair of electrons on nitrogen atom d) high electronegativity
of nitrogen
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29.
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Two amino acids say A, B- react to give
a) two dipeptides b)
three dipeptides c) four dipeptides d) only one
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30.
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Ultimate products of hydrolysis of proteins is
a) aniline
b) aliphatic acid c) amino acid d) aromatic acid
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PART – II
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Note : i) Answer any
fifteen questions.
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ii) Each
answer should be in one or two sentences. 15 x 3 = 45
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31.
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What is bond order?
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32.
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Arrange the following elements in the increasing order of their
first ionisation potentials; give
proper explanation for your answer. C, N,
O, F
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33.
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Write a note on plumbo solvency.
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34.
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Why do noble gases form compounds with fluorine and oxygen only?
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35.
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[Ti (H2O)6]3+ is coloured,
while [Sc (H2O)6]3+ is colourless.
Explain
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36.
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Why transition elements form complexes?
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37.
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What is binding energy of Nucleus?
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38.
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What is coordination number?
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39.
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State
Troutan’s rule.
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40.
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What happens when Δng = 0,
Δng = –ve, Δng = +ve in a
gaseous reaction.
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41.
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Write a note on activation energy?
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42.
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What is a
consecutive reaction? Give example.
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43.
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What is
meant by catalytic poison? Give an example.
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44.
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State Ostwald’s dilution law.
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45.
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Meso-tartaric
acid is an optically inactive compound with chiral carbon atoms. Justify.
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46.
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Give
chemical tests to distinguish between propan-2-ol and 2-methyl- propan-2-ol.
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47.
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How does
glycerol react with i) PCl5 ii) KHS O4
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48.
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Distinguish acetaldehyde and acetone.
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49.
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Write a note on esterification reaction with an example.
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50.
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Give the reduction of nitromethane in a) acid medium, b)
neutral medium.
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51.
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In what way antipyretics are important.
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PART – III
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Note:Answer any seven questions choosing at least two
questions from each Section.7x5 =
35
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SECTION – A
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52.
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Briefly explain Molecular Orbital Theory.
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53.
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List the ores of silver. How silver is extracted from its chief
ore?
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54.
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What is lanthanide contraction? Discuss its causes and
consequences.
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55.
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Taking [FeF6]4- as an example, discuss
geometry and magnetic property using VB theory.
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SECTION – B
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56.
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Write the characteristics of entropy.
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57.
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In 1 litre volume reaction vessel, the equilibrium constant Kc
of the reaction
PCl5 ⇌ PCl3 + Cl2
is 2 x 10-4 lit-1 What will be the degree of
dissociation assuming only
a small extent of 1 mole of PCl5 has
dissociated?
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58.
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Differentiate simple and complex reactions.
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59.
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Write
notes on IUPAC conventions of representation of a cell with suitable
examples.
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SECTION – C
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60.
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Mention the methods of preparation of anisole. Explain the
reaction of HI with anisol.
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61.
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Explain the mechanism of Cannizaro’s reaction.
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62.
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How is benzoic acid obtained from (a) ethyl benzene (b) phenyl cyanide (c) carbon dioxide
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63.
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Write briefly on antibiotics? In what way antipasmodics are
helpful?
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PART – IV
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Note : i) Question No. 70 is compulsory and answer
any three from the remaining questions.
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ii) Answer four
questions in all. 4 x 10 = 40
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64.a.
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Explain the various factors that affect electron affinity.
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b.
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What are
silicones? How are they prepared?
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65.a.
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Explain coordination and ionisation isomerism with suitable
examples.
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b.
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How do nuclear fusion reactions differ from fission reactions?
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66.a.
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What is
super conductivity? Give its uses.
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b.
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Write
briefly about the preparation of colloids by dispersion methods.
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67.a.
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Explain
buffer action with example.
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b.
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Derive
Nernst equation of reversible cell.
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68.a.
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Write a short account on cis-trans isomerism.
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b.
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Account for reducing nature of Formic acid.
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69.a.
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How
are i) phenol ii) chloro benzene iii) biphenyl prepared
by using benzene diazonium chloride?
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b.
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How is
the structure of fructose determined?
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70.a.
