Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2019 (Series BVM/4) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2019 (Series BVM/4) with Solutions
SET I Code No. 56/4/1
Section ‘A’
Question 1.
Name the defect in the following crystal: [1]
Answer:
Not in Syllabus.
Question 2.
When a coordination compound CrCl
3
.6H
2
0 is mixed with AgNO
3
, two moles of AgCl are precipitated per mole of the compound. What is the structural formula of the coordination compound? [1]
Answer:
The complex formed on mixing a coordination compound CrCl
3
.6H
2
O with AgNO
3
is as follows:
CrCl
3
.6H
2
O + AgNO
3
→ [Cr(H
2
O)
5
Cl]Cl
2
.H
2
O
Or
What is the difference between a complex and a double salt?
Answer:
Double salt dissociates cornplctel’ into its constituent ions in their aqueous solution.
Example: KCl.MgCl
2
.6H
2
O dissociates into K
+
, Cl
+
, Mg2 and H.O.
Complex does not dissociate into its constituent ions.
Example: K
4
[Fe(CN)
6
] → 4K
+
+ [Fe(CN)
6
]
4-
Question 3.
Define associated colloid with an example. [1]
Answer:
Colloids which act as electrolyte at low concentration and show colloidal behaviour at high concentration are called Associated colloids.
Example : Soap solution, Detergents.
Question 4.
Why is t-butyl bromide more reactive towards S
N
1 reaction as compared to n-butvl
bromide? [1]
Answer:
t—butyl bromide is more reactive towards S
N
1 reaction as compared to n-butyl bromide because it is a tertiary halide and produces higher stability of 3° carbocation.
Question 5.
Write the reaction involved in the Hoffmani’bromamide degradation reaction.
Answer:
Hofmann’s bromanide reaction: Primary amines can be prepared by treating an amide with Br
2
in an aqueous or alcoholic soin of NaOH.
Or
Propanamine and N, N-dimethylmethanamine contain the same number of carbon atoms, even though Propanamine has higher boiling point than N, N-dimethylmethanamine. Why?
Answer:
Propanamine has higher boiling point than N, N-dimethylmethanamine because it has intermolecular hydrogen-bonding due’to two hydrogen atoms of CH
3
CH
2
CH
2
NH
2
. Whereas dimethylmethanamine has no such intermolecular bonding.
Section – ‘B’
Question 6.
Give reasons for the following: [2]
(a) Aquatic species are more comfortable in cold water than in warm water.
(b) At higher altitudes people suffer from anoxia resulting in inability to think.
Answer:
(a) Aquatic species need dissolved oxygen for breathing. As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lake. Hence the aquatic species feel more comfortable in winter (low temperature) when the solubility of oxygen is higher.
(b) At higher altitude people suffer from anoxia because the partial pressure of oxygen is less than at the ground level due to which the concentration of oxygen in the blood and tissues decreases. .
Or
What type of azeotropic mixture will be formed by a solution of acetone and chloroform? Justify on the basis of strength of intermolecular interactions that develop in the solution.
Answer:
Minimum boiling azeotropes will be formed by a solution of acetone and chloroform. In positive deviation, the liquid pairs A-B have weak mtpraction forces than that of A – A or B – B molecular interaction forces.
Question 7.
Explain with a graph, the variation of molar conductivity of a strong electrolyte with dilution. [2]
Answer:
In case of strong electrolytes there is a small increase in conductance with dilution because it is completely dissociated in solution and number of ions remains constant.
\(\Lambda_{\mathrm{m}}\) increases with increases in dilution.
Question 8.
When dilute ferrous sulphate solution is added to an aqueous solution containing nitrate ion followed by careful addition of concentrated sulphuric acid along the sides of test tube, a brown ring is formed at the interface between the solution and sulphuric acid layers. Which anion is confirmed by the appearance of brown ring? What is the composition of the brown ring? [2]
Answer:
Not in syllabus.
