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Chemistry (Section A)

Question 1

250 g solution of D-glucose in water contains 10.8% of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are H, 1u ; C, 12u ; O, 16u)
(A) 1.03
(B) 2.06
(C) 3.09
(D) 5.40
▶️ Answer/Explanation
Detailed solution

C6H12O6 → Glucose

We know: \(\frac{\text{mass of C}}{\text{mass of glucose}} = \frac{72}{180}\)

Given: %C = 10.8 = \(\frac{\text{mass of C}}{\text{mass of solution}} \times 100\)

\(\text{mass of C} = \frac{10.8 \times 250}{100} = 27 \text{ gm}\)

∴ mass of glucose = 67.5 gm

∴ moles of glucose = 0.375 moles

Mass of solvent = 250 – 67.5 gm = 182.5 gm

∴ Molality = \(\frac{0.375}{0.1825} = 2.055 \approx 2.06\)

✅ Answer: (B)

Question 2

Given below are two statements.
Statement I : O2, Cu2+ and Fe3+ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.
Statement II : NaCl and H2O are weakly magnetized in opposite direction to magnetic field.
In the light of the above statements, choose the most appropriate answer form the options given below :
(A) Both Statement I and Statement II are correct.
(B) Both Statement I and Statement II are incorrect.
(C) Statement I is correct but Statement II is incorrect.
(D) Statement I is incorrect but Statement II is correct.
▶️ Answer/Explanation
Detailed solution

O2, Cu2+ and Fe3+ are paramagnetic,

∴ Weakly attracted by magnetic field.

NaCl and H2O are diamagnetic,

∴ Weakly repelled by magnetic field.

✅ Answer: (A)

Question 3

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium.
Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
▶️ Answer/Explanation
Detailed solution

Energy of orbitals decreases on increasing the atomic number.

✅ Answer: (A)

Question 4

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Activated charcoal adsorbs SO2 more efficiently than CH4.
Reason R : Gases with lower critical temperatures are readily adsorbed by activated charcoal.
In the light of the above statements, choose the correct answer from the options given below.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is NOT the correct explanation of A.
(C) A is correct but R is not correct.
(D) A is not correct but R is correct.
▶️ Answer/Explanation
Detailed solution

SO2 is absorbed to a greater extent than CH4 on activated charcoal under same conditions.

Gases with higher critical temperature are readily absorbed by activated charcoal.

✅ Answer: (C)

Question 5

Boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of 8% aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is.
(A) MA = 4MB
(B) MB = 4MA
(C) MA = 8MB
(D) MB = 8MA
▶️ Answer/Explanation
Detailed solution

For A : 100 gm solution → 2 gm solute A

∴ Molality = \(\frac{2 / M_A}{0.098}\)

For B : 100 gm solution → 8 gm solute B

∴ Molality = \(\frac{8 / M_B}{0.092}\)

∴ (ΔTB)A = (ΔTB)B

∴ Molality of A = Molality of B

∴ \(\frac{2}{0.098M_A} = \frac{8}{0.092M_B}\)

\(\frac{2}{98} \times \frac{92}{8} = \frac{M_A}{M_B}\)

\(\frac{1}{4.261} = \frac{M_A}{M_B}\)

∴ MB = 4.261 × MA

✅ Answer: (B)

Question 6

The incorrect statement is
(A) The first ionization enthalpy of K is less than that of Na and Li
(B) Xe does not have the lowest first ionization enthalpy in its group
(C) The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.
(D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.
▶️ Answer/Explanation
Detailed solution

Ionization enthalpy order :

Li > Na > K

He > Ne > Ar > Kr > Xe > Rn

Sr > Rb

Zn > Ga

✅ Answer: (D)

Question 7

Which of the following methods are not used to refine any metal?
(A) Liquation (B) Calcination
(C) Electrolysis (D) Leaching
(E) Distillation
Choose the correct answer from the options given below:
(A) B and D only
(B) A, B, D and E only
(C) B, D and E only
(D) A, C and E only
▶️ Answer/Explanation
Detailed solution

Calcination and leaching are the methods of concentration of ore and not that of refining.

✅ Answer: (A)

Question 8

Given below are two statements:
Statement I : Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions.
Statement II: Density of hydrogen peroxide at 298 K is lower than that of D2O.
In the light of the above statements. Choose the correct answer from the options.
(A) Both statement I and Statement II are true
(B) Both statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
▶️ Answer/Explanation
Detailed solution

Depending on the nature of reducing agent H2O2 can act as an oxidising agent in both acidic as well as basic medium.