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An
organic compound (A) C2H6O liberates hydrogen on
treatment with metallic sodium. (A) on
mild oxidation gives (B) C2H4O
which answers iodoform test. (B) when treated with
Conc. H2SO4
undergoes polymerisation to give (C), a cyclic compound.
Identify (A), (B)
and (C) and explain the reactions.
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b.
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A
compound of chromium, in which chromium exists in +6 oxidation state. Its
chief ore (A) on
roasting with molten alkali gives compound (B). This compound on acidification gave compound C. Compound C on treatment with KCl gave compound D. Identify the compounds A, B, C and D. Explain with proper chemical reactions. |
OR
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c.
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Compound (A) with molecular formula C2H4O
reduces Tollen's reagent. (A) on treatment with
HCN gives compound (B).
Compound (B) on hydrolysis with an acid gives compound (C) with
molecular
formula C3H6O3. Compound (C) is optically
active. Compound (C) on treatment with
Fenton's reagent gives compound (D)
with molecular formula C3H4O3. Compounds (C)
and (D)
give effervescence with NaHCO3 solution. Identify the
compounds (A), (B), (C) and (D) and
explain the reactions.
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d.
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0.1978 g
of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the
electrochemical equivalent of copper?
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BEST WISHES
Ex. No.: 1 COMPARISON OF POTASSIUM PERMANGANATE solutions
Date:
AIM
To compare the normalities of the two given Potassium permanganate solutions A and B and to estimate the amount of crystalline Potassium permanganate present in 500 ml of the weaker solution. You are provided with a standard solution of Ferrous ammonium sulphate containing 39.2 grams in one litre.
PROCEDURE
TITRATION - I: STANDARDISATION OF KMnO4 (A)
The burette is filled with the given KMnO4 (A) solution and titrated against 20ml of Ferrous ammonium sulphate solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of KMnO4 (A) is calculated.
TITRATION - I I: STANDARDISATION OF KMnO4 (B)
The burette is filled with the given KMnO4 (B) solution and titrated against 20ml of Ferrous ammonium sulphate solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted and the titration is repeated for concordant value. From the titer value, the normality of KMnO4 (B) is calculated. From the normalities of Potassium permanganate solutions A and B the weaker solution and the amount of KMnO4 present in 500ml is calculated, knowing that the equivalent mass of Potassium permanganate is 31.6.
SIMPLE PROCEDURE
Sl No.
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CONTENTS
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TITRATION - I
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TITRATION - II
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1
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Burette Solution
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Potassium permanganate (A)
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Potassium permanganate (B)
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2
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Pipette Solution
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20 ml of Ferrous ammonium sulphate
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20 ml of Ferrous ammonium sulphate
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3
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Acid Added
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20 ml of 2N Sulphuric acid
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20 ml of 2N Sulphuric acid
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4
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Temperature
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Lab temperature
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Lab temperature
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5
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Indicator
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Self-indicator (KMnO4)
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Self-indicator (KMnO4)
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6
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Endpoint
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Appearance of permanent pale pink colour
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Appearance of permanent pale pink colour
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7
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Equivalent mass of KMnO4 = 31.6
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NORMALITY OF STANDARD SOLUTION
Amount of Ferrous ammonium sulphate present in 1 litre = 39.2 g
Normality of Standard Ferrous ammonium sulphate = Mass per litre / Equivalent mass
= 39.2 / 392
\ Normality of Standard Ferrous ammonium sulphate = 0.1 N
TITRATION - I : KMnO4 (A) Vs STANDARD Ferrous ammonium sulphate
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
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Volume of
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Burette Readings
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Volume of
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Concordant
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No
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Ferrous ammonium sulphate
ml
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Initial
ml
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Final
ml
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KMnO4 (A)
ml
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Titer Value
ml
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1
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20
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0
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2
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20
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0
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CALCULATION
Volume of Ferrous ammonium sulphate = V1 = 20 ml
Normality of Ferrous ammonium sulphate = N1 = 0.1 N
Volume of KMnO4 (A) = V2 =
Normality of KMnO4 (A) = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 (A) = N
TITRATION - II : KMnO4 (B) Vs STANDARD Ferrous ammonium sulphate
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
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Volume of
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Burette Readings
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Volume of
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Concordant
