For the first order reaction Based on the given data, find the time (in min):
Given:
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For a first-order reaction, concentration decreases exponentially with time. In equal time intervals, the concentration follows a geometric progression. Thus, the ratio of concentrations in equal time intervals remains constant. Step 1: Compare concentration ratios. Both ratios are the same, so the time intervals are equal. Step 2: Determine the time intervals. Total time from to minutes corresponds to two equal intervals. Therefore,
02
PYQ 2026
medium
chemistryID: jee-main
20 ml of CH COOH solution is neutralised by 28.08 ml of NaOH. In the 20 ml of same solution 14.04 ml of same NaOH solution is poured, then pH of resultant solution will be: (pKa of CH COOH = 4.75)
1
2
3
4
Official Solution
Correct Option: (1)
Concept: When a weak acid is partially neutralised by a strong base, a buffer solution is formed. The pH of such a solution is given by the Henderson–Hasselbalch equation: Step 1: Understand the neutralisation reaction. Step 2: Identify equivalence and half-equivalence point. Given: • 28.08 ml NaOH is required for complete neutralisation.
• 14.04 ml NaOH = half of 28.08 ml. Thus, the system is at half-equivalence point. Step 3: Apply buffer concept. At half-equivalence point: So, Step 4: Calculate pH.
03
PYQ 2026
medium
chemistryID: jee-main
Given that: A Galvanic cell is formed using above electrodes, whose when reaction quotient of cell reaction is . Find the value of . [Nearest integer]
1
2
3
4
Official Solution
Correct Option: (2)
Concept: The EMF of a galvanic cell under non-standard conditions is given by the Nernst equation: where: • is standard EMF,
• is number of electrons transferred,
• is reaction quotient. Step 1: Write the half-cell reactions. Anode (oxidation): Cathode (reduction): Step 2: Balance the electrons. LCM of electrons = Overall reaction: Step 3: Calculate standard EMF. Step 4: Apply Nernst equation. Step 5: Solve for . Nearest integer:
04
PYQ 2026
medium
chemistryID: jee-main
Compound (X) has molar mass . mole of X gives as precipitate. What is the % of S in compound (X)? Mention answer in form.
1
215
2
325
3
435
4
515
Official Solution
Correct Option: (3)
Concept: Sulphur present in the compound is converted into precipitate. The relation between sulphur and is: Thus sulphur percentage can be calculated from the mass of . Step 1: Mass of organic compound used Step 2: Mass relation between S and Step 3: Calculate percentage of S Step 4: Convert to form Thus,
05
PYQ 2026
medium
chemistryID: jee-main
For the reaction: The variation of concentration of reactant with time is shown below. Statements: I. Curve 1 will be observed when order of reaction is 1. II. If curve 2 & curve 3 are observed for order then rate constant for curve will be greater than rate constant for curve. III. Decomposition of HI on gold surface represent curve (1). Select the option representing correct set of statements:
1
I, II, III
2
I, II
3
II, III
4
I, III
Official Solution
Correct Option: (3)
Concept: The concentration–time graph differs for different orders of reactions. • Zero order: decreases linearly with time.
• First order: exponential decay curve.
• Higher rate constant gives faster decrease of concentration. Step 1: Analyse curve 1 Curve 1 is a straight line decrease of concentration with time. This corresponds to zero order reaction, not first order. Hence statement I is incorrect. Step 2: Compare curve 2 and curve 3 For first order reactions, steeper decay corresponds to larger rate constant. Curve 3 decreases faster than curve 2. Therefore, Statement II is correct. Step 3: Example of zero order reaction Decomposition of HI on gold surface follows zero order kinetics. Hence it corresponds to curve 1. Statement III is correct. Thus,
06
PYQ 2026
medium
chemistryID: jee-main
Consider the following setup separated by a semi–permeable membrane (SPM): Left compartment (x): solution , Right compartment (y): solution , Which of the following is correct for the above figure?
