Class XII_Delhi_Chemistry_Set-2
9.
Ans.
10.
How will you convert the following?
(i)
(ii)
Propan-2-ol to propanone.
Phenol to 2, 4, 6 – tribromophenol?
(i)
Propan-2-ol to propanone:
(ii)
Phenol to 2, 4, 6-tribromophenol:
Explain the mechanism of the following reaction:
H
+
..
2CH3  CH2  OH 
 CH3CH2  O
 CH2  CH3  H2O
..
413K
Ans.
The mechanism of the reaction is given below:
2
2
Class XII_Delhi_Chemistry_Set-2
11.
What is the difference between oil/water (O/W) type and water/oil (W/O) type
emulsions? Give an example of each type.
2
Ans.
Emulsion of oil-in-water has oil as dispersed phase and water as dispersion medium. For
example, Milk etc.
Emulsion of water-in-oil has water as dispersed phase and oil as dispersion medium. For
example, Cod liver oil etc.
12.
18 g of glucose, C6H12O6 (Molar Mass = 180 g mol–1) is dissolved in 1 kg of water in a
sauce pan. At what temperature will this solution boil?
2
(Kb for water = 0.52 K kg mol–1, boiling point of pure water = 373.15 K)
Ans.
w1 = weight of solvent (H2O) = 1 kg and w2 = weight of solute (C6H12O6) = 18 gm
M2 = Molar mass of solute (C6H12O6) = 180 g mol–1
Kb = 0.52 K Kg mol–1
T b  373.15K
K 1000  w 2 0.52 1000 18
Tb  b

 0.052K
M 2  w1
180 1000
Tb  Tb  T b  0.052  Tb  373.15  Tb  373.202K
13.
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm–1. Calculate its molar
conductivity.
2
Ans.
κ (S cm–1) = 0.025 S cm–1 and molarity (mol L–1) = 0.20 M
K
0.025
Molar conductivity   m  =

 1.25 104 S cm2 mol1
1000  molarity 1000  0.20
14.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i)
Smoke
(ii)
Milk
2
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated
on the addition of small amounts of electrolytes?
Ans.
(i)
(ii)
Dispersed phase in smoke: Solid and dispersion medium in smoke: Gas
Dispersed phase in milk: Liquid Fat and dispersion medium in milk: Water
OR
Lyophillic colloids: It is made up of two words; ‘Lyo’ meaning liquid and ‘Phillic’
meaning loving, so those colloids which are attracted by the liquid (solvent), are called as
lyophillic colloids. These are also called reversible sols. These are quite stable and cannot
be easily coagulated.
Class XII_Delhi_Chemistry_Set-2
Lyophobic colloids: It is made up of two words; ‘Lyo’ meaning liquid and ‘Phobic’
meaning repelling, so those colloids which are repelled by the liquid ( solvent), are called
as lyophobic colloids. These are also called irreversible sols and these are unstable and
can be easily coagulated due to lack of protecting layer around charged colloidal
particles, they easily form cluster. Hence, they got easily coagulated on addition of small
amount of electrolyte.
15.
What happens when
(i)
PCl5 is heated?
(ii)
H3PO3 is heated?
Write the reactions involved.
Ans.
(i)
(ii)
PCl5 on heating gives PCl3 and Cl2: PCl5  PCl3  Cl2
H3PO3 on heating gives orthophosphoric acid and phosphine:
4H3PO3  3H3PO4  PH3
16.
(a)
Which metal in the first transition series (3d series) exhibits + 1 oxidation state
most frequently and why?
2
Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
(b)
Ans.
(a)
(b)
2
Cu is the only metal in the first transition series (3d series) which shows +1
oxidation state most frequently. This is because the electronic configuration of Cu
is 3d10 4s1 and after losing one electron it acquires a stable 3d10 configuration.
The color of cations is dependent on the number of unpaired electrons present in
d-orbital. The electronic configuration of the following cations is as follows:
Sc (Atomic number 21) = 3d1 4s2 and Sc3+ = 3d0 4s0. As d-orbital is empty, it is
colourless.
V (Atomic number 23) = 3d3 4s2 and V3+ = 3d2 4s0. As d-orbital is having 2
unpaired electrons, it undergoes d-d transition and shows green colour.
Ti = (Atomic number 22) = 3d2 4s2 and Ti4+ = 3d0 4s0. As d-orbital is empty, it is
colourless.
Mn = (Atomic number 25) = 3d5 4s2 and Mn2+ = 3d5 4s0. As d-orbital is having 5
unpaired electrons, it shows pink color.
Class XII_Delhi_Chemistry_Set-2
17.
(a)
(b)
Ans.
(a)
(b)
Which of the following ores can be concentrated by froth floatation method and
why?
Fe2O3, ZnS, Al2O3
What is the role of silica in the metallurgy of Copper?
2
Sulphide ores are concentrated by froth flotation method therefore, ZnS will be
concentrated by this method.
Silica is added in the reverberatory furnance during the extraction of copper to
remove iron oxide present in the ore. Iron oxide reacts with silica and is removed
as slag of iron silicate.
FeO  SiO2  FeSiO3
Slag
18.
(a)
(b)
Why does p-dichlorobenzene have a higher m.p than its o – and m – isomers?
Why is () – Butan-2-ol is optically inactive?
2
Ans.
(a)
p-dichlorobenzene have higher melting point than ortho and meta isomer. This is
because the para isomer is having a symmetrical structure and therefore, its
packing is more efficient as compared to the ortho and meta isomer, therefore, it
shows higher melting point.
(b)
The (  ) - Butan-2-ol is optically inactive because it exist in two enantiomeric
forms which are non-superimposable mirror images of each other. Both the
isomers are present in equal amounts therefore, it does not rotate the plane of
polarized light and is optically inactive.