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No
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Ferrous ammonium sulphate
ml
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Initial
ml
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Final
ml
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KMnO4 (B)
ml
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Titer Value
ml
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1
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20
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0
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2
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20
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0
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CALCULATION
Volume of Ferrous ammonium sulphate = V1 = 20 ml
Normality of Ferrous ammonium sulphate = N1 = 0.1 N
Volume of KMnO4 (B) = V2 =
Normality of KMnO4 (B) = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 (B) = N
COMPARISON
1. Normality of KMnO4 (A) =
2. Normality of KMnO4 (B) =
\ The Weaker KMnO4 solution is = KMnO4 (____)
AMOUNT CALCULATION
Amount of KMnO4 present in 500ml of the weaker solution
= Equivalent mass x Normality x 500 / 1000
= Equivalent mass x Normality x 500 / 1000
= 31.6 x x 500 / 1000
= gram
RESULT
1. Normality of KMnO4 (A) = N
2. Normality of KMnO4 (B) = N
3. The Weaker KMnO4 solution is = KMnO4 (____)
4. Amount of crystalline KMnO4 present in 500ml of the weaker solution = gram
Ex. No: 2 COMPARISON OF Ferrous ammonium sulphate solutions
Date:
AIM
To compare the normalities of the two given Ferrous ammonium sulphate solutions A and B, and to estimate the amount of crystalline Ferrous ammonium sulphate present in 750 ml of the stronger solution. You are provided with a standard solution of KMnO4 containing 3.16 grams in one litre.
PROCEDURE
TITRATION - I : STANDARDISATION OF Ferrous ammonium sulphate (A)
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous ammonium sulphate solution (A) pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of Ferrous ammonium sulphate (A) is calculated.
TITRATION - I I: STANDARDISATION OF Ferrous ammonium sulphate (B)
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous ammonium sulphate solution (B) pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titer value, the normality of Ferrous ammonium sulphate (B) is calculated. From the normalities of Ferrous ammonium sulphate solutions A and B the stronger solution and the amount of Ferrous ammonium sulphate present in 750ml is calculated knowing that the equivalent mass of Ferrous ammonium sulphate is 392.
SIMPLE PROCEDURE
Sl No
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CONTENTS
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TITRATION - I
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TITRATION - I I
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1
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Burette Solution
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Potassium permanganate
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Potassium permanganate
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2
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Pipette Solution
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20 ml of Ferrous ammonium sulphate (A)
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20 ml of Ferrous ammonium sulphate (B)
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3
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Acid Added
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20 ml of 2N Sulphuric acid
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20 ml of 2N Sulphuric acid
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4
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Temperature
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Lab temperature
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Lab temperature
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5
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Indicator
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Self-indicator (KMnO4)
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Self-indicator (KMnO4)
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6
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Endpoint
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Appearance of permanent pale pink colour
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Appearance of permanent pale pink colour
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7
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Equivalent mass of Ferrous ammonium sulphate = 392
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NORMALITY OF STANDARD SOLUTION
Amount of KMnO4 present in 1 litre = 3.16 g
Normality of Standard KMnO4 = Mass per litre / Equivalent mass
= 3.16 / 31.6
\ Normality of Standard KMnO4 = 0.1 N
TITRATION - I : STANDARD KMnO4 Vs Ferrous ammonium sulphate (A)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
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Volume of
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Burette Readings
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Volume of
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Concordant
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No
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Ferrous ammonium sulphate (A)
ml
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Initial
ml
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Final
ml
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KMnO4
ml
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Titer Value
ml
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1
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20
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0
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2
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20
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0
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CALCULATION
Volume of Ferrous ammonium sulphate (A) = V1 = 20 ml
Normality of Ferrous ammonium sulphate (A) = N1 = ?
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 = 0.1 N
V1 X N1 = V2 X N2
Therefore, N1 = V2 X N2 / V1
=
Normality of Ferrous ammonium sulphate (A) = N
TITRATION - I I : STANDARD KMnO4 Vs Ferrous ammonium sulphate (B)
INDICATOR: KMnO4 (SELF-INDICATOR)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
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Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
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Ferrous ammonium sulphate (B)
ml
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Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Ferrous ammonium sulphate (B) = V1 = 20 ml
Normality of Ferrous ammonium sulphate (B) = N1 = ?