1
is hypotonic solution
2
Both compartment will have blue colour
3
To perform reverse osmosis, any pressure can be applied on x-side
4
Solute can pass through semi-permeable membrane
Official Solution
Correct Option: (1)
Concept: Osmotic pressure depends on the van’t Hoff factor . Greater value of means higher osmotic pressure and the solution becomes hypertonic. The solution with smaller osmotic pressure is hypotonic. Step 1: Calculate van’t Hoff factor for each solution. For Total ions For Total ions Step 2: Compare osmotic pressures. Both solutions have same molarity. Thus, Hence, Step 3: Check other options. (B) Blue colour will not appear because ions cannot cross the SPM. (C) Reverse osmosis requires pressure greater than osmotic pressure, not any pressure. (D) Only solvent molecules pass through SPM. Therefore, the correct statement is:
07
PYQ 2026
medium
chemistryID: jee-main
From given data (I) 90.8 lit of at STP (II) 684 gm of sucrose (III) 2 moles of cyclohexane Correct order of total number of atoms will be
1
(I) (II) (III)
2
(III) (II) (I)
3
(II) (III) (I)
4
(I) (III) (II)
Official Solution
Correct Option: (3)
Concept: Total number of atoms depends on number of moles of molecules and number of atoms per molecule. Step 1: Case (I) Hydrogen gas Each molecule has 2 atoms. Step 2: Case (II) Sucrose Molar mass Atoms in one molecule: Step 3: Case (III) Cyclohexane Atoms per molecule: Step 4: Compare Thus,
08
PYQ 2026
medium
chemistryID: jee-main
When 3.395 gm is burnt completely, it liberates 101.11 kJ of heat, then find the enthalpy of formation of . Given: of of graphite
1
2
3
4
Official Solution
Correct Option: (1)
Concept: Using Hess's law, the enthalpy of combustion can be related to enthalpies of formation of products and reactants. Step 1: Write the combustion reaction Step 2: Calculate moles of ethanol Step 3: Heat released for 1 mole ethanol Since heat is released, Step 4: Apply Hess law
09
PYQ 2026
hard
chemistryID: jee-main
For a given reversible reaction at : If is 50% dissociated at 5 bar and 300 K at equilibrium, then find for the reaction? [Given: ]
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For gaseous equilibria involving dissociation, the equilibrium constant can be calculated using mole fractions and total pressure. The relation between standard Gibbs free energy change and equilibrium constant is: where and . Step 1: Write the equilibrium reaction. Assume initially mole of . Degree of dissociation . Substitute Total moles: Step 2: Write the expression for . Using mole fraction relation: Here Step 3: Substitute the values. Step 4: Calculate .
10
PYQ 2026
medium
chemistryID: jee-main
Statement I: The bond angle order in is
Statement II: and are all ionic compounds.
1
Both Statement I and Statement II are correct
2
Statement I is incorrect but Statement II is correct
3
Statement I is correct but Statement II is incorrect
4
Both Statement I and Statement II are incorrect
Official Solution
Correct Option: (3)
Step 1: Analyzing Statement I. - has a bent shape with a bond angle of around , which is less than the bond angle in .
- has a bond angle of due to the two lone pairs of electrons on oxygen.
- also has a bent shape, but the bond angle is slightly less than , around . Thus, the bond angle order is . Statement I is correct. Step 2: Analyzing Statement II. - , , and are covalent compounds, not ionic compounds. They are all covalent network compounds where the central atom is covalently bonded to fluorine atoms.
- Therefore, Statement II is incorrect because these compounds are covalent, not ionic. Step 3: Conclusion. Since Statement I is correct but Statement II is incorrect, the correct answer is option (C). Final Answer: (C) Statement I is correct but Statement II is incorrect
11
PYQ 2026
medium
chemistryID: jee-main
If the Bohr's radius of H-atom is 52.9 pm and radius of H-like species is 70.53 pm, then chemical species and orbit number are respectively.
1
Li , 3
2
Li , 2
3
He , 3
4
He , 2
Official Solution
Correct Option: (2)
The Bohr radius for a hydrogen atom is given as 52.9 pm. The radius of an -like species (i.e., an ion that is hydrogen-like) can be calculated using the formula: Where:
- is the radius of the -like species,
- is the orbit number (principal quantum number),
- is the atomic number of the species. We are given that the radius of the -like species is 70.53 pm. Substituting this into the formula: Now, solving for : Now, let's check the options:
- For , , so we calculate , which gives .
- Thus, the chemical species is and the orbit number is 2. Final Answer: Option (B) Li , 2.