Density of D2O = 1.1 g/cc

Density of H2O2 = 1.45 g/cc

✅ Answer: (C)

Question 9

Given below are two statements:
Statement I : The chlorides of Be and Al have Cl-bridged structure. Both are soluble in organic solvents and act as Lewis bases.
Statement II: Hydroxides of Be and Al dissolve in excess alkali to give beryllate and aluminate ions. In the light of the above statements. Choose the correct answer from the options given below.
(A) Both statement I and Statement II are true
(B) Both statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
▶️ Answer/Explanation
Detailed solution
Be2Cl4 is lewis acid and Al2Cl6 has complete octet.
Be and Al are amphoteric metals therefore dissolve
in acid as well as alkaline solution and form
beryllate and aluminate ions in excess alkali.

✅ Answer: (D)

Question 10

Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula?
(A) Pyrophosphorous acid
(B) Hypophosphoric acid
(C) Phosphoric acid
(D) Pyrophosphoric acid
▶️ Answer/Explanation
Detailed solution
Pyrophosphorous acid  → H4P2O5.
Hypophosphoric acid  → H4P2O6.
Phosphoric acid  → H3PO4.
Pyrophosphoric acid   → H4P2O7.

✅ Answer: (D)

Question 11

Given below are two statements:
Statement I : Iron (III) catalyst, acidified K2Cr2O7 and neutral KMnO4 have the ability to oxidise I to I2 independently.
Statement II: Manganate ion is paramagnetic in nature and involves pπ-dπ bonding. In the light of the above statements, choose the correct answer from the options.
(A) Both statement I and Statement II are true
(B) Both statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
▶️ Answer/Explanation
Detailed solution

Neutral KMnO4 oxidises I– to IO3–

Manganate ion has dπ-pπ bonding.

✅ Answer: (B)

Question 12

The total number of Mn = O bonds in Mn2O7 is ______
(A) 4
(B) 5
(C) 6
(D) 3
▶️ Answer/Explanation
Detailed solution

 

 

 

✅ Answer: (C)

Question 13

Match List I with List II
List I PollutantList II Disease/sickness
A. Sulphate (>500 ppm)I. Methemoglobinemia
B. Nitrate (>50 ppm)II. Brown motting of teeth
C. Lead (>50 ppb)III. Laxative effect
D. Fluoride (>2 ppm)IV. Kidney damage
Choose the correct answer from the options given below:
(A) A-IV, B -I, C-II, D-III
(B) A-III, B -I, C-IV, D-II
(C) A-II, B -IV, C-I, D-III
(D) A-II, B -IV, C-III, D-I
▶️ Answer/Explanation
Detailed solution

A. Sulphate (>500 ppm) – Causes Laxative effect that leads to dehydration

B. Nitrate (>50 ppm) – Causes Methemoglobinemia, skin appears blue

C. Lead (> 50 ppb) – It damage kidney and RBC

D. Fluoride (>2 ppm) – It Causes Brown mottling of teeth

✅ Answer: (B)

Question 14

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : [6] Annulene, [8] Annulene and cis–[10] Annulene, are respectively aromatic, not-aromatic and aromatic.
 
 
 
 
 
 
 
 
 
 
 
Reason R : Planarity is one of the requirements of aromatic systems. In the light of the above statements, choose the most appropriate answer from the options given below.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is NOT the correct explanation of A.
(C) A is correct but R is not correct.
(D) A is not correct but R is correct.
▶️ Answer/Explanation
Detailed solution

Assertion A : Not correct , Reason R : correct

 

 

 

 

 

 

In [10] –Annulene – the hydrogen atoms in the 1
and 6 position interfere with each other and force
the molecule out of planarity

 

 

 

 

If this annulene with five cis double bonds were
planar, each internal angle would be 144°. Since a
normal double bond has bond angle of 120°, this
would be from ideal. This compound can be made
but it does not adopt a planar conformation and
therefore is not aromatic even though it has ten 
electrons.