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 = 0.1 N
V1 X N1 = V2 X N2
Therefore, N1 = V2 X N2 / V1
=
Normality of Ferrous ammonium sulphate (B) = N
COMPARISON
1.Normality of Ferrous ammonium sulphate (A) = N
2.Normality of Ferrous ammonium sulphate (B) = N
\ The Stronger Ferrous ammonium sulphate solution is = Ferrous ammonium sulphate (__)
AMOUNT CALCULATION
present in 750ml of the stronger solution = Equivalent mass x Normality x 750 /1000
= 392 x x 750 / 1000
= gram
RESULT
1. Normality of Ferrous ammonium sulphate (A) = N
2. Normality of Ferrous ammonium sulphate (B) = N
3. The Stronger Ferrous ammonium sulphate solution = Ferrous ammonium sulphate (__)
4. Amount of crystalline Ferrous ammonium sulphate present in 750ml of the Stronger solution
= gram
= gram
Ex No.: 3 COMPARISON OF Ferrous sulphate solutions
Date:
AIM
To compare the normalities of the two given Ferrous sulphate solutions A and B, and to estimate the amount of crystalline Ferrous sulphate present in 1250 ml of the weaker solution. You are provided with a standard solution of KMnO4 containing 1.6 grams in 500 ml.
PROCEDURE
TITRATION - I : STANDARDISATION OF Ferrous sulphate (A)
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous sulphate (A) solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of Ferrous sulphate (A) is calculated.
TITRATION - I I: STANDARDISATION OF Ferrous sulphate (B)
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous sulphate (B) solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titer value, the normality of Ferrous sulphate (B) is calculated. From the normalities of Ferrous sulphate solutions A and B the weaker solution and the amount of Ferrous sulphate present in 1250ml is calculated knowing that the equivalent mass of Ferrous sulphate is 278.
SIMPLE PROCEDURE
S.
No
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CONTENTS
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TITRATION - I
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TITRATION - I I
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1
|
Burette Solution
|
Potassium permanganate
|
Potassium permanganate
|
2
|
Pipette Solution
|
20 ml of Ferrous sulphate (A)
|
20 ml of Ferrous sulphate (B)
|
3
|
Acid Added
|
20 ml of 2N Sulphuric acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
Lab temperature
|
Lab temperature
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent mass of Ferrous sulphate = 278
| ||
NORMALITY OF STANDARD SOLUTION
Amount of KMnO4 present in 500ml = 1.6 g
Amount of KMnO4 present in 1 litre = 1.6 x 2 = 3.2 g
Normality of Standard KMnO4 = Mass per litre / Equivalent mass
= 3.2 / 31.6
\ Normality of Standard KMnO4 = 0.1013 N
TITRATION - I : STANDARD KMnO4 Vs Ferrous sulphate (A)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Ferrous sulphate (A)
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Ferrous sulphate (A) = V1 = 20 ml
Normality of Ferrous sulphate (A) = N1 = ?
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 = 0.1013 N
V1 X N1 = V2 X N2
Therefore, N1 = V2 X N2 / V1
=
Normality of Ferrous sulphate (A) = N
TITRATION - I I : STANDARD KMnO4 Vs Ferrous sulphate (B)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Ferrous sulphate (B)
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Ferrous sulphate (B) = V1 = 20 ml
Normality of Ferrous sulphate (B) = N1 = ?
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 = 0.1013 N
V1 X N1 = V2 X N2
Therefore, N1 = V2 X N2 / V1
=
Normality of Ferrous sulphate (B) = N
COMPARISON
1. Normality of Ferrous sulphate (A) = N
2. Normality of Ferrous sulphate (B) = N
\ The Weaker Ferrous sulphate solution is = Ferrous sulphate (____)
AMOUNT CALCULATION
present in 1250ml of the weaker solution = gram
RESULT
1. Normality of Ferrous sulphate (A) = N
2. Normality of Ferrous sulphate (B) = N
3. The Weaker Ferrous sulphate solution is = Ferrous sulphate (____)
4. Amount of crystalline Ferrous sulphate present in 1250 ml of the weaker solution
= gram
Qn. No: 1, 2 & 3 COMPARISON OF __________________________Solutions
AIM
To compare the normalities of the two given KMnO4 / Ferrous ammonium sulphate / Ferrous sulphate solutions A and B, and to estimate the amount of crystalline ___________________________ present in ________ ml of the weaker / stronger solution. You are provided with a standard solution of ________________containing ________ grams in ____________.