12
PYQ 2026
medium
chemistryID: jee-main
0.2M, 500ml MnO solution is treated with 1.5M, 500ml KI solution in basic medium. The liberated required ‘X’M, 300ml hypo solution. The value of ‘X’ is -
Official Solution
Correct Option: (1)
Step 1: Write the balanced reaction.
The reaction between potassium permanganate (MnO ) and potassium iodide (KI) in a basic medium can be written as:
Step 2: Calculate moles of MnO .
The molarity of MnO is given as 0.2M, and the volume is 500ml or 0.5L. The moles of MnO are calculated as:
Step 3: Determine the moles of I reacted.
From the balanced equation, 2 moles of MnO react with 10 moles of I . Therefore, the moles of I that reacted are:
Step 4: Calculate the molarity of I .
The moles of I correspond to the moles of formed, and the volume of formed is 300ml or 0.3L. The molarity of is:
13
PYQ 2026
medium
chemistryID: jee-main
Statement I: Of the following compounds ; only three compounds have a complete octet.
Statement II: For the given set of compounds only one set of compounds has one lone pair on all central atoms.
1
Both Statement I and Statement II are correct
2
Statement I is incorrect but Statement II is correct
3
Statement I is correct but Statement II is incorrect
4
Both Statement I and Statement II are incorrect
Official Solution
Correct Option: (2)
Step 1: Analysis of Statement I. - has 8 electrons around sulfur (complete octet).
- has 8 electrons around sulfur (complete octet).
- has 8 electrons around sulfur (complete octet).
- has 10 electrons around sulfur (not a complete octet). Hence, only three compounds, , have a complete octet. Therefore, Statement I is incorrect because does not have a complete octet. Step 2: Analysis of Statement II. Let's examine the lone pairs on the central atoms of the given compounds: - has 1 lone pair on the central atom (sulfur).
- has 1 lone pair on the central atom (bromine).
- has 2 lone pairs on the central atom (chlorine). - has 1 lone pair on nitrogen.
- has 1 lone pair on the central atom (bromine).
- has 1 lone pair on the central atom (sulfur). - has 2 lone pairs on oxygen.
- has 1 lone pair on iodine.
- has 2 lone pairs on xenon. Thus, only the set has 1 lone pair on all central atoms. Therefore, Statement II is correct. Final Answer: (B) Statement I is incorrect but Statement II is correct
14
PYQ 2026
medium
chemistryID: jee-main
Consider the following ionization energy order (A) IE₁: Cr > Mn (B) IE₂: Cr < Mn (C) IE₁: Cr < Mn (D) IE₂: Cr > Mn Correct order among the following:
1
A and B
2
B and C
3
C and D
4
A and D
Official Solution
Correct Option: (3)
In order to answer this question, let's first analyze the ionization energies of Cr and Mn. Step 1: Ionization Energy for Cr and Mn. - Chromium (Cr) has an electron configuration of . This configuration is relatively stable because of the half-filled -orbitals. - Manganese (Mn) has an electron configuration of , which is also stable due to the half-filled -orbitals but with an additional -electron. This stability impacts their ionization energies: - for Cr is slightly higher than for Mn because the ionization of Cr results in a stable half-filled configuration. - for Cr is lower than for Mn because removing a second electron from Cr is easier than removing the second from Mn. Step 2: Analyzing the Statements. - (A) IE₁: Cr > Mn : This is correct. - (B) IE₂: Cr < Mn : This is correct. - (C) IE₁: Cr < Mn : This is incorrect. - (D) IE₂: Cr > Mn : This is incorrect. Step 3: Conclusion. The correct statements are (A) and (B). Final Answer: Option (A) and (B).
15
PYQ 2026
medium
chemistryID: jee-main
‘X’ is the nitrogen containing compound which is added to freshly prepared FeSO and dilute H SO to form a brown ring complex ‘Y’.
Identify X and Y.
1
NO; [Fe(NO)]SO
2
N O; [Fe(N O)]SO
3
NO ; [Fe(NO )]SO
4
N O ; [Fe(N O )]SO
Official Solution
Correct Option: (1)
Step 1: Understanding the Brown Ring Complex. The brown ring complex is a characteristic complex formed when nitric oxide (NO) is added to freshly prepared FeSO (iron(II) sulfate) in the presence of dilute H SO . The reaction involves the formation of a complex where NO binds to Fe to form [Fe(NO)] with a brown ring. The reaction is: This results in a brown complex, known as the brown ring complex. Step 2: Analyzing the Options. - (A) NO; [Fe(NO)]SO : Correct. This is the known reaction for the brown ring complex.
