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Chemistry Section-A

Question 1

A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is :
(Atomic mass : Cl = 35.5 amu, H = 1 amu)
(A) 10.2 M
(B) 12.5 M
(C) 14.0 M
(D) 18.2 M
▶️ Answer/Explanation
Detailed solution

Let total volume = 1000 mL = 1L
Total mass of solution = 1460 g
Mass of HCl = \( \frac{35}{100} \times 1460 \) g
Moles of HCl = \( \frac{35 \times 1460}{100 \times 36.5} \)
So molarity = \( \frac{35 \times 1460}{100 \times 36.5} = 14 \text{ M} \)

✅ Answer: (C)

Question 2

An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL–1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol–1 is :
(Given : R = 0.082 L atm K–1 mol–1)
(A) 35
(B) 50
(C) 75
(D) 125
▶️ Answer/Explanation
Detailed solution

Mass of liquid = 135 – 40 = 95 g
Volume of liquid = \( \frac{95}{0.95} \) mL = 100 mL = 0.1 L
Mass of ideal gas = 40.5 – 40 = 0.5 g
Using PV = nRT:
\( 0.82 \times 0.1 = \frac{0.5}{M} \times 0.082 \times 250 \)
Solving for M: M = 125

✅ Answer: (D)

Question 3

If the radius of the 3rd Bohr’s orbit of hydrogen atom is r3 and the radius of 4th Bohr’s orbit is r4. Then :
(A) \( \frac{r_4}{r_3} = \frac{9}{16} \)
(B) \( \frac{r_4}{r_3} = \frac{16}{9} \)
(C) \( \frac{r_4}{r_3} = \frac{3}{4} \)
(D) \( \frac{r_4}{r_3} = \frac{4}{3} \)
▶️ Answer/Explanation
Detailed solution

Radius formula: \( r = \frac{n^2}{z} \times 0.529 \) Å
\( r_3 = 0.529 \times \frac{3^2}{1} \)
\( r_4 = 0.529 \times \frac{4^2}{1} \)
\( \frac{r_4}{r_3} = \frac{4^2}{3^2} = \frac{16r_2}{9} \)

✅ Answer: (B)

Question 4

Consider the ions/molecule O2+, O2, O2–, O22–
For increasing bond order the correct option is :
(A) O22– < O2– < O2 < O2+
(B) O2– < O22– < O2 < O2+
(C) O2– < O22– < O2+ < O2
(D) O2– < O2+ < O22– < O2
▶️ Answer/Explanation
Detailed solution
ion/moleculeNumber of e⁻ in BMONumber of e⁻ in ABMOBond order
O₂⁺1052.5
O₂1062
O₂⁻1071.5
O₂²⁻1081

Bond order \( O_2^{2-} < O_2^- < O_2 < O_2^+ \)

✅ Answer: (A)

Question 5

The \( \left( \frac{\partial E}{\partial T} \right)_P \) of different types of half cells are as follows:
A: \( 1 \times 10^{-4} \), B: \( 2 \times 10^{-4} \), C: \( 0.1 \times 10^{-4} \), D: \( 0.2 \times 10^{-4} \)
(Where E is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?
(A) A
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Detailed solution

A cell with less variation in EMF with temperature is preferred as reference electrode because it can be used for wider range of temperature without much derivation from standard value so a cell with less \( \left( \frac{\partial E}{\partial T} \right)_P \) is preferred.

✅ Answer: (C)

Question 6

Choose the correct stability order of group 13 elements in their +1 oxidation state.
(A) Al < Ga < In < Tl
(B) Tl < In < Ga < Al
(C) Al < Ga < Tl < In
(D) Al < Tl < Ga < In
▶️ Answer/Explanation
Detailed solution

Moving down the group stability of lower oxidation state increases: Al < Ga < In < Tl

✅ Answer: (A)

Question 7

Given below are two statements:
Statement I : According to the Ellingham diagram, any metal oxide with higher ΔG° is more stable than the one with lower ΔG°.
Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.
In the light of the above statements, choose the most appropriate answer from 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

Metal oxide with lower ΔG° is more stable. Statement I is incorrect.
Statement II is correct: The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.