Or
How can you prepare Cl
2
from HCl and HCl from CI
2
? Write reactions only.
Answer:
Cl
2
from HCl. Cl
2
can be prepared from HCl by Deacon’s Process.
HCl from Cl
2
. Chlorine (Cl
2
) reacts with hydrogen (H) in presence of diffused sunlight to form HCl
Or, Cl
2
on treating with water gives HCl
Question 9.
Use the data to answer the following and also justify giving reason: [1 × 2 = 2]
(a) Which is a stronger reducing agent in aqueous medium, Cr
2+
or Fe
2+
and why?
(b) Which is the most stable ion in +2 oxidation and why?
Answer:
(a) Cr
2+
is a stronger reducing agent than Fe
2+
because of its more negative standard reduction potential and Cr
2+
(3d
4
) after losing one electron forms Cr
3+
(3d
3
) which has \(t_2^3 g\)g eg° configuration which is a stable configuration.
(b) Manganese is the most stable ion in +2 oxidation state because of its most negative standard reduction potential value and stable half filled configuration (d
5
).
Question 10.
Define with equation: [1 × 2 = 2]
(a) Riemer-Tiemann Reaction
(b) Williamson’s Synthesis
Answer:
(a) Riemer-Tiemann Reaction:
(b) Williamson ether synthesis: Alkyl Halide reacts with Alkoxide
Question 11.
Give the structures of monomers of the following polymers: [1 × 2 = 2]
(a) Nylon-6,6
(b) Buna-S
Answer:
(a) Nylon-6,6 : It has two repeating monomers
(b) Buna-S.
Question 12.
Classify the following as addition and condensation polymers giving reason:
(а) Teflon
(b) PHBV
Answer:
Not in syllabus.
Section – ‘C’
Question 13.
Chromium crystallises in bcc structure. If its edge length is 300 pm, find its density.
Atomic mass of chromium is 52 u. [N
A
= 6.022 × 10
23
mol
-1
] [3]
Answer:
Not in Syllabus.
Question 14.
At 300 K, 30 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of a glucose solution is 1.52 bar at the same temperature, what would be its concentration? [3]
Answer:
Given: π
1
= 4.98 bar
π
2
= 1.52 bar
C
1
= 30/180
C
2
= ?
Using van’t Hoff equation
π = CRT
Putting the values in above equation, we get
4.98 = \(\frac{30 \mathrm{RT}}{180}\) ….. (i)
Now dividing equation (ii) by (i), we get
Question 15.
Calculate ∆
r
G° and log K
c
for the following reaction:
Cd
+2
(aq) + Zn(s) → Zn
2+
(aq) + Cd(s)
Given \(\mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^0\) = -0.403 V,
\(\mathbf{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0\) = -0.763 V
Answer:
Or
Chromium metal is electroplated using an addic solution containing CrO
3
according to the following equation:
CrO
3
(aq) + 6H
+
+ 6e
–
→ Cr(s) + 3H
2
O
Calculate how many grams of chromium will be electroplated by 24,000 coulombs.
How long will it take to electroplate 1.5g chromium using 12.5A current?
[Atomic mass of Cr 52g mol’, 1F 96500 C mol
-1
Answer:
Given Equation: Cr0
3
+ 6H
+
+ 6e
–
→ Cr + 3H
2
O
6F, i.e. 6 × 96500 C deposit Cr = 1 mol = 52g
∴ 24,000C deposit Cr = \(\frac{52}{6 \times 96500}\) × 24000
r = \(\frac{1248}{579}\) = 2.15g
∴ 52g of Cr uses charge = 6 × 96500 C
1.5 g of Cr uses charge = \(\frac{6 \times 96500}{52}\) × 1.5
= \(\frac{868500}{52}\) = 16701.92 C
Q = It
t = \(\frac{Q}{I}\) = \(\frac{16701.92}{12.5}\)
∴ t = 1336.15 sec.
Question 16.