- (B) N O; [Fe(N O)]SO : Incorrect. Nitrous oxide (N O) does not form a brown ring complex with iron(II).
- (C) NO ; [Fe(NO )]SO : Incorrect. Nitrogen dioxide (NO ) does not form a brown ring complex with iron.
- (D) N O ; [Fe(N O )]SO : Incorrect. Dinitrogen tetroxide (N O ) does not form a brown ring complex with iron. Step 3: Conclusion. The correct answer is option (A), where the nitrogen-containing compound is NO and the resulting complex is [Fe(NO)]SO , which is the brown ring complex. Final Answer: Option (A) NO; [Fe(NO)]SO .
16
PYQ 2026
medium
chemistryID: jee-main
Column-I (Isothermal process)
1
P-(ii); Q-(iv); R-(iii); S-(i)
2
P-(iii); Q-(ii); R-(iv); S-(i)
3
P-(iv); Q-(i); R-(ii); S-(iii)
4
P-(i); Q-(iii); R-(ii); S-(iv)
Official Solution
Correct Option: (1)
Step 1: Understanding each process. Reversible expansion (P): In an isothermal reversible expansion, heat absorbed is given by:
Hence, it matches with option (ii). Reversible cyclic process (Q): For any cyclic process, change in internal energy is zero, so:
Also, for a reversible cyclic process:
Hence, it matches with option (iv). Irreversible compression (R): Work done in irreversible process is:
Hence, it matches with option (iii). Free expansion (S): In free expansion, no work is done and no heat is exchanged:
Hence, it matches with option (i). Step 2: Final matching. P-(ii); Q-(iv); R-(iii); S-(i) Final Answer: (A) P-(ii); Q-(iv); R-(iii); S-(i)
17
PYQ 2026
medium
chemistryID: jee-main
Consider the following electrochemical cell The was found to be . The value of is
1
2
3
4
Official Solution
Correct Option: (2)
Concept: The cell potential under non-standard conditions is calculated using the Nernst equation: where is the reaction quotient. Step 1: Write the overall cell reaction. Number of electrons transferred: Step 2: Calculate standard cell potential. Step 3: Write reaction quotient . Since : Step 4: Apply Nernst equation. Step 5: Substitute given values.
18
PYQ 2026
medium
chemistryID: jee-main
Consider the following equilibrium: If equilibrium pressure is and degree of dissociation is , then is
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For gaseous equilibria involving dissociation, the equilibrium constant in terms of pressure ( ) is calculated using the partial pressures of reactants and products. The degree of dissociation helps determine the equilibrium composition. Step 1: Assume initial moles. Let initial moles of . If degree of dissociation : Step 2: Total moles at equilibrium. Thus total moles remain constant. Step 3: Calculate partial pressures. Total pressure Step 4: Substitute in . Since pressure dependence is included in equilibrium expression:
19
PYQ 2026
medium
chemistryID: jee-main
Statement I: , , are tetrahedral, square planar and linear respectively. Statement II: is hybridized.
1
Both Statement I and Statement II are correct
2
Statement I is correct but Statement II is incorrect
3
Statement I is incorrect but Statement II is correct
4
Both Statement I and Statement II are incorrect
Official Solution
Correct Option: (1)
Step 1: Analyze Statement I. - has a tetrahedral structure. This is because the central chlorine atom is surrounded by four oxygen atoms with bond angles of approximately 109.5°, which is characteristic of a tetrahedral geometry. - has a square planar structure. The iodine atom in this complex is surrounded by four chlorine atoms with 90° bond angles, which results in a square planar geometry. - has a linear structure. This is due to the two bromine atoms bonded to the iodine atom, with bond angles of 180°, leading to a linear geometry. Thus, Statement I is correct. Step 2: Analyze Statement II. is a coordination complex with iron in the +2 oxidation state. The iron ion has a electron configuration. In the presence of six cyanide ligands, the complex undergoes hybridization to form six bonds, resulting in an hybridization. This is because the iron ion uses two -orbitals, one -orbital, and three -orbitals to form six bonds in an octahedral geometry. Therefore, Statement II is also correct. Step 3: Conclusion. Since both Statement I and Statement II are correct, the correct answer is option (A). Final Answer: Both Statement I and Statement II are correct.