✅ Answer: (A)

Question 15

 
 
 
 
 
 
In the above reaction product B is:
 
▶️ Answer/Explanation
Detailed solution

 

 

 

 

✅ Answer: (A)

Question 16

Match List I with List II
List IList II
A. Phenol-formaldehyde resinI. Glyptal
B. Copolymer of 1,3-butadiene and styreneII. Novolac
C. Polyester of glycol and phthalic acidIII. Buna-S
D. Polyester of glycol and terephthalic acidIV. Dacron
Choose the correct answer from the options given below:
(A) A-II, B -III, C-IV, D-I
(B) A-II, B -III, C-I, D-IV
(C) A-II, B -I, C-III, D-IV
(D) A-III, B -II, C-IV, D-I
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

 

 

✅ Answer: (B)

Question 17

A sugar ‘X’ dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar ‘X’ is
(A) Aldopentose
(B) Aldotetrose
(C) Oxalic acid
(D) Ketotetrose
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

✅ Answer: (A)

Question 18

Match List I with List II
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Choose the correct answer from the options given below:
(A) A-IV, B -III, C-II, D-I
(B) A-III, B -I, C-II, D-IV
(C) A-III, B -IV, C-I, D-II
(D) A-III, B -I, C-IV, D-II
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

 

✅ Answer: (C)

Question 19

In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound.
(Molar masses : AgBr = 188 g mol-1; Br = 80 g mol-1)
(A) 34.04%
(B) 40.04%
(C) 36.03%
(D) 38.04%
▶️ Answer/Explanation
Detailed solution

Mass of organic compound = 0.45 gm

Mass of AgBr obtained = 0.36 gm

Moles of AgBr = \(\frac{0.36}{188}\)

Mass of Bromine = \(\frac{0.36}{188} \times 80 = 0.1532\) gm

% Br in compound = \(\frac{0.1532}{0.45} \times 100 = 34.04\%\)

✅ Answer: (A)

Question 20

Match List I with List II
List IList II
A. Benzenesulphonyl chlorideI. Test for primary amines
B. Hoffmann bromamide reactionII. Anti Saytzeff
C. Carbylamine reactionIII. Hinsberg reagent
D. Hoffmann orientationIV. Known reaction of Isocyanates

Choose the correct answer from the options given below:

(A) A-IV, B -III, C-II, D-I
(B) A-IV, B -II, C-I, D-III
(C) A-III, B -IV, C-I, D-II
(D) A-IV, B -III, C-I, D-II
▶️ Answer/Explanation
Detailed solution

A.

 

                               

 

→ Hinsberg reagent
Benzenesulphonyl chloride

B. Hoffmann bromamide reaction → Known reaction of isocyanates \(R-CO-NH_2 + X_2 + 4 NaOH → R-NH_2 + 2NaX + Na_2CO_3 + 2H_2O\)
Intermediate : R- N = C = O (isocyanate)

C. Carbylamine reaction → Test for primary amine
\(R-NH_2 or Ar – NH_2 + CHCl_3 + 3KOH → RNC \) or \(Ar – NC+ 3KCl + 3H_2O\)

D. Hoffmann orientation → Anti Saytzeff
(Formation of less substituted alkene as major product)

✅ Answer: (C)

(Section B)

Question 1

20 mL of 0.02 M \(\mathrm{K_2Cr_2O_7}\) solution is used for the titration of 10 mL of \(\mathrm{Fe^{2+}}\) solution in the acidic medium. The molarity of \(\mathrm{Fe^{2+}}\) solution is ______ \(\times 10^{-2}\) M. (Nearest Integer)
▶️ Answer/Explanation
Detailed solution

Eq. of \(\mathrm{K_2Cr_2O_7}\) = Eq. of \(\mathrm{Fe^{2+}}\)
\(\Rightarrow\) (Molarity \(\times\) volume \(\times\) n.f) of \(\mathrm{K_2Cr_2O_7}\) = (molarity \(\times\) volume \(\times\) n.f ) of \(\mathrm{Fe^{2+}}\)
\(\Rightarrow 0.02 \times 20 \times 6 = M \times 10 \times 1\)
\(\Rightarrow M = 0.24\)
\(\Rightarrow\) Molarity = \(24 \times 10^{-2}\)

✅ Answer: 24

Question 2

\(\mathrm{2NO + 2H_2 \rightarrow N_2 + 2H_2O}\)
The above reaction has been studied at 800°C. The related data are given in the table below
Reaction serial numberInitial pressure of \(\mathrm{H_2}\) / kPaInitial Pressure of NO / kPaInitial rate \(\left( -\frac{dp}{dt} \right)\)/(kPa/s)
165.640.00.135
265.620.10.033
338.665.60.214
419.265.60.106
The order of the reaction with respect to NO is______
▶️ Answer/Explanation
Detailed solution

On decreasing pressure of NO by a factor of ‘2’ the rate of reaction decreases by a factor of ‘4’.
∴ Order of reaction w.r.t. ‘NO’ = 2

✅ Answer: 2

Question 3

Amongst the following the number of oxide(s) which are paramagnetic in nature is
\(\mathrm{Na_2O, KO_2, NO_2, N_2O, ClO_2, NO, SO_2, Cl_2O}\)
▶️ Answer/Explanation
Detailed solution

\(\mathrm{KO_2, NO_2, ClO_2, NO}\) are paramagnetic.

