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…

Year	2026	2025	2024	Total
Questions	8	15	7	30

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 6

Q jee_main_2024_29_january_evening Volumetric Titration and Molarity

50 mL

0.5 M

oxalic acid is required to neutralise

25 mL

NaOH

solution, the amount of

NaOH

50 mL

of given

NaOH

solution is ________ g.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Equivalents of Acid = Equivalents of Base

N₁ V₁ = N₂ V₂ (M₁ \times n₁) \times V₁ = (M₂ \times n₂) \times V₂

Core Logic

For oxalic acid ( $H₂C₂O₄$ $\text{H}_2\text{C}_2\text{O}_4$), the valence factor ( $n$ $n$-factor) is $2$ $2$. For $NaOH$ $\text{NaOH}$, the $n$ $n$-factor is $1$ $1$. Substituting the values into the normality equivalence expression:

50 \times 0.5 \times 2 = 25 \times MNaOH \times 1

50 = 25 \times MNaOH MNaOH = 2 M

Step 1: Mass Isolation

To find the mass of $NaOH$ $\text{NaOH}$ present in $50 mL$ $50\text{ mL}$ of this solution:

Mass = Molarity \times Volume (in L) \times Molar Mass

Mass = 2 \times ((50)/(1000)) \times 40 = 2 \times 0.05 \times 40 = 4 g

Pattern Recognition

Remember to use the correct $n$ $n$-factor ( $2$ $2$) for dibasic oxalic acid during equivalence matching to avoid calculation errors.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q83 jee_main_2024_27_jan_morning Stoichiometry

Mass of methane required to produce

22 g

CO₂

after complete combustion is

g. (Given Molar mass in

g mol⁻¹

C=12.0

H=1.0

O=16.0

)

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Balanced combustion chemical equation:

CH₄ + 2O₂ arrow CO₂ + 2H₂O

Moles = MassMolar Mass

Step 1: Determine moles of product generated

Molar Mass of CO₂ = 12 + (2 \times 16) = 44 g mol⁻¹

Moles of CO₂ produced = (22)/(44) = 0.5 moles

Step 2: Relate to input mass via stoichiometry metrics

From the balanced equation, $1 mole of CH₄$ $1\text{ mole of CH}_4$ produces $1 mole of CO₂$ $1\text{ mole of CO}_2$.

Required Moles of CH₄ = 0.5 moles

Molar Mass of CH₄ = 12 + (4 \times 1) = 16 g mol⁻¹

Mass of CH₄ = 0.5 \times 16 = 8 g

Pattern Recognition

$22 g of CO₂$ $22\text{ g of CO}_2$ is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to $8 g$ $8\text{ g}$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q88 jee_main_2024_30_jan_morning Mole Concept

0.05 cm

thick coating of silver is deposited on a plate of

0.05 m²

area. The number of silver atoms deposited on plate are ________

\times 10²³

. (At mass

Ag=108, d=7.9 g cm⁻³

)

Numerical Answer. Answer: 11 to 11

Solution

Related Formula

Volume = Area \times Thickness

Mass = Density \times Volume

Moles = MassMolar Mass

Number of Atoms = Moles \times NA

Step 1: Calculate Volume of Coating

Area = $0.05 m² = 0.05 \times 10⁴ cm² = 500 cm²$ Thickness = $0.05 cm$ $0.05 \text{ cm}$ Volume = $500 cm² \times 0.05 cm = 25 cm³$

Step 2: Calculate Mass and Moles

Mass = Volume $\times$ $\times$ Density = $25 cm³ \times 7.9 g/cm³ = 197.5 g$ Moles of Ag = $(197.5)/(108) = 1.8287 moles$

Step 3: Calculate Number of Atoms

Number of Atoms = 1.8287 \times 6.022 \times 10²³

= 11.01 \times 10²³

Rounding to nearest integer, we get $11$ $11$.