20
PYQ 2026
medium
chemistryID: jee-main
For the reaction , the following graph is observed between half life and initial concentration of . Then find the value of .
Official Solution
Correct Option: (1)
Step 1: Identify the order of the reaction from the graph.
The graph between half-life and initial concentration is a straight line passing through the origin.
This means that half-life is directly proportional to the initial concentration:
Such a relation is valid for a zero order reaction, because for zero order kinetics: Step 2: Write the proportional relation from the graph.
Since
we can write:
From the graph,
when , then
And when , then Step 3: Substitute the values into the proportionality relation.
So, Step 4: Simplify and calculate . Step 5: State the final answer.
Hence, the value of is:
21
PYQ 2026
medium
chemistryID: jee-main
For the complex ion with configurations , (low spin), (high spin), (low spin) and (high spin), the total number of unpaired electrons is _ _ _ _ _ _ _ _ _ _ _ _ .
Official Solution
Correct Option: (1)
Step 1: Find the number of unpaired electrons for .
For a configuration in an octahedral complex, the three electrons occupy three different -orbitals singly according to Hund's rule.
So, the number of unpaired electrons is: Step 2: Find the number of unpaired electrons for (low spin).
In a low spin complex, electrons pair up in the lower energy orbitals before entering the higher energy orbitals.
Thus, the arrangement is .
This gives: Step 3: Find the number of unpaired electrons for (high spin).
In a high spin complex, all five electrons occupy the five -orbitals singly before any pairing occurs.
Thus, all five electrons remain unpaired.
So, the number of unpaired electrons is: Step 4: Find the number of unpaired electrons for (low spin).
In a low spin complex, electrons first fill and pair in the orbitals and then start entering the orbitals.
Thus, the arrangement is .
Hence, only one electron remains unpaired.
So, the number of unpaired electrons is: Step 5: Find the number of unpaired electrons for (high spin).
In a high spin complex, the arrangement is .
In this case, four electrons remain unpaired.
So, the number of unpaired electrons is: Step 6: Add all the unpaired electrons.
Now, total number of unpaired electrons is: Step 7: State the final answer.
Hence, the total number of unpaired electrons is:
22
PYQ 2026
medium
chemistryID: jee-main
Molar conductivity and conductance of an electrolytic solution are 123.5 Scm /mole and 0.19 S respectively. Concentration of solution is % w/w, then find the value of x.
(cell constant = 1.3 S cm ; density of solution = 1 gm/ml; molar mass of electrolyte = 75 g/mole)
Official Solution
Correct Option: (1)
Step 1: Write the relation for conductance.
The conductance ( ) of an electrolytic solution is given by the relation:
Where:
is the molar conductivity, is the length, and is the area of the electrode. Step 2: Define the relationship for molar conductivity.
Molar conductivity ( ) is defined as the conductance of the solution for a given concentration:
Where is the volume of the solution and is the number of moles of the solute. Step 3: Relate given data.
We are given:
,
,
cell constant = ,
density = ,
molar mass = . Step 4: Use the formula for concentration.
We use the formula for mass fraction concentration:
Where is the mass of solute, is the density, and is the volume of the solution. Step 5: Calculate the value of x.
By substituting the known values and solving, we get:
23
PYQ 2026
medium
chemistryID: jee-main
Silver rod is dipped in aqueous solution of of unknown concentration and Zn rod is dipped in aqueous solution. Both these containers are connected to form a galvanic cell showing emf of . Calculate the value of .
[Use ]
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Write the cell reaction and calculate standard emf. The given half-cells are: Since silver has a higher reduction potential, it acts as the cathode and zinc acts as the anode. Therefore, the overall cell reaction is: Now, the standard emf of the cell is: Step 2: Apply the Nernst equation. For the reaction
the number of electrons transferred is . The reaction quotient is:
Given that , we get:
Now use the Nernst equation:
Step 3: Simplify the logarithmic term. We know that:
Substituting this into the equation:
Now subtract from both sides:
Step 4: Find the value of . Hence,
So, the correct option is . Final Answer: .
24
PYQ 2026
medium
chemistryID: jee-main
For the given reaction: The following data table is provided at : & & & 80 & 140 & 8 & 250 & 42 & 200 \endtabular Calculate at .
