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

Q33 jee_main_2025_03_april_morning Mole Concept and Atomic Mass

Among

10⁻⁹~g

(each) of the following elements, which one will have the highest number of atoms? Element:

Pb

,

Po

,

Pr

and

Pt

A. $Po$ $\mathrm{Po}$
B. $Pr$ $\mathrm{Pr}$
C. $Pb$ $\mathrm{Pb}$
D. $Pt$ $\mathrm{Pt}$

Solution

Related Formula

Number of atoms = Mass (in g)Molar Mass (g/mol) \times NA

Core Logic

For equal mass ( $10⁻⁹~g$ $10^{-9}\mathrm{~g}$), the number of atoms is inversely proportional to the molar mass ( $M$ $M$):

Number of atoms \propto (1)/(M)

Step 1: Comparison of Molar Masses

Molar masses of given elements:

$MPo = 209~g/mol$ $M_{\mathrm{Po}} = 209\mathrm{~g/mol}$
$MPr = 141~g/mol$ $M_{\mathrm{Pr}} = 141\mathrm{~g/mol}$
$MPb = 207~g/mol$ $M_{\mathrm{Pb}} = 207\mathrm{~g/mol}$
$MPt = 195~g/mol$ $M_{\mathrm{Pt}} = 195\mathrm{~g/mol}$

Praseodymium ( $Pr$ $\mathrm{Pr}$) has the lowest molar mass ( $141~g/mol$ $141\mathrm{~g/mol}$), so it contains the maximum number of atoms.

Pattern Recognition

Same mass $arrow$ $\rightarrow$ lowest molar mass has highest number of atoms.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q47 jee_main_2025_04_april_evening Stoichiometry

The amount of calcium oxide produced on heating

150~kg

limestone (75% pure) is _________________ kg. (Nearest integer) Given : Molar mass (in g mol

⁻¹

) of Ca-40, O-16, C-12

Numerical Answer. Answer: 62.5 to 63.5

Solution

Related Formula

CaCO₃ Δ CaO + CO₂

Pure mass = Total mass \times Purity %100

Core Logic

Find the pure mass of calcium carbonate ( $CaCO₃$ $CaCO_3$) present:

Mass of CaCO₃ = 150 \times (75)/(100) = 112.5 ~kg = 112500 ~g

Convert this mass into moles (Molar mass of $CaCO₃ = 40+12+48 = 100 ~g\cdot mol⁻¹$ ):

moles of CaCO₃ = (112500)/(100) = 1125 moles

From the stoichiometry of the reaction, 1 mole of $CaCO₃$ $CaCO_3$ yields 1 mole of $CaO$ $CaO$:

moles of CaO = 1125 moles

Mass of CaO = 1125 \times 56 ~g = 63000 ~g = 63 ~kg

Pattern Recognition

Always multiply by the purity fraction first before entering regular stoichiometric conversion chains. Since the formula weight of limestone is exactly $100$ $100$, tracking percentages directly mirrors mole factors seamlessly.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q48 jee_main_2025_04_april_morning Concentration Terms

Fortification of food with iron is done using

FeSO₄\cdot 7H₂O

. The mass in grams of the

FeSO₄\cdot 7H₂O

required to achieve

12~ppm

of iron in

150~kg

of wheat is _______. (Nearest integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and

16 ~g~mol⁻¹

]

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

ppm = Mass of solute (g)Total mass of solution/mixture (g) \times 10⁶

Core Logic

Let the required mass of pure iron be $w~g$ $w\mathrm{~g}$. The total mass of the wheat mixture is $150~kg = 150 \times 10³~g$ . Applying the parts-per-million concentration condition:

12 = (w)/(150 \times 10³) \times 10⁶ 12 = w \times 6.666

w = (12 \times 150 \times 10³)/(10⁶) = 1.8~g of Iron

Now, determine the molar mass of the complete green vitriol salt crystal template, $FeSO₄ \cdot 7H₂O$ $\mathrm{FeSO_4 \cdot 7H_2O}$:

M = 56 + 32 + (4 \times 16) + (7 \times 18) = 56 + 32 + 64 + 126 = 278 ~g~mol⁻¹

Set up a stoichiometric mass balance proportion to find the total salt mass $w₁$ $w_1$:

Moles of Fe = (1.8)/(56) = (w₁)/(278)

w₁ = (1.8 \times 278)/(56) = (500.4)/(56) \approx 8.935~g

Rounding off to the nearest integer value gives 9.