Give reasons for the following: [1 × 3 = 3]
(a) Leather gets hardened after tanning.
(b) FeCl
3
is preferred over KCI in case of a cut leading to bleeding.
(c) Freundlich isotherm becomes independent of pressure at high pressure for a gas absorbed on a solid.
Answer:
(b) FeCl
3
is preferred over KCI in case of bleeding from a cut because its coagulation
efficiency (Fe
3+
) is more than K ion due to its high valency. Moreover it is non-irritable to skin.
(c) Because at high pressure, amount of gas adsorbed on a solid (x/m) becomes independent of pressure as \(\frac{1}{n}\) becomes zero and \(\frac{x}{m}\) = constant × P°
Question 17.
What is the role of [1 × 3 = 3]
(a) Depressants in froth floatation?
(b) Carbon monoxide in Mond’s process?
(c) Concentrated sodium hydroxide in leaching of alumina from bauxite?
Answer:
(a) To selectively prevent the formation of froth by one of the sulphide ore present in a mixture of sulphide ores.
(b) Nickel is heated in a stream of carbon monoxide forming a volatile complex, nickeltetracarbonyl which is further subjected to higher temperature so that it is decomposed to give pure nickel.
(c) The powdered bauxite ore s heated with conc. NaOH to remove impurities of Fe
2
O
3
, SiO
2
and TiO
2
.
Or
Write chemical reactions taking place in the extraction of Aluminium from Bauxite ore.
Answer:
Aluminium can be extracted by leaching method followed by Hall-Heroult electrolytic reduction in the following ways:
(i) The pulverized bauxite ore is treated with the hot concentrated solution of sodium hydroxide to produce soluble sodium aluminate leaving behind insoluble impurities after filteration.
(ii) The filterate is cooled and neutralized with CO
2
to precipitate aluminium hydroxide.
(iii) Hydrated alumina is then filtered, washed and heated to about 1473K to get pure alumina
(iv) Alumina then reduced by carbon to get pure Al.
2Al
2
O
3
+ 3C → 4Al + 3CO
2
Question 18.
Explain the method of preparation of sodium dichromate from chromite ore. Give the equation representing oxidation of ferrous salts by dichromate ion. [1 × 3 = 3]
Answer:
Preparation of Sodium Dichromate from chromite ore:
(i) The powdered chromite ore is mixed with Na
2
CO
3
and quick lime and then roasted in a reverberatory furnace to produce yellow sodium chromate.
(ii) Sodium chromate filterate is treated with conc. H
2
SO
4
to produce Sodium Dichromate and the by-product Na
2
SO
4
crystallized out to leave behind sodium dichromate which is further cooled and crystallized out.
2Na
2
CrO
4
+ H
2
SO
4
(conc.) → Na
2
Cr
2
O
7
+ Na
2
SO
4
+ H
2
O
Dichromate ion oxidises ferrous salts to ferric salt.
\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) + 6Fe
2+
+ 14H
+
→ 2Cr
3+
+ 6Fe
3+
+ 7H
2
O
Or
Complete the following reactions:
(a) MnO
2
+ KOH + O
2
→
(b) I
–
+ Mn\(\mathrm{O}_{\mathbf{4}}^{-}\) + H
+
→
(c) Cr
2
\(\mathrm{O}_7^{2-}\) + Sn
2+
+ H
+
→
Answer:
(a) 2MnO
2
+ 4KOH + O
2
→ 2KM
2
O
4
+ 2H
2
O
(b) 2Mn\(\mathrm{O}_4^{-}\) + 10I
–
+ 16H
+
) → 2Mn
2+
+ 8H
2
O + 5I
2
(c) Cr
2
\(\mathrm{O}_7^{2-}\) + \(3 S_{n^{2+}}\) + 14H
+
→ 2Cr
3+
+ 3Sn
4+
+ 7H
2
O
Question 19.
