A 20 mL sample of a sodium iodide solution yields 4.74 g of silver iodide precipitate when treated with an excess of…

Year	2026	2025	2024	Total
Questions	8	15	7	30

Q65 jee_main_2026_22_january_evening Primary Standards in Volumetric Analysis

Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below:

A. A, B, C and E only
B. A, B, and C only
C. A, B and E only
D. D and E only

Solution

Related Formula

Primary Standard Criteria: High molar mass, stable in air, completely soluble, stoichiometric reaction.

Core Logic

Statement A: TRUE - Certain stable hydrated salts (e.g. oxalic acid dihydrate) are used as primary standards.

Statement B: TRUE - Primary standards must be stable in air and not hygroscopic or oxidized by air.

Statement C: TRUE - Reactions must be rapid, complete, and strictly stoichiometric.

Statement D: FALSE - Primary standards MUST be highly soluble in water to prepare volumetric standard solutions.

Statement E: FALSE - Primary standards should ideally have high relative molar mass to minimize weighing errors.

Hence, Statements A, B, and C are correct.

Pattern Recognition

Sees: Primary standard characteristics. Shortcut: Soluble in water (eliminates D) and high molar mass required (eliminates E).

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q73 jee_main_2026_23_january_morning Acid-Base Titration and Stoichiometry

x

mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1M

Ba(OH)₂

solution is used when the HCl solution was titrated against it. The numerical value of

x

is ____

\times 10⁻¹

(Nearest integer) Given : Molar mass of HCl and

Ba(OH)₂

are 36.5 and

171.0 g mol⁻¹

respectively.

Numerical Answer. Answer: 1825 to 1825

Solution

Related Formula

N₁ V₁ = N₂ V₂

Equivalents of Acid = Equivalents of Base

Core Logic

Write the balanced chemical equation to find the molar ratio, or use the concept of equivalents directly. Barium hydroxide is a diacidic base ( $n$ $n$-factor = 2) and hydrochloric acid is a monobasic acid ( $n$ $n$-factor = 1).

$Ba(OH)2(aq) + 2HCl(aq) arrow BaCl2(aq) + 2H₂O(l)$

Step 1: Calculating Equivalents

Millimoles of $Ba(OH)₂ = M \times V = 0.1 M \times 25.0 mL = 2.5 mmoles$ . Since 1 mole of $Ba(OH)₂$ $Ba(OH)_2$ neutralizes 2 moles of $HCl$ $HCl$: Millimoles of $HCl = 2 \times 2.5 = 5.0 mmoles$ .

Step 2: Calculating Mass

Weight of $HCl = mmoles \times molar mass (mg/mmol)$ Weight = $5.0 \times 36.5 = 182.5 mg$ $5.0 \times 36.5 = 182.5 \text{ mg}$.

Step 3: Matching Format

We need the answer in the form of $x \times 10⁻¹$ $x \times 10^{-1}$ mg. $182.5 mg = 1825 \times 10⁻¹ mg$ . Thus, $x = 1825$ $x = 1825$.

Acid-Base Titration and Stoichiometry diagram for Q73 - JEE Main 2026 Morning

Pattern Recognition

Always multiply the Molarity by the n-factor to get Normality when performing titrations. $0.1 M Ba(OH)₂ = 0.2 N$ . Equivalents = $0.2 \times 25 = 5 meq$ $0.2 \times 25 = 5 \text{ meq}$. Mass = $5 \times 36.5 = 182.5 mg$ $5 \times 36.5 = 182.5 \text{ mg}$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry Class 12 Chemistry: Solutions

Q68 jee_main_2026_28_january_evening Stoichiometry And Limiting Reagent

For the given reaction;

CaCO₃ + 2HCl arrow CaCl₂ + H₂O + CO₂

If

90 g CaCO₃

is added to

300 mL

of HCl which contains

38.55% HCl

by mass and has density

1.13 g mL⁻¹

then which of the following option is correct? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and

16 g mol⁻¹

respectively.

A. $(1) 64.97 g of HCl remains unreacted$
B. $(2) 32.85 g of CaCO₃ remains unreacted$
C. $(3) 97.30 g of HCl reacted$ $(3)\text{ 97.30 g of HCl reacted}$
D. $(4) 60.32 g of HCl reamains unreacted$

Solution

Core Logic

Step 1: Calculate moles of HCl available. Density of solution ( $d$ $d$) $= 1.13 g/mL$ $= 1.13\text{ g/mL}$. Volume ( $V$ $V$) $= 300 mL$ $= 300\text{ mL}$. Mass of solution $= 300 \times 1.13 = 339 g$ $= 300 \times 1.13 = 339\text{ g}$. Mass of pure HCl $= 339 \times (38.55)/(100) = 130.68 g$ . Molar mass of HCl $= 36.5 g/mol$ $= 36.5\text{ g/mol}$. Moles of HCl initially $= (130.68)/(36.5) = 3.58 moles$ .

Step 2: Calculate moles of $CaCO₃$ $CaCO_3$ available. Molar mass of $CaCO₃ = 100 g/mol$ $CaCO_3 = 100\text{ g/mol}$. Moles of $CaCO₃ = (90)/(100) = 0.90 mole$ .

Step 1: Determine Limiting Reagent and Reaction

Reaction: $CaCO₃ + 2HCl arrow CaCl₂ + H₂O + CO₂$ From the stoichiometry, 1 mole of $CaCO₃$ $CaCO_3$ requires 2 moles of $HCl$ $HCl$. $0.90 mole$ $0.90\text{ mole}$ of $CaCO₃$ $CaCO_3$ requires $0.90 \times 2 = 1.80 moles$ $0.90 \times 2 = 1.80\text{ moles}$ of $HCl$ $HCl$. Since 3.58 moles of $HCl$ $HCl$ are available, $CaCO₃$ $CaCO_3$ is the limiting reagent (LR) and $HCl$ $HCl$ is in excess.