SIMPLE PROCEDURE
Sl. No
|
CONTENTS
|
TITRATION - I
|
TITRATION - I I
|
1
|
Burette Solution
|
Potassium permanganate__
|
Potassium permanganate__
|
2
|
Pipette Solution
|
20 ml of ______________
|
20 ml of _____________
|
3
|
Acid Added
|
20 ml of 2N Sulphuric acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
Lab temperature
|
Lab temperature
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent mass of _______________________ = ______________
(solution to be estimated)
| ||
NORMALITY OF STANDARD SOLUTION
Amount of ______________ present in 1 litre = _________ g x 1000 ml / _________ ml
= _________ g
Normality of Standard _________________ = Mass per litre / Equivalent mass
= ______ / ______
\ Normality of Standard ______________ = _____________ N
TITRATION - I : STANDARD __________________ Vs ___________________ (A)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
_____________
ml
|
Initial
ml
|
Final
ml
|
KMnO4 ___
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of _____________ = V1 =
Normality of_______________ = N1 =
Volume of KMnO4 ___ = V2 =
Normality of KMnO4 ___ = N2 =
V1 X N1 = V2 X N2
Therefore, _____ = ________ / ______
=
Normality of _______________ (A) = N
TITRATION - I I : STANDARD_________________ Vs ___________________(B)
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
_____________
ml
|
Initial
ml
|
Final
ml
|
KMnO4 ___
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of _____________ = V1 =
Normality of_______________ = N1 =
Volume of KMnO4 ___ = V2 =
Normality of KMnO4 ___ = N2 =
V1 X N1 = V2 X N2
Therefore, _____ = ________ / ______
=
Normality of ____________________ (B) = N
COMPARISON
1. Normality of ______________ (A) = N
2. Normality of ______________ (B) = N
\ The Weaker / Stronger _________________solution is = ______________ (____)
AMOUNT CALCULATION
= Equivalent mass x Normality x ___/1000
= ________x ________ x _____ / 1000
= gram
RESULT
1. Normality of ______________ (A) = N
2. Normality of ______________ (B) = N
3. The Weaker / Stronger ________________ solution is = __________ (___)
4. Amount of crystalline ___________________ present in _______ml of the weaker / stronger solution = gram
Ex. No.: 4 ESTIMATION OF OXALIC ACID
Date:
AIM
To estimate the amount of crystalline Oxalic acid present in 500 ml of the given solution. You are provided with 0.1 N Ferrous ammonium sulphate as standard solution and Potassium permanganate as the link solution.
PROCEDURE
TITRATION - I : STANDARDISATION OF KMnO4
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous ammonium sulphate solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of KMnO4 is calculated.
TITRATION - I I : STANDARDISATION OF OXALIC ACID
The burette is filled with the given KMnO4 solution and titrated against 20ml of Oxalic acid solution pipetted out into a clean conical flask, mixed with equal volume of 2N dilute Sulphuric acid and heated to 700C. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the normality of Oxalic acid solution, the amount of Oxalic acid present in 500ml is calculated, knowing the equivalent mass of Oxalic acid is 63.
SIMPLE PROCEDURE
S.
No
|
CONTENTS
|
TITRATION - I
|
TITRATION - I I
|
1
|
Burette Solution
|
Potassium permanganate
|
Potassium permanganate
|
2
|
Pipette Solution
|
20 ml of Ferrous ammonium sulphate
|
20 ml of Oxalic acid
|
3
|
Acid Added
|
20 ml of 2N Sulphuric acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
Lab temperature
|
60 - 700 C
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent mass of Oxalic acid = 63
| ||
Normality of Standard Ferrous ammonium sulphate = 0.1 N
TITRATION - I : KMnO4 Vs STANDARD Ferrous ammonium sulphate
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Ferrous ammonium sulphate
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Ferrous ammonium sulphate = V1 = 20 ml
Normality of Ferrous ammonium sulphate = N1 = 0.1 N
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 = N
TITRATION - I I : STANDARD KMnO4 Vs OXALIC ACID
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Oxalic acid
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of KMnO4 = V1 =
Normality of KMnO4 = N1 =
Volume of Oxalic acid = V2 = 20 ml
Normality of Oxalic acid = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality Of Oxalic Acid = N
AMOUNT CALCULATION
Amount of Oxalic acid present in 500ml of the solution =
Equivalent mass x Normality x 500/1000
= 63 x x 500 /1000
= gram.