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Write the formula for Gibbs free energy change. For a reaction, Gibbs free energy change is calculated by the relation: So, we first need to calculate the reaction enthalpy change and the reaction entropy change , and then substitute . Step 2: Calculate the enthalpy change of reaction . Given reaction is: Using the relation: Substituting the given values: Thus, the reaction enthalpy change is: Step 3: Calculate the entropy change of reaction . Using the relation: Substituting the given values: Thus, Step 4: Convert entropy term into proper units. Since is in , the term must also be converted into . First calculate: Now convert joules to kilojoules: So, Step 5: Calculate . Now use: Step 6: Compare with the options. • (A) : Correct. This matches the calculated value. • (B) : Incorrect. This is much larger in magnitude than the actual value. • (C) : Incorrect. This ignores the proper contribution of the entropy term. • (D) : Incorrect. The sign is wrong because the overall Gibbs free energy change is negative.
Step 7: Conclusion. Therefore, the Gibbs free energy change for the reaction at is: Final Answer:
25
PYQ 2026
medium
chemistryID: jee-main
For the reaction: Initial moles of A(g) is . At equilibrium, moles of A decompose at total pressure . Calculate for the given reaction.
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Write the equilibrium changes in moles. The reaction is: Initially, moles of A , and moles of B and C are zero. If moles of A decompose, then at equilibrium: So, the total number of moles at equilibrium is: Step 2: Find the partial pressures of all gases. If the total pressure at equilibrium is , then partial pressure of each gas is given by: Therefore, Step 3: Write the expression for . For the reaction the equilibrium constant in terms of pressure is: Substituting the partial pressures: Step 4: Simplify the expression. Cancelling one and one : Now use the identity: Hence, Step 5: Compare with the given options. • (A) : Correct. This matches the derived expression. • (B) : Incorrect. The denominator should come from , which gives , not . • (C) : Incorrect. The numerator should contain , not . • (D) : Incorrect. One power of is missing in the numerator.
Step 6: Conclusion. Therefore, the correct value of for the given reaction is: Final Answer:
26
PYQ 2026
medium
chemistryID: jee-main
On heating of , some solid residue is left behind. Determine the mass of residue left.
1
2
3
4
Official Solution
Correct Option: (1)
Concept: When silver carbonate is heated, it decomposes to form solid silver along with gaseous carbon dioxide and oxygen. The residue left after heating will therefore be metallic silver. Using stoichiometry and molar mass relations, we can determine the amount of residue formed. Step 1: Calculate moles of . Molar mass of : Step 2: Use stoichiometry of the reaction. From the balanced equation: Therefore, Step 3: Calculate mass of residue (Ag). Molar mass of Ag Thus, the mass of solid residue left after heating is:
27
PYQ 2026
medium
chemistryID: jee-main
If solubility of sparingly soluble salt is gm/litre and is the molar mass (in gm/mole) of the salt, then determine the value of .
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For sparingly soluble salts, solubility is first converted into molar solubility. Using the dissociation equation, concentrations of ions are written in terms of solubility and substituted into the solubility product expression. Step 1: Write the dissociation reaction. Step 2: Convert solubility into molarity. Given solubility g/L Step 3: Write the expression. Step 4: Calculate required ratio.
28
PYQ 2026
medium
chemistryID: jee-main
For reaction Calculate (in Kelvin) based on above data. (Nearest integer)
Official Solution
Correct Option: (1)
Concept: The relation between equilibrium constant and temperature is given by the Van’t Hoff equation: Step 1: Substitute given values. Using first two data points: Step 2: Simplify the equation.
29
PYQ 2026
medium
chemistryID: jee-main
Three solutions (A, B and C) are prepared according to given diagrams If pH of solutions A, B and C are respectively , and , then correct option will be:
1
2
3
4
Official Solution
Correct Option: (1)
Concept: pH comparison can be determined by calculating milli-equivalents of acids and bases. The nature of the resulting solution (acidic, basic, or neutral) depends on which component is in excess. Step 1: Solution A analysis. Milli-equivalent of HCl: Milli-equivalent of : Since base acid, the solution is basic. Step 2: Solution B analysis. Milli-equivalent of : Milli-equivalent of : Acid = Base, so the solution is neutral. Step 3: Solution C analysis. Milli-equivalent of : Milli-equivalent of NaOH: Acid Base, so the solution is acidic. Step 4: Final comparison.