Write the hybridization and magnetic character of the following complexes: [3]
(i) [Fe(H
2
O
6
]
3+
(ii) [Ni(CN)
4
]
2-
[Atomic number: Fe = 26, Ni = 28]
Answer:
(i) [Fe(H
2
O
6
]
3+
: The element Fe is in +3 oxidation state. As H
2
O is a weak field ligand, so electron pairing is not possible in this case.
(ii)
Question 20.
Give reasons for the following: [1 × 3 = 3]
(a) The presence of —NO
2
group at ortho or para position increases the reactivity of
haloarenes towards nucleophilic substitution reactions.
(b) p-dicholorobenzene has higher melting point than that of ortho or meta isomer.
(c) Thionyl chloride method is preferred for preparing alkyl chlorides from alcohols.
Answer:
(b) p-isomers are comparatively more symmetrical and fit closely in the crystal lattice thus require more heat to break these strong forces of attraction. Therefore higher melting point than o- and m-isomers.
(c) Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the gaseous by-products like SO
2
and HCl are formed which escape into the atmosphere leaving behind pure alkyl chlorides.
CH
3
CH
2
OH + SOCl
2
→ CH
3
CH
2
Cl + SO
2
↑ + HCl
Or
(a) Write equation for preparation of 1-iodobutane from 1-chiorobutane.
(b) Out of 2-bromopentane, 2-bromo-2-methylbutane and 1-bromopentane, which compound is most reactive towards elimination reaction and why?
(c) Give IUPAC name of
Answer:
Out of above 2-bromo-2-methyl butane is most reactive towards elimination reaction because it gives highly substituted alkene than others.
(c) TUPAC name: 4-Bromo 4-methylbut-2-ene
Question 21.
(a) How will you synthesise the following alcohol from appropriate alkene: [3]
(b) Write the mechanism of the followsing reaction:
Answer:
(a) The alkene must be but-2-ene
(b)
Question 22.
(a) Give one chemical test to distinguish between the compounds of the following pairs: [3]
(i) CH
3
NH
2
and (CH
3
)
2
NH
(ii) Aniline and Ethanamine
(b) Why aniline does not undergo Friedel-Crafts reaction?
Answer:
(a) (i) CH
3
NH
2
with Hinsberg’s teagent (Benzene Suiphonyl Chloride) forms a precipitate which is soluble in base.
(CH
3
)
2
NH with Hinsherg’s Reagent (Benzene Suiphonyl Chloride) forms a
precipitate which is insoluble in base.
(ii) Add benzene diazoniumchloride to both the compounds, aniline forms yellow dye while ethanamine does not.
(b) Aniline being a Lewis base reacts with Lewis acid AlCl
3
to form a salt.
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
Question 23.
(a) Give anyone property of glucose that cannot be explained by the open chain structure.
(b) Compare amylase with amylopectin in terms of constituting structure.
(c) Why do amino acids show amphoteric behaviour? [1 × 3 = 3]
Answer:
(a) Limitations of the open chain structure of glucose:
(i) Glucose does not form NaHSO3 addition product. Despite having aldehyde ammonia group, it does not give test 2,4-DNP and does not respond to Schiff’s reagent test.
(ii) Glucose penta acetate does not react with NH
2
OH due to absence of aldehydic group.
(b) Amylose is a linear polymer of α-D-glucose in which C 1 of one glucose unit is attached to C 4 of the other through α-glvcosidic linkage while Amylopectin is a highly branched polymer containing 20-25 glucose units which are joined together through α-glycosidic linkages between C 1 of one and C 4 of the other which is further linked to C 6 of other glucose.
(c) The amphoteric behaviour of Amino acids is due to the presence of carboxylic
(-COOH) group and Amino (-NH
2
) on it, in their aqueous state. The —COOH group releases proton and -NH
2
group accepts it to form a dipolar ion with both positive and negative ion known as Zwitter ion.
In Zwitter ionic form, the amino acid can exhibit both acidic and basic properties.