RESULT
1. Normality of KMnO4 = N
2. Normality of Oxalic acid = N
3. Amount of crystalline Oxalic acid present in 500ml of the solution = gram
Ex. No.: 5 ESTIMATION OF FERROUS AMMONIUM SULPHATE
Date:
AIM
To estimate the amount of crystalline Ferrous ammonium sulphate present in 750 ml of the given solution. You are provided with 0.1025 N Oxalic acid as standard solution and Potassium permanganate as the link solution.
PROCEDURE
TITRATION - I : STANDARDISATION OF KMnO4
The burette is filled with the given KMnO4 solution and titrated against 20ml of Oxalic acid solution pipetted out into a clean conical flask, mixed with equal volume of 2N dilute Sulphuric acid and heated to 700C. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of KMnO4 is calculated.
TITRATION - I I : STANDARDISATION OF Ferrous ammonium sulphate
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous ammonium sulphate solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of Ferrous ammonium sulphate is calculated. From the normality of Oxalic acid solution, the amount of Oxalic acid present in 500ml is calculated, knowing the equivalent mass of Ferrous ammonium sulphate is 392.
SIMPLE PROCEDURE
S. No
|
CONTENTS
|
TITRATION - I
|
TITRATION - II
|
1
|
Burette Solution
|
Potassium permanganate
|
Potassium permanganate
|
2
|
Pipette Solution
|
20 ml of Oxalic acid
|
20 ml of Ferrous ammonium sulphate
|
3
|
Acid Added
|
20 ml of 2N Sulphuric Acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
60 - 700 C
|
Lab temperature
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent Mass of Ferrous ammonium sulphate = 392
| ||
Normality of Standard Oxalic acid = 0.1025 N
TITRATION - I : KMnO4 Vs Std. OXALIC ACID
INDICATOR: KMnO4(SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Oxalic acid
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Oxalic acid = V1 = 20 ml
Normality of Oxalic acid = N1 = 0.1025 N
Volume of KMnO4 = N1 =
Normality of KMnO4 = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 = N
TITRATION - I I: STANDARD KMnO4 Vs FERROUS ammonium sulphate
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Ferrous ammonium sulphate
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of KMnO4 = V1 =
Normality of KMnO4 = N1 =
Volume of Ferrous ammonium sulphate = V2 = 20 ml
Normality of Ferrous ammonium sulphate = N2 =?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of Ferrous Ammonium Sulphate = N
Amount of crystalline Ferrous ammonium sulphate present in 750ml
= Equivalent mass x Normality x 750 / 1000
= Equivalent mass x Normality x 750 / 1000
= 392 x x 750 / 1000
= gram
= gram
RESULT
1. Normality of KMnO4 = N
2. Normality of Ferrous ammonium sulphate = N
3. Amount of crystalline Ferrous ammonium sulphate present in 750 ml of the solution
= gram
Ex. No.: 6 ESTIMATION OF FERROUS SULPHATE
Date:
AIM
To estimate the amount of crystalline Ferrous sulphate present in 250 ml of the given solution. You are provided with 0.0952 N Ferrous ammonium sulphate as standard solution and Potassium permanganate as the link solution.
PROCEDURE
TITRATION - I : STANDARDISATION OF KMnO4
The burette is filled with the given KMnO4 solution and titrated against 20ml of Ferrous ammonium sulphate solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of KMnO4 is calculated.
TITRATION - II : STANDARDISATION OF FeSO4
The burette is filled with the given KMnO4 solution and titrated against 20ml of FeSO4solution pipetted out into a clean conical flask and mixed with equal volume of 2N dilute Sulphuric acid. The endpoint is the appearance of permanent pale pink colour. The final burette reading is noted, and the titration is repeated for concordant value. From the titre value, the normality of FeSO4 is calculated. From the normality of Ferrous Sulphate solution, the amount of FeSO4 present in 250ml it is calculated knowing the equivalent mass of Ferrous Sulphate is 278.