30
PYQ 2026
medium
chemistryID: jee-main
Half life of first order reaction is min. What is the time required (in min.) to complete of reaction?
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For a first order reaction: and Step 1: Calculate the rate constant. Step 2: Determine concentration ratio for completion. If reaction is completed: Step 3: Calculate required time.
31
PYQ 2026
medium
chemistryID: jee-main
Determine mole fraction of water in an aqueous solution of urea having w/w urea.
1
2
3
4
Official Solution
Correct Option: (1)
Concept: Weight percent composition can be converted to mole fraction by first determining the moles of each component using their molar masses and then applying the mole fraction formula: Step 1: Assume of solution.} Since the solution contains w/w urea: Step 2: Calculate moles of each component. Molar mass of urea Molar mass of water Step 3: Find mole fraction of water.
32
PYQ 2026
medium
chemistryID: jee-main
For a first order reaction : If initial pressure is and total pressure at time is , then the expression of rate constant is:
1
2
3
4
Official Solution
Correct Option: (2)
Concept: For a first-order reaction: When pressure is used instead of concentration, the same formula applies since pressure is proportional to concentration. Step 1: Set up the reaction and pressures. At : At time : Step 2: Express total pressure. Step 3: Apply first-order rate law. Substitute :
33
PYQ 2026
easy
chemistryID: jee-main
of an organic compound is mixed with of and ignited.
It gives of gaseous mixture at NTP.
After passing through KOH, of gas remains.
The organic compound is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: After combustion, total gaseous mixture . After passing through KOH (which absorbs ):
Step 2: Oxygen consumed:
Step 3: Let the compound be . From combustion:
Step 4: Oxygen required per mole:
Step 5: Hence the compound is:
34
PYQ 2026
easy
chemistryID: jee-main
Consider the following reaction:If of calcium reacts with excess , choose the {incorrect} option.
1
Mass of produced is
2
Moles of produced is
3
Volume of produced at STP is
4
Mass of produced is
Official Solution
Correct Option: (4)
Step 1: Calculate moles of calcium: Step 2: From the balanced equation: So, Step 3: Mass of : Step 4: Volume of hydrogen gas at STP: Step 5: Hence, option (D) is incorrect.
35
PYQ 2026
hard
chemistryID: jee-main
of is dissolved in of a solvent such that . When of is dissolved in of the same solvent, . Find the molar mass of . Given . Both and are non-electrolytes.
Official Solution
Correct Option: (1)
Given:
for 1g of in 50g of solvent.
for 1g of in 50g of solvent.
(Cryoscopic constant).
Both and are non-electrolytes.
Step-by-Step Solution:
1. Formula for Freezing Point Depression:
The freezing point depression is given by: Where: - is the freezing point depression, - is the van't Hoff factor (number of particles the solute dissociates into), - is the cryoscopic constant (5 K kg/mol), - is the molality of the solution.
2. Molarity and Van't Hoff Factor:
For , the van't Hoff factor (since dissociates into 3 ions). Thus, the equation for becomes:
For , the van't Hoff factor (since dissociates into 2 ions). The equation for becomes:
3. Solving for Molar Mass of :
For :
Solving for the molar mass of , we get:
Final Answer: The molar mass of is approximately 145 g/mol.
36
PYQ 2026
medium
chemistryID: jee-main
25 ml, 0.1 M Ba(OH) react completely with HCl. Find the weight of HCl (in milligram) required?
Official Solution
Correct Option: (1)
Step 1: Write the chemical equation. Ba(OH) (aq) + 2HCl (aq) → BaCl (aq) + 2H O (l)
Step 2: Find the number of moles of Ba(OH) . Moles of Ba(OH) = 0.1 M × 0.025 L = 2.5 × 10 moles
Step 3: Calculate moles of HCl required. From the equation, 1 mole of Ba(OH) reacts with 2 moles of HCl. So, moles of HCl = 2 × 2.5 × 10 = 5 × 10 moles.
Step 4: Calculate the mass of HCl. Molar mass of HCl = 36.5 g/mol Mass of HCl = moles × molar mass = 5 × 10 × 36.5 = 182.5 milligrams.
Step 5: Conclusion. The required weight of HCl is 182.5 milligrams.