✅ Answer: (D)

Question 8

Consider the following reaction:
 
 
The dihedral angle in product A in its solid phase at 110 K is:
(A) 104°
(B) 111.5°
(C) 90.2°
(D) 111.0°
▶️ Answer/Explanation
Detailed solution


 

 

 

✅ Answer: (C)

Question 9

The correct order of melting point is:
(A) Be > Mg > Ca > Sr
(B) Sr > Ca > Mg > Be
(C) Be > Ca > Mg > Sr
(D) Be > Ca > Sr > Mg
▶️ Answer/Explanation
Detailed solution

Melting points:
Be: 1560 K
Mg: 924 K
Ca: 1124 K
Sr: 1062 K
Order: Be > Ca > Sr > Mg

✅ Answer: (D)

Question 10

The correct order of melting points of hydrides of group 16 elements is:
(A) H2S < H2Se < H2Te < H2O
(B) H2O < H2S < H2Se < H2Te
(C) H2S < H2Te < H2Se < H2O
(D) H2Se < H2S < H2Te < H2O
▶️ Answer/Explanation
Detailed solution

Melting points:
H2O: 273 K
H2S: 188 K
H2Se: 208 K
H2Te: 222 K
Order: H2S < H2Se < H2Te < H2O

✅ Answer: (A)

Question 11

Consider the following reaction:
A + alkali → B (Major Product)
If B is an oxoacid of phosphorus with no P–H bond, then A is:
(A) White P4
(B) Red P4
(C) P2O3
(D) H3PO3
▶️ Answer/Explanation
Detailed solution

Red P4 + Alkali → H4P2O6 (No P–H bond)

✅ Answer: (B)

Question 12

Polar stratospheric clouds facilitate the formation of:
(A) CIONO2
(B) HOCl
(C) ClO
(D) CH4
▶️ Answer/Explanation
Detailed solution

Polar stratospheric clouds provide surface on which hydrolysis of CIONO2 takes place to form HOCl (Hypochlorous acid)
CIONO2 (g) + H2O(g) → HOCl(g) + HNO3(g)

✅ Answer: (B)

Question 13

Given below are two statements:
Statement I : In ‘Lassaigne’s Test, when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed.
Statement II : If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give NaCN and Na2S.
In the light of the above statements, choose the most appropriate answer from 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

Both statement I & statement II are correct.

✅ Answer: (A)

Question 14

\( \left( C_7H_5O_2 \right)_2 \xrightarrow{\text{hv}} [X] + 2C_6H_5 + 2CO_2 \)
Consider the above reaction and identify the intermediate ‘X’
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

✅ Answer: (D)

Question 15

 
▶️ Answer/Explanation
Detailed solution

Although Acetyl Acetone predominantly gives Acid base reaction with G.R due to Active methylene group but according to given option ans should be based on nucleophilic addition reaction (NAR).

 

 

 

 

 

 

 

✅ Answer: (A)

Question 16

Which will have the highest enol content ?
 
▶️ Answer/Explanation
Detailed solution

 

 

 

 

Which is aromatic in nature.

✅ Answer: (C)

Question 17

Among the following structures, which will show the most stable enamine formation ?
▶️ Answer/Explanation
Detailed solution

All these enamines are interconvertible through their resonating structures. So most stable form is ‘C’ due to steric factor.

✅ Answer: (C)

Question 18

Which of the following sets are correct regarding polymer ?
(A) Copolymer : Buna–S
(B) Condensation polymer : Nylon–6,6
(C) Fibre : Nylon–6,6
(D) Thermosetting polymer : Terylene
(E) Homopolymer : Buna–N

Choose the correct answer from given options below:
(A) (A), (B) and (C) are correct
(B) (B), (C) and (D) are correct
(C) (A), (C) and (E) are correct
(D) (A), (B) and (D) are correct
▶️ Answer/Explanation
Detailed solution

Buna-S is a copolymer of butadiene and styrene. Nylon-6,6 is a condensation polymer of adipic acid and hexamethylenediamine. Nylon-6,6 is a fiber. Terylene is a fiber, not a thermosetting polymer. Buna-N is a copolymer, not a homopolymer.