Chapter Mix

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

Q62 jee_main_2024_31_jan_evening Stoichiometry and Calculations

A sample of

CaCO₃

and

MgCO₃

weighed

2.21 g

is ignited to constant weight of

1.152 g

. The composition of mixture is: (Given molar mass in

g mol⁻¹

CaCO₃:100, MgCO₃:84

)

A. $1.187~g~CaCO₃ + 1.023~g~MgCO₃$
B. $1.023~g~CaCO₃ + 1.023~g~MgCO₃$
C. $1.187~g~CaCO₃ + 1.187~g~MgCO₃$
D. $1.023~g~CaCO₃ + 1.187~g~MgCO₃$

Solution

Related Formula

CaCO₃(s) Δ CaO(s) + CO₂(g)

MgCO₃(s) Δ MgO(s) + CO₂(g)

Core Logic

Let the weight of $CaCO₃$ $\mathrm{CaCO}_3$ be $x g$ $x\text{ g}$. Then, the weight of $MgCO₃ = (2.21 - x) g$ $\mathrm{MgCO}_3 = (2.21 - x)\text{ g}$.

Moles of $CaCO₃$ $\mathrm{CaCO}_3$ decomposed = Moles of $CaO$ $\mathrm{CaO}$ formed. $(x)/(100) = Moles of CaO formed$ $Weight of CaO formed = (x)/(100) \times 56$

Moles of $MgCO₃$ $\mathrm{MgCO}_3$ decomposed = Moles of $MgO$ $\mathrm{MgO}$ formed. $((2.21 - x))/(84) = Moles of MgO formed$ $Weight of MgO formed = (2.21 - x)/(84) \times 40$

Step 1: Setting up the Equation

The total weight of the residue ( $CaO$ $\mathrm{CaO}$ + $MgO$ $\mathrm{MgO}$) is given as $1.152 g$ $1.152\text{ g}$.

(2.21 - x)/(84) \times 40 + (x)/(100) \times 56 = 1.152

Step 2: Solving for x

(88.4 - 40x)/(84) + 0.56x = 1.152

1.0523 - 0.4761x + 0.56x = 1.152

0.0839x = 0.0997

x = 1.188 g

So, weight of $CaCO₃ \approx 1.187 g$ (accounting for rounding) Weight of $MgCO₃ = 2.21 - 1.188 = 1.022 g \approx 1.023 g$ .

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q83 jee_main_2024_31_jan_morning Stoichiometry

Number of moles of methane required to produce 22g

CO2(g)

after combustion is

x \times 10⁻²

moles. The value of x is

Numerical Answer. Answer: 50 to 50

Solution

Step 1: Stoichiometric Equation

CH4(g) + 2O2(g) arrow CO2(g) + 2H₂O(l)

1 mole of $CH₄$ $CH_4$ produces 1 mole of $CO₂$ $CO_2$.

Step 2: Moles Calculation

Molar mass of $CO₂ = 12 + 2(16) = 44 g/mol$ $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$

nCO₂ = MassMolar mass = (22)/(44) = 0.5 moles

Since 1 mole of $CH₄$ $CH_4$ produces 1 mole of $CO₂$ $CO_2$, the moles of $CH₄$ $CH_4$ required is $0.5$ $0.5$ moles.

Step 3: Finding x

0.5 moles = 50 \times 10⁻² moles

$x = 50$ $x = 50$

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculating Equivalents

Step 2: Calculating Mass

Step 3: Matching Format

Pattern Recognition

Chapter Mix

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Mass Isolation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Determine moles of product generated

Step 2: Relate to input mass via stoichiometry metrics

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate Volume of Coating

Step 2: Calculate Mass and Moles

Step 3: Calculate Number of Atoms

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Equation

Step 2: Solving for x

Chapter Mix

Solution

Step 1: Stoichiometric Equation

Step 2: Moles Calculation

Step 3: Finding x

Chapter Mix

More Some Basic Concepts of Chemistry Questions — jee_main_2026_23_january_morning

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