Therefore, amino acids show an amphoteric behaviour.
Question 24.
Define the following with suitable example of each:
(a) Antiseptics
(b) Non-narcotic analgesics
(c) Cationic detergents [1 × 3]
Answer:
(a) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues, e.g., Penicillin.
(b) Non-narcotic analgesics. They inhibit the synthesis of prostaglandins which stimulate inflammation in tissues and cause pain, effective in relieving skeletal pain, reducing fever and preventing blood platelet coagulation. Example. Aspirin, Ibuprofen.
(c)
Cationic detergents. They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the çationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example: Cetylitrimethyl ammonium bromide
Section ‘D’
Question 25.
(a) Consider the reaction R → P for which the change in concentration of R with time is shown by the following graph:
(i) Predict the order of reaction.
(ii) What does the slope of the curve indicate?
(b) The rate of reaction quadruples when temperature changes from 293 K to 313 K. Calculate Ea assuming that it does not change with time. [R = 8.314 JK
-1
mol
-1
]. [2 + 3 = 5]
Answer:
(a)
(i) It is a zero order reaction.
(ii) Slope of the curve = -K.
(b)
Or
(a) Draw the plot of in k vs 1/T for a chemical reaction. What does the intercept represent? What is the relation between slope and E
a
?
(b) A first order reaction takes 30 minutes for 20% decomposition. Calculate t
1/2
[log 2 = 0.3010]
Answer:
(a) The intercept represents ln
Slope represents = \(\frac{-E_a}{R}\)
ln K = \(\frac{-E_a}{R_T}\) + ln A
(b)
Given:
t = 30 minutes, t
1/2
= ?
Let a = 100
x = 20% of 100 = 20
Using formula
Question 26.
(a) Draw the the following: [2 + 3 = 5]
(i) HClO
3
(ii) H
2
S
2
O
8
(b) Give reasons for the following:
(i) Above 1000 K sulphur shows paramagnetism.
(ii) Although electron gain enthalpy of fluorine is less negative than that of chlorine, yet flourine is a better oxidising agent than chlorine.
(iii) In solid state PCl
5
exists as an ionic compound.
Answer:
(a) (i) structure of HClO
3
(Chloric acid):
(ii) Structure of H
2
S
2
O
8
(b)
(i) In vapour state sulphur partly exists as S
2
molecule which has two unpaired electrons in the antihonding π* orbitais like O
2
and hence exhibits paramagnetism.
(ii) Because of small size of fluorine atom and strong electron-electron repulsions in its compact 2p orbitals.
(iii) Not in Syllabus.
Or
(a) Complete the following reactions:
(i) PbS(s) + O
3
→
(ii) XeF
6
+ NaF →
(b) Arrange the following in increasing order of property indicated, giving reason:
(i) Hydrides of group 15 — boiling points
(ii) Hydrides of group 17 — acidic strength
(iii) Hydrides of group 16 — reducing character
Answer:
(a) (i) PbS + 4O
3
→ PbSO + 4O
2
(ii)
Or
XeF
6
+ NaF → Na
+
[XeF
7
]
–
(b)
(i) Not in Syllabus.
(ii) HF < HCl < HBr < HI
The higher the bond dissociation energy, lower is the degree of ionization and hence weaker is the acid. HF has highest bond dissociation energy which further decreases in group.
(iii) H
2
O < H
2
S < H
2
Se < H
2
Te < H
2
P
O
This is due to decrease in thermal stability of above hydrides as H – E becomes weaker with increase in size of E. On moving down the group bond dissociation enthalpy decreases.
Question 27.
(a) Carry out the following conversions: [2]
(i) P-nitrotoluene to 2-bromobenzoic acid
(ii) Propanoic acid to acetic acid
(b) An alkene with molecular formula C
5
H
10
on ozonolysis gives a mixture of two compounds, B and C. Compound B gives positive Fehiing test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C. [3]
Answer:
(a) (i) P-nitrotoluene to 2-bromobenzoic acid
(ii) Propanoic acid to acetic acid
(b) The most probable alkene i.e. (Compound A) with molecular formula C
5
H
10
is
2-methyl but-2-ene.