✅ Answer: 4

Question 4

The molar heat capacity for an ideal gas at constant pressure is 20.785 J \(\mathrm{K^{-1}}\) mol\(^{-1}\). The change in internal energy is 5000 J upon heating it from 300K to 500K. The number of moles of the gas at constant volume is ___ [Nearest integer]
(Given: R = 8.314 J \(\mathrm{K^{-1}}\) mol\(^{-1}\))
▶️ Answer/Explanation
Detailed solution

\(\mathrm{C_{p,m} = C_{v,m} + R}\)
\(\Rightarrow \mathrm{C_{v,m} = 20.785 – 8.314 = 12.471 \, J \, K^{-1} \, mol^{-1}}\)
\(\Delta U = nC_{v,m}\Delta T\)
\(\Rightarrow n = \frac{5000}{12.471 \times 200} = \frac{25}{12.471} \approx 2\)

✅ Answer: 2

Question 5

According to MO theory, number of species/ions from the following having identical bond order is ___ :
\(\mathrm{CN^- , NO^+ , O_2 , O_2^+ , O_2^{2+}}\)
▶️ Answer/Explanation
Detailed solution

\(\mathrm{CN^- , NO^+ , O_2^{2+}}\) have bond order = 3

✅ Answer: 3

Question 6

At 310 K, the solubility of \(\mathrm{CaF_2}\) in water is \(2.34 \times 10^{3}\) g /100 mL. The solubility product of \(\mathrm{CaF_2}\) is ___ \(\times 10^{8}\) (mol/L\(^3\)). (Given molar mass : \(\mathrm{CaF_2 = 78}\) g mol\(^{-1}\))
▶️ Answer/Explanation
Detailed solution

Solubility of \(\mathrm{CaF_2}\) = S mole/L
\(S = \frac{2.34 \times 10^{-3}}{0.1 \times 78} = \frac{2.34}{78} \times 10^{-2} = 3 \times 10^{-4}\) mol/L
\(K_{sp} (\mathrm{CaF_2}) = 4S^3 = 4(3 \times 10^{-4})^3\)
\(= 108 \times 10^{-12}\)
\(= 0.0108 \times 10^{-8}\) (mol/L)\(^3\)

✅ Answer: 0

Question 7

The conductivity of a solution of complex with formula \(\mathrm{CoCl_3(NH_3)_4}\) corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is ___
▶️ Answer/Explanation
Detailed solution

\(\mathrm{[Co(NH_3)_4 Cl_2]Cl}\)
Primary valency = oxidation no. = +3

✅ Answer: 1

Question 8

In the titration of \(\mathrm{KMnO_4}\) and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ___
▶️ Answer/Explanation
Detailed solution

Oxidation state of carbon changes from +3 to +4.
\(\mathrm{2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4(dil.) \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O}\)

✅ Answer: 1

Question 9

Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of (+) isomer is +30°. The optical purity is ___%
▶️ Answer/Explanation
Detailed solution

% optical purity = \(\frac{\text{observed rotation of mixture} \times 100}{\text{rotation of pure enantiomer}}\)
\(= \frac{+12.6°}{+30°} \times 100 = 42\)

✅ Answer: 42

Question 10

In the following reaction
 
 
 
 
 
 
The % yield for reaction I is 60% and that of reaction II is 50%. The overall yield of the complete reaction is ___% [nearest integer]
▶️ Answer/Explanation
Detailed solution

(I)

 

 

 

 

 

Let initial moles of reactant taken = n
Total moles obtained for benzene sulphonic acid (with % yield = 60%) = 0.6n
(II)

 

 

 

 

 

 

Moles of benzene sulphonic acid before reaction II = 0.6n
Moles obtained for phenol (with % yield = 50%) = 0.6×0.5n = 0.3n
So over all % yield of complete reaction = \(\frac{0.3n}{n}\) × \(100 = 30\)

✅ Answer: 30

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