✅ Answer: (A)

Question 19

A chemical which stimulates the secretion of pepsin is :
(A) Anti histamine
(B) Cimetidine
(C) Histamine
(D) Zantac
▶️ Answer/Explanation
Detailed solution

Histamine stimulates the secretion of pepsin and HCl in the stomach.

✅ Answer: (C)

Question 20

Which statement is not true with respect to nitrate ion test ?
(A) A dark brown ring is formed at the junction of two solutions.
(B) Ring is formed due to nitroferrous sulphate complex.
(C) The brown complex is [Fe(H\(_2\)O)\(_5\)(NO)]SO\(_4\).
(D) Heating the nitrate salt with conc. H\(_2\)SO\(_4\), light brown fumes are evolved.
▶️ Answer/Explanation
Detailed solution

The brown ring is formed due to the formation of nitrosoferrous sulphate, not nitroferrous sulphate.

✅ Answer: (B)

Section-B

Question 1

For complete combustion of methanol \[ CH_3OH(1)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(1) \] the amount of heat produced as measured by bomb calorimeter is 726 kJ mol\(^{-1}\) at 27°C. The enthalpy of combustion for the reaction is \(-\text{x kJ mol}^{-1}\), where x is ______. (Nearest integer)
(Given : R = 8.3 JK\(^{-1}\) mol\(^{-1}\))
▶️ Answer/Explanation
Detailed solution

\(\Delta U = -726 \, \text{kJ/mol}\)
\(\Delta n_g = 1 – \frac{3}{2} = -\frac{1}{2}\)
\(\Delta H = \Delta U + \Delta n_gRT\)
\(= -726 – \frac{1}{2} \times \frac{8.3 \times 300}{1000}\)
\(= -727.245\)
Nearest integer is 727.

✅ Answer: 727

Question 2

A 0.5 percent solution of potassium chloride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is ______. (Nearest integer)
(Molal depression constant for water is 1.80 K kg mol\(^{-1}\) and molar mass of KCl is 74.6 g mol\(^{-1}\))
▶️ Answer/Explanation
Detailed solution

0.5% solution of KCl means 0.5 g KCl in 100 g solution, so mass of solvent ≈ 99.5 g = 0.0995 kg.
Molality, \(m = \frac{0.5}{74.6} \times \frac{1}{0.0995}\)
\(\Delta T_f = i \times m \times K_f\)
\(0.24 = i \times \frac{0.5}{74.6 \times 0.0995} \times 1.80\)
\(i \approx 1.979\)
For KCl, \(i = 1 + \alpha\)
\(1.979 = 1 + \alpha\)
\(\alpha = 0.979\)
Percentage dissociation = \(97.9\% \approx 98\%\)

✅ Answer: 98

Question 3

50 mL of 0.1 M CH\(_3\)COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be ______ × 10\(^{-2}\). (Nearest integer)
(Given : pK\(_a\) (CH\(_3\)COOH) = 4.76)
\(\log 2 = 0.30\), \(\log 3 = 0.48\), \(\log 5 = 0.69\), \(\log 7 = 0.84\), \(\log 11 = 1.04\)
▶️ Answer/Explanation
Detailed solution

Moles of CH\(_3\)COOH = \(50 \times 0.1 = 5\) mmol
Moles of NaOH = \(25 \times 0.1 = 2.5\) mmol
After reaction:
CH\(_3\)COOH left = \(5 – 2.5 = 2.5\) mmol
CH\(_3\)COONa formed = 2.5 mmol
This is a buffer solution.
pH = pK\(_a\) + \(\log \frac{[\text{Salt}]}{[\text{Acid}]}\) = 4.76 + \(\log \frac{2.5/75}{2.5/75}\) = 4.76
So, pH = \(476 \times 10^{-2}\)

✅ Answer: 476

Question 4

A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______ s.
(Given : ln 2 = 0.693)
▶️ Answer/Explanation
Detailed solution

\(k_A = \frac{\ln 2}{100}\), \(k_B = \frac{\ln 2}{50}\)
\(A_t = A_0 e^{-k_A t}\), \(B_t = B_0 e^{-k_B t}\)
Given \(A_0 = B_0\) and \(A_t = 4B_t\)
So, \(A_0 e^{-k_A t} = 4 B_0 e^{-k_B t}\)
\(e^{-k_A t} = 4 e^{-k_B t}\)
\(e^{(k_B – k_A)t} = 4\)
\((\frac{\ln 2}{50} – \frac{\ln 2}{100})t = \ln 4\)
\(\frac{\ln 2}{100}t = 2 \ln 2\)
\(t = 200\) s