Compound B i.e. Ethanal gives Todoforni test and Fehiing Test.
Compound C i.e. Acetone doesn’t give Fehling’s test but gives haloform test.
Or
(a) Carry out the following conversions:
(i) Benzoic acid to aniline
(ii) Bromomethane to ethanol
(b) Write the structure of major product(s) in the following:
Answer:
Set II Code No. 56/4/2
Question 1.
What is the coordination number of atoms in a
(i) bcc structure, and
(ii) fcc structure? [1]
Answer:
Not in Syllabus.
Question 2.
Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances are more effective because the surface area of powdered form is more than crystalline form, which also further increases physiosorption.
Question 4.
Which alkyl halide from the following pair would you expect to react more rapidly by an S
N
2 mechanism? [1]
Answer:
reacts more rapidly by an S
N
2 mechanism because of 2° carbocation.
Question 7.
Define electrochemical cell. What happens when applied external potential becomes greater than E°
cell
of electrochemical cell? [2]
Answer:
Electrochemical cell is a device in which chemical energy is converted into electrical energy. When applied external potential becomes greater than E°
cell
of electrochemical cell, then the reaction starts in opposite direction and current flows in the opposite direction and cell acts like an electrolytic cell.
Question 9.
Write the equations involved in the following reactions: [2]
(a) Kolbe’s reaction
(b) Friedel-Crafts alkylation of anisole
Answer:
(a) Kolbe’s reaction : Phenol reacts with CO
2
at 390-400 K under pressure 4-7 giving Selicyle Acid
(b) Friedel Crafts alkylation of anisole
Question 12.
Write the structures of monomers of the following polymers: [1 × 2 = 2]
(a) Terylene
(b) Buna-N
Answer:
(a)
(b)
Buna-N
Question 16.
Give reasons for the following: [1 × 3 = 3]
(a) Brownian movement provides stability to the colloidal solution.
(b) True solution does not show Tyndall effect.
(c) Addition of alum purifies the water.
Answer:
(a) Brownian movement provides stability to the colloidal solution because it opposes the gravitational force acting on colloidal particles and prevents them from settling down.
(b) In true solution, the diameter of the dispersed particles is much smaller than the wavelength of the light used, hence there is no scattering of light.
(c) Addition of alum purifies the water as alum coagulates the colloidal impurities present in water so that these impurities get settled down and removed by decantation or filtration.
Question 21.
(a) Show how you will synthesise the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? [3]
(b) Write the mechanism of the following reaction:
Answer:
(a)
(b)
Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols.
Mechanism : It involves the following three steps:
Question 23.
What happens when D-Glucose is treated with the following reagents:
(a) Br
2
water
(b) HCN
(c) (CH
3
CO)
2
O
Answer:
(a)
(b) Glucose reacts with HCN togiveacvanohdrin, or with a Schiffs base to give an oxime. It shows the presence of a carbonyl group.
(c) D-glucose reacts with acetic anhydride to give penta-acetate.
Question 24.
Define the following terms with a suitable example of each:
(a) Antacids
(b) Artificial sweeteners
(c) Anionic detergents
Answer:
(a) Antacids. Those substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example: Sodium bicarbonate, Mg(OH)
2
(b) Artificial sweeteners: Those chemical substances which are sweet in taste but do not add any calories to our body are called artificial sweetening agents. These are excreted as such through urine.
Example: Saccharin, aspartame etc.
(c) Anionic detergents: Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example: Sodium alkyl sulphates
SET III Code No. 56/4/3
Question 1.
What would be the nature of solid if there is no energy gap between Valence band and
Conduction band? [1]
Answer:
Not in Syllabus.
Question 2.