Moles of $HCl$ $HCl$ remained unreacted $= 3.58 - 1.80 = 1.78 moles$ $= 3.58 - 1.80 = 1.78\text{ moles}$. Mass of $HCl$ $HCl$ remained $= 1.78 \times 36.5 = 64.97 g$ $= 1.78 \times 36.5 = 64.97\text{ g}$.

Pattern Recognition

Always convert volume, density, and mass percentage into pure mass. Identify Limiting Reagent before finding the leftover.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q33 jee_main_2025_02_april_morning Stoichiometry and Limiting Reagent

CaCO₃(s) + 2HCl(aq) arrow CaCl₂(aq) + CO₂(g) + H₂O(l)

Consider the above reaction, what mass of

CaCl₂

will be formed if

250~mL

of

0.76~M

HCl reacts with

1000~g

of

CaCO₃

? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and

35.5~g \cdot mol⁻¹

, respectively)

A. $(1) 3.908~g$ $(1)\ 3.908\mathrm{~g}$
B. $(2) 2.636~g$ $(2)\ 2.636\mathrm{~g}$
C. $(3) 10.545~g$ $(3)\ 10.545\mathrm{~g}$
D. $(4) 5.272~g$ $(4)\ 5.272\mathrm{~g}$

Solution

Related Formula

Molarity conversion relation matrix:

Moles = Molarity (M) \times Volume (L)

Mass = Moles \times Molar Mass

Core Logic

Let's perform molar quantities verification row-by-row:

Molar mass properties: $CaCO₃ = 100~g/mol$ $\mathrm{CaCO_3} = 100\mathrm{~g/mol}$, $CaCl₂ = 40 + (35.5 \times 2) = 111~g/mol$ .
Initial chemical moles calculated:

Moles of CaCO₃ = (1000)/(100) = 10~mol

Moles of HCl = 0.76 \times (250)/(1000) = 0.19~mol

Determine the limiting reactant via stoichiometric ratios: $HCl$ $\mathrm{HCl}$ acts as the Limiting Reagent (L.R.) because its proportional structural requirement is much smaller.
Moles of product $CaCl₂$ $\mathrm{CaCl_2}$ formed based on L.R. configuration:

Moles of CaCl₂ = (0.19)/(2) = 0.095~mol

Mass of CaCl₂ = 0.095 \times 111 = 10.545~g

Pattern Recognition

Always compare the available moles divided by the respective stoichiometric coefficients to quickly find the Limiting Reagent: $10/1 gg 0.19/2$ $10/1 \gg 0.19/2$. This trick saves execution seconds during complex numeric problems.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q27 jee_main_2025_03_april_evening Molarity and Stoichiometry of Neutralization

10~mL

of

2~M~NaOH

solution is added to

20~mL

of

1~M~HCl

solution kept in a beaker. Now,

10~mL

of this mixture is poured into a volumetric flask of

100~mL

containing 2 moles of

HCl

and made the volume upto the mark with distilled water. The solution in this flask is :

A. $0.2~M~NaCl$ $0.2\mathrm{~M~NaCl}$ solution
B. $20~M~HCl$ $20\mathrm{~M~HCl}$ solution
C. $10~M~HCl$ $10\mathrm{~M~HCl}$ solution
D. Neutral solution

Solution

Related Formula

Number of millimoles ( $n$ $n$) is given by:

n = M \times VmL

Molarity ( $M$ $M$) of a diluted mixture:

M = Total molesTotal Volume in Liters

Core Logic

Evaluate the first mixing step to determine the net acid-base state:

Millimoles of $NaOH = 10~mL \times 2~M = 20~mmol$
Millimoles of $HCl = 20~mL \times 1~M = 20~mmol$

Since millimoles are equal, $HCl$ $\mathrm{HCl}$ and $NaOH$ $\mathrm{NaOH}$ completely neutralize each other, producing a neutral aqueous salt solution.

Step 1: Analyze transfer to volumetric flask

Taking $10~mL$ $10\mathrm{~mL}$ of this neutralized solution provides no excess $H^+$ $\mathrm{H}^+$ or $OH^-$ $\mathrm{OH}^-$ ions. It is added to a volumetric flask containing $2~mol$ $2\mathrm{~mol}$ of pure $HCl$ $\mathrm{HCl}$.

Step 2: Calculate final molarity of $HCl$ $\mathrm{HCl}$

The volume of the flask is made up to $100~mL = 0.1~L$ $100\mathrm{~mL} = 0.1\mathrm{~L}$:

M = 2~mol0.1~L = 20~M

Hence, the resulting solution is $20~M~HCl$ $20\mathrm{~M~HCl}$.

Pattern Recognition

Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence ( $MVacid = MVbase$ $MV_{\text{acid}} = MV_{\text{base}}$) is reached, any sub-aliquot of that solution remains completely neutral.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Chemistry: Ionic Equilibrium

Solution & Explanation

Related Formula

Execution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Equivalents

Step 2: Calculating Mass

Step 3: Matching Format

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Determine Limiting Reagent and Reaction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyze transfer to volumetric flask

Step 2: Calculate final molarity of HCl$\mathrm{HCl}$

Pattern Recognition

Chapter Mix

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Step 2: Calculate final molarity of $HCl$ $\mathrm{HCl}$