✅ Answer: 200

Question 5

2.0 g of \( H_2 \) gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______ mL.
(Given : \( R = 0.083 \, \text{L bar K}^{-1} \, \text{mol}^{-1} \))
▶️ Answer/Explanation
Detailed solution

Moles of H\(_2\) = \(\frac{2.0}{2} = 1.0\) mol
Volume of H\(_2\) = \(\frac{nRT}{P} = \frac{1.0 \times 0.083 \times 300}{1} = 24.9\) L = 24900 mL
Volume adsorbed per gram of adsorbent = \(\frac{24900}{2.5} = 9960\) mL

✅ Answer: 9960

Question 6

The spin-only magnetic moment value of the most basic oxide of vanadium among \( V_2O_3 \), \( V_2O_4 \) and \( V_2O_5 \) is ______ B.M. (Nearest Integer)
▶️ Answer/Explanation
Detailed solution

The most basic oxide is \( V_2O_3 \).
V\(^{3+}\) has electronic configuration [Ar] 3d\(^2\).
Number of unpaired electrons, n = 2.
Spin-only magnetic moment, \(\mu = \sqrt{n(n+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83\) B.M. ≈ 3 B.M.

✅ Answer: 3

Question 7

The spin-only magnetic moment value of an octahedral complex among CoCl\(_3\).4NH\(_3\), NiCl\(_2\).6H\(_2\)O and PtCl\(_4\).2HCl, which upon reaction with excess of AgNO\(_3\) gives 2 moles of AgCl is ______ B.M. (Nearest Integer)
▶️ Answer/Explanation
Detailed solution

CoCl₃, 4NH₃ → [Co(NH₃)₄ Cl₂]Cl
NiCl₂,6H₂O → [Ni(H₂O)₆]Cl₂
PtCl₄, 2HCl → H₂[PtCl₆]

[Ni(H₂O)₆]Cl₂ → \( ^{2AgNO₃} \) → 2AgCl ↓ + [Ni(H₂O)₆](NO₃)₂

 

 

\( \mu = \sqrt{2(2+2)} \text{B.M} = 2.84 \text{B.M} \approx 3 \) 

✅ Answer: 3

Question 8

On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is ______ (Nearest Integer)
▶️ Answer/Explanation
Detailed solution

\( C_xH_YO_z + \left( x + \frac{y}{4} – \frac{z}{2} \right) O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O \) 0.3g  0.2g  0.1g \( \frac{n_{CO_2}}{n_{H_2O}} = \frac{x}{y/2} = \frac{0.2/44}{0.1/18} \) \( \frac{2x}{y} = \frac{36}{44} = \frac{9}{11} \) \( x = \frac{9y}{22} \) \( \frac{n_{C_xH_YO_z}}{n_{CO_2}} = \frac{1}{x} \) \( \frac{0.3}{12x + y + 16z} \times \frac{44}{0.2} = \frac{1}{x} \) 66x = 12x + y + 16z
54x = y + 16z
54 × \(\frac{9y}{22}\) – y = 16z \( \frac{464y}{22} = 16z \)
\( z = \frac{29y}{22} \) \( C_xH_yO_z = C_xH_yO_z \) \( C_{9y}H_yO_{29y} \) \( C_9H_{22}O_{29} \) % of \( C = \frac{12 \times 9}{(12 \times 9 + 22 + 29 \times 16)} \times 100 = \frac{108}{594} \times 100 \) 18.1%

✅ Answer: 18

Question 9

Compound ‘P’ on nitration with dil. HNO\(_3\) yields two isomers (A) and (B). These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. HNO\(_3\) yields a yellow compound ‘C’, a strong acid. The number of oxygen atoms is present in compound ‘C’ ______.
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

 

✅ Answer: 7

Question 10

The number of oxygens present in a nucleotide formed from a base, that is present only in RNA is ______.
▶️ Answer/Explanation
Detailed solution

The base present only in RNA is uracil.

 

 

 

 

 

✅ Answer: 9

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