Write the main reason for the stability of colloidal sols. [1]
Answer:
The stability of the colloid is due to charges on it which make them repel each other so that they cannot combine together to form large particles and remain dispersed in the medium and makes colloidal solution stable.
Question 4.
Write the IUPAC name of
Answer:
TUPAC Name: 1-Chloro-2-phenylethane
Question 7.
Using the E° values of X and Y, predict which is better for coating the surface of iron to prevent rust and why? [2]
Given: [\(\mathrm{E}_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^{\mathrm{o}}\) = -0.44V; \(\mathbf{E}_{\left(X^2 / X\right)}\) = -2.36V; \(\mathbf{E}_{\left(\mathrm{Y}^{2+} / \mathrm{Y}\right)}\) = -0.14V]
Answer:
Since Y has lower reduction potential (\(E^{\circ}{ }_{Y^{2+} / Y}\) = – 0.14V) than Iron (\(\mathrm{E}^{\circ} \mathrm{Fe}^{2+} / \mathrm{Fe}\) = 0.44V) so it is more reactive metal than iron and can be easily oxidized in preference to X which has higher negative reduction potential. Thus X is better for coating the surface of iron to prevent rust.
Question 9.
Write the name of monomers and their structures for the following polymers: [2]
(a) Neoprene
(b) Nylon-6
Answer:
(a) Neoprene:
(b)
Question 12.
What happens when
(a) Phenol reacts with Conc. HNO
3
?
(b) Ethyl chloride reacts with NaOC
2
H
5
?
Write the chemical equations involved in the above reactions.
Answer:
(a) Picric acid (2, 4, 6 – trinitrophenol) formed when Phenol reacts with Conc. HNO
3
(b) Ethyl chloride reacts with NaOC
2
H
5
Question 15.
Define the following terms with a suitable example of each: [3]
(a) Sol
(b) Aerosol
(c) Hydrosol
Answer:
(a) Sol. The colloids in which a solid is dispersed in the liquid. In other words, in a colloidal sol, the dispersed phase is a solid and the dispersion medium is a liquid.
(b) Aerosol. Aerosol is a colloidal solution of solid or liquid (dispersed phase) in gas (dispersion medium). Example. smoke, dust, fog, mist, cloud.
(C) A hydrosol. The colloidal solution where water is used as the dispersion medium is called hydrosol or aquasol. Example. Starch sol.
Question 20.
(a) What are antidepressant drugs? Give an example. [3]
(b) Name the sweetening agent used in preparation of sweets for a diabetic patient.
(c) Why are detergents non-biodegradable?
Answer:
(a) Those drugs which are used for treatment of depression by inhibiting the enzymes which catalyse the degradation of noradrenaline are called Antidepressants.
For example. Iproniazid and Pheneizine.
(b) The sweetening agent used in the preparation of sweets for a diabetic patient is
Saccharin.
(c) Soaps are sodium potassium salts of long chain fatty acids and being soluble in water can easily be decomposed by microorganisms while detergents are NH
4
or sulphate salts which can’t be decemposed by microo4ganisms easily hence are non-biodegradable.
Question 21.
(a) What is the difference between native protein and denatured protein?
(b) Which one of the following is a disaccharide:
Glucose, Lactose, Amylose, Fructose
(c) Write the name of the vitamin responsible for the coagulation of blood. [3]
Answer:
(a) Native protein is the protein which is found in a biological system having a unique three-dimensional structure and specific biological activity.
Denatured protein is the protein which is formed by coagulation of native protein when it is subjected to change in pH or in temperature.
(b) Lactose is a disaccharide
(c) Vitamin K is responsible for the coagulation of blood.
Question 22.
(a) Butan-1-ol has a higher boiling point than diethyl ether. Why?
(b) Write the mechanism of the following reaction: [3]
Answer:
(a) Due to presence of inter-molecular H-bonding, associated molecules are formed. Hence butanol has high boiling point while diethyl ether does not have inter molecular H-bonding.
(b)