Pattern Recognition

Always convert concentration metrics back to absolute molar mass equivalence values before distributing across full hydrated molecular templates.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q48 jee_main_2025_07_april_evening Stoichiometry and Limiting Reagent

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below: [cite: 461, 463]

C₄H10(g) + (13)/(2)O₂(g) arrow 4CO₂(g) + 5H₂O(l)

If

174.0 kg

of butane is mixed with

320.0 kg

of

O₂

, the volume of water formed in litres is

. (Nearest integer) [cite: 465, 466] [Given: (a) Molar mass of

C

,

H

,

O

are

12

,

1

,

16 g mol⁻¹

respectively, (b) Density of water

= 1 g mL⁻¹

] [cite: 467, 468]

Numerical Answer. Answer: 137.5 to 138.5

Solution

Related Formula

Moles (n) = Mass in gramsMolar Mass

Volume of water (V) = Mass of waterDensity of water

Core Logic

First, calculate initial molar quantities for both reactants:

Molar mass of $C₄H₁₀ = 4(12) + 10(1) = 58 g/mol$
Initial Moles of butane $= 174.0 \times 10³ g58 g/mol = 3000 mol = 3 \times 10³ mol$
Molar mass of $O₂ = 32 g/mol$ $\text{O}_2 = 32\text{ g/mol}$
Initial Moles of oxygen $= 320.0 \times 10³ g32 g/mol = 10000 mol = 10 \times 10³ mol$

Step 1: Identify Limiting Reagent

Let's test the stoichiometric requirements via calculation ratios:

Ratio for $C₄H₁₀ = (3000)/(1) = 3000$
Ratio for $O₂ = (10000)/(13/2) = 1538.46$

Since the ratio for $O₂$ $\text{O}_2$ is lower, oxygen behaves as the limiting reagent and commands the output steps.

Step 2: Compute Water Yield

Using stoichiometric proportions defined by the balanced reaction field:

Moles of H2O = 5 \times Moles of O213/2 = 5 \times (2)/(13) \times 10000 = (100000)/(13) mol

Convert moles to mass ( $MH2O = 18 g/mol$ ):

Mass of water = (100000)/(13) \times 18 = 138461.5 g \approx 138.46 kg

Since density $= 1 g/mL = 1 kg/L$ $= 1\text{ g/mL} = 1\text{ kg/L}$, the net volume is exactly:

Vwater = 138.46 Litres \approx 138 Litres

Pattern Recognition

Limiting reagent shortcut: Always check ratios (moles / stoichiometric coefficient) right away. Do not spend time calculating theoretical products based on butane before establishing whether oxygen runs out first.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q36 jee_main_2025_24_jan_evening Empirical and Molecular Formulae

The elemental composition of a compound is

54.2% C

,

9.2% H

and

36.6% O

. If the molar mass of the compound is

132 g mol⁻¹

, the molecular formula of the compound is: [Given : The relative atomic mass of

C : H : O = 12 : 1 : 16

]

A. $C₄H₉O₃$
B. $C₆H₁₂O₆$
C. $C₆H₁₂O₃$
D. $C₄H₈O₂$

Solution

Core Logic

Let's find the empirical formula by calculating the relative molar ratios of each element:

Carbon (C):

Moles = (54.2)/(12) = 4.516

Hydrogen (H):

Moles = (9.2)/(1) = 9.200

Oxygen (O):

Moles = (36.6)/(16) = 2.287

Next, divide each value by the smallest number of moles ( $2.287$ $2.287$) to get the simplest whole-number ratio:

C: $(4.516)/(2.287) \approx 1.97 \approx 2$
H: $(9.200)/(2.287) \approx 4.02 \approx 4$
O: $(2.287)/(2.287) = 1$ $\frac{2.287}{2.287} = 1$

Thus, the Empirical Formula is $C₂H₄O$ .

Step 2: Molecular Formula Determination

Calculate the empirical formula mass:

Empirical Mass = (2 \cdot 12) + (4 \cdot 1) + 16 = 24 + 4 + 16 = 44 g/mol

Find the multiplier ( $n$ $n$):

n = Molar MassEmpirical Mass = (132)/(44) = 3

Determine the molecular formula:

Molecular Formula = (C₂H₄O)₃ = C₆H₁₂O₃

This matches Option (3).

Pattern Recognition

Always determine the empirical formula ratio first. Then, compare the empirical mass with the given molar mass to find the whole-number multiplier ( $n$ $n$). Multiplication gives the definitive molecular formula.

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 4

Solution

Related Formula

Core Logic

Step 1: Comparison of Molar Masses

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identify Limiting Reagent

Step 2: Compute Water Yield

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 2: Molecular Formula Determination

Pattern Recognition

Chapter Mix

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