SIMPLE PROCEDURE
S. No
|
CONTENTS
|
TITRATION - I
|
TITRATION - II
|
1
|
Burette Solution
|
Potassium permanganate
|
Potassium permanganate
|
2
|
Pipette Solution
|
20 ml of Ferrous ammonium sulphate
|
20 ml of FeSO4
|
3
|
Acid Added
|
20 ml of 2N Sulphuric acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
Lab temperature
|
Lab temperature
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent mass of FeSO4 = 278
| ||
Normality of Standard Ferrous ammonium sulphate = 0.0952 N
TITRATION - I : KMnO4 Vs STANDARD Ferrous ammonium sulphate
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
Ferrous ammonium sulphate
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of Ferrous ammonium sulphate = V1 = 20 ml
Normality of Ferrous ammonium sulphate = N1 = 0.0952 N
Volume of KMnO4 = V2 =
Normality of KMnO4 = N2 =?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 = N
TITRATION - II : STANDARD KMnO4 Vs FeSO4
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
FeSO4
ml
|
Initial
ml
|
Final
ml
|
KMnO4
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of KMnO4 = V1 =
Normality of KMnO4 = N1 =
Volume of FeSO4 = V2 = 20 ml
Normality of FeSO4 = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of FeSO4 = N
Amount CALCULATION
Amount of crystallineFeSO4 present in 250ml = Equivalent mass x Normality x 250 / 1000
= 278 x x 250 / 1000
= gram.
RESULT
1. Normality of KMnO4 = N
2. Normality of FeSO4 = N
3. Amount of crystalline Ferrous sulphate present in 250ml of the solution = gram
Qn. No.: 4, 5 & 6.
ESTIMATION OF OXALIC ACID / FERROUS AMMONIUM SULPHATE / FERROUS SULPHATE
AIM
To estimate the amount of crystalline _____________________________________ present in ________ml of the given solution. You are provided with ____________N _________________ as standard solution and Potassium permanganate as the link solution.
SIMPLE PROCEDURE
S. No
|
CONTENTS
|
TITRATION - I
|
TITRATION - II
|
1
|
Burette Solution
|
Potassium permanganate
|
Potassium permanganate
|
2
|
Pipette Solution
|
20 ml of _______________
|
20 ml of ______________
|
3
|
Acid Added
|
20 ml of 2N Sulphuric acid
|
20 ml of 2N Sulphuric acid
|
4
|
Temperature
|
_______________________
|
_____________________
|
5
|
Indicator
|
Self-indicator (KMnO4)
|
Self-indicator (KMnO4)
|
6
|
Endpoint
|
Appearance of permanent pale pink colour
|
Appearance of permanent pale pink colour
|
7
|
Equivalent mass of _______________________________ = _________
(solution to be estimated)
| ||
NORMALITY OF STANDARD SOLUTION
Normality of Standard _________________ = ___________________N
TITRATION - I : KMnO4 Vs STANDARD _____________________________
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
_____________
ml
|
Initial
ml
|
Final
ml
|
KMnO4 ___
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of ____________________ = V1 = 20 ml
Normality of _______________________ = N1 = N
Volume of KMnO4 = V2 = ml
Normality of KMnO4 = N2 = ?
V1 X N1= V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of KMnO4 = N
TITRATION - II : STANDARD KMnO4 Vs __________________________
INDICATOR: KMnO4 (SELF-INDICATOR)
Sl
|
Volume of
|
Burette Readings
|
Volume of
|
Concordant
| |
No
|
_____________
ml
|
Initial
ml
|
Final
ml
|
KMnO4 ___
ml
|
Titer Value
ml
|
1
|
20
|
0
| |||
2
|
20
|
0
| |||
CALCULATION
Volume of KMnO4 = V1 = ml
Normality of KMnO4 = N1 = N
Volume of _________________ = V2 = 20 ml
Normality of ________________ = N2 = ?
V1 X N1 = V2 X N2
Therefore, N2 = V1 X N1 / V2
=
Normality of ____________________ = N
Amount CALCULATION
_________ml of the given solution = Equivalent mass x Normality x ______ / 1000
= ________x_____x _______ / 1000
= gram.
RESULT
1. Normality of KMnO4 = N
2. Normality of ____________________ = N
3. Amount of crystalline __________ present in ____ml of the solution = gram
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