Some Basic Concepts of Chemistry Previous Year Questions — JEE Main Chemistry
30 past-year Some Basic Concepts of Chemistry questions from JEE Main (Chemistry) — page 6.
Q56 (2026)
In the reaction,
$$2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^{-}(aq) + 3H_{2}(g)$$
- $\text{11.2 L } H_{2}(g) \text{ at STP is produced for every mole of HCl consumed.}$
- $\text{67.2 L } H_{2}(g) \text{ at STP is produced for every mole of Al that reacts.}$
- $\text{12 L HCl(aq) is consumed for every 6L } H_{2}(g) \text{ produced.}$
- $\text{33.6 L } H_{2}(g) \text{ is produced regardless of temperature and pressure for every mole of Al that reacts.}$
### Related Formula
$$\text{Volume of gas at STP} = \text{Moles} \times 22.4 \text{ L}$$
### Core Logic
From the balanced chemical equation:
$$2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^{-}(aq) + 3H_{2}(g)$$
6 moles of HCl produce 3 moles of $H_2$.
Therefore, 1 mole of HCl produces $\frac{3}{6} = 0.5$ moles of $H_2$.
Volume of $H_2$ produced at STP for 1 mole of HCl:
$$V = 0.5 \times 22.4\text{ L} = 11.2\text{ L}$$
### Step 1: Check other options
Option (2): 2 moles of Al produce 3 moles of $H_2$. 1 mole Al produces 1.5 moles $H_2 = 1.5 \times 22.4 = 33.6\text{ L}$ (Incorrect).
Option (4) is incorrect because volume depends on temperature and pressure.
### Pattern Recognition
Standard stoichiometry. Directly map the mole ratio from the balanced equation to molar volume at standard conditions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q53 (2026)
$\text{A} + 2\text{B} \rightarrow \text{AB}_2$
$36.0\text{ g}$ of 'A' (Molar mass: $60\text{ g mol}^{-1}$) and $56.0\text{ g}$ of 'B' (Molar mass: $80\text{ g mol}^{-1}$) are allowed to react. Which of the following statements are correct?
(A) 'A' is the limiting reagent
(B) $77.0\text{ g}$ of $\text{AB}_2$ is formed
(C) Molar mass of $\text{AB}_2$ is $140\text{ g mol}^{-1}$
(D) $15.0\text{ g}$ of A is left unreacted after the completion of reaction.
Choose the correct answer from the options given below:
- C and D only
- A and C only
- B and D only
- A and B only
### Related Formula
$$\text{Moles } (n) = \frac{\text{Given Mass}}{\text{Molar Mass}}$$
$$\text{Molar Mass of } \text{AB}_2 = M_A + 2 M_B$$
### Core Logic
Step 1: Calculate initial moles:
$$n_A = \frac{36}{60} = 0.6\text{ mol}$$
$$n_B = \frac{56}{80} = 0.7\text{ mol}$$
Step 2: Identify Limiting Reagent (LR):
$$\text{Ratio for A} = \frac{0.6}{1} = 0.6, \quad \text{Ratio for B} = \frac{0.7}{2} = 0.35$$
Since ratio of B is smaller, B is the limiting reagent.
Step 3: Evaluate product formed and remaining reactant:
$$\text{Molar mass of } \text{AB}_2 = 60 + 2(80) = 220\text{ g mol}^{-1}$$
$$\text{Moles of } \text{AB}_2 \text{ formed} = 0.35\text{ mol}$$
$$\text{Mass of } \text{AB}_2 \text{ formed} = 0.35 \times 220 = 77.0\text{ g} \quad \text{(Statement B is correct)}$$
$$\text{Moles of A reacted} = 0.35\text{ mol}$$
$$\text{Moles of A remaining} = 0.6 - 0.35 = 0.25\text{ mol}$$
$$\text{Mass of A remaining} = 0.25 \times 60 = 15.0\text{ g} \quad \text{(Statement D is correct)}$$
### Pattern Recognition
Sees: Initial masses of reactants in stoichiometric equation.
Shortcut: Compare $n/\text{coefficient}$ to find LR, then compute formed product mass and unreacted mass directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q65 (2026)
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, B, C and E only
- A, B, and C only
- A, B and E only
- D and E only
### Related Formula
$$\text{Primary Standard Criteria: High molar mass, stable in air, completely soluble, stoichiometric reaction.}$$
### 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).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q73 (2026)
$x$ 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 $171.0 \text{ g mol}^{-1}$ respectively.
### Related Formula
$$N_1 V_1 = N_2 V_2$$
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 ($n$-factor = 2) and hydrochloric acid is a monobasic acid ($n$-factor = 1).
$Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_2O_{(l)}$
### Step 1: Calculating Equivalents
Millimoles of $Ba(OH)_2 = M \times V = 0.1 \text{ M} \times 25.0 \text{ mL} = 2.5 \text{ mmoles}$.
Since 1 mole of $Ba(OH)_2$ neutralizes 2 moles of $HCl$:
Millimoles of $HCl = 2 \times 2.5 = 5.0 \text{ mmoles}$.
### Step 2: Calculating Mass
Weight of $HCl = \text{mmoles} \times \text{molar mass (mg/mmol)}$
Weight = $5.0 \times 36.5 = 182.5 \text{ mg}$.
### Step 3: Matching Format
We need the answer in the form of $x \times 10^{-1}$ mg.
$182.5 \text{ mg} = 1825 \times 10^{-1} \text{ mg}$.
Thus, $x = 1825$.
{{SOL_IMG1}}
### Pattern Recognition
Always multiply the Molarity by the n-factor to get Normality when performing titrations. $0.1 \text{ M } Ba(OH)_2 = 0.2 \text{ N}$. Equivalents = $0.2 \times 25 = 5 \text{ meq}$. Mass = $5 \times 36.5 = 182.5 \text{ mg}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Class 12 Chemistry: Solutions
Q68 (2026)
For the given reaction;
$$CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$$
If $90\text{ g CaCO}_{3}$ is added to $300\text{ mL}$ of HCl which contains $38.55\%\text{ HCl}$ by mass and has density $1.13\text{ g mL}^{-1}$ then which of the following option is correct?
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and $16\text{ g mol}^{-1}$ respectively.
- $(1)\text{ 64.97 g of HCl remains unreacted}$
- $(2)\text{ 32.85 g of CaCO}_3\text{ remains unreacted}$
- $(3)\text{ 97.30 g of HCl reacted}$
- $(4)\text{ 60.32 g of HCl reamains unreacted}$
### Core Logic
Step 1: Calculate moles of HCl available.
Density of solution ($d$) $= 1.13\text{ g/mL}$.
Volume ($V$) $= 300\text{ mL}$.
Mass of solution $= 300 \times 1.13 = 339\text{ g}$.
Mass of pure HCl $= 339 \times \frac{38.55}{100} = 130.68\text{ g}$.
Molar mass of HCl $= 36.5\text{ g/mol}$.
Moles of HCl initially $= \frac{130.68}{36.5} = 3.58\text{ moles}$.
Step 2: Calculate moles of $CaCO_3$ available.
Molar mass of $CaCO_3 = 100\text{ g/mol}$.
Moles of $CaCO_3 = \frac{90}{100} = 0.90\text{ mole}$.
### Step 1: Determine Limiting Reagent and Reaction
Reaction: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$
From the stoichiometry, 1 mole of $CaCO_3$ requires 2 moles of $HCl$.
$0.90\text{ mole}$ of $CaCO_3$ requires $0.90 \times 2 = 1.80\text{ moles}$ of $HCl$.
Since 3.58 moles of $HCl$ are available, $CaCO_3$ is the limiting reagent (LR) and $HCl$ is in excess.
Moles of $HCl$ remained unreacted $= 3.58 - 1.80 = 1.78\text{ moles}$.
Mass of $HCl$ remained $= 1.78 \times 36.5 = 64.97\text{ g}$.
### Pattern Recognition
Always convert volume, density, and mass percentage into pure mass. Identify Limiting Reagent before finding the leftover.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
