Some Basic Concepts of Chemistry Previous Year Questions — JEE Main Chemistry
30 past-year Some Basic Concepts of Chemistry questions from JEE Main (Chemistry).
Q33 (2025)
$$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$$
Consider the above reaction, what mass of $\mathrm{CaCl}_2$ will be formed if $250\mathrm{~mL}$ of $0.76\mathrm{~M}$ HCl reacts with $1000\mathrm{~g}$ of $\mathrm{CaCO_3}$?
(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and $35.5\mathrm{~g \cdot mol}^{-1}$, respectively)
- $(1)\ 3.908\mathrm{~g}$
- $(2)\ 2.636\mathrm{~g}$
- $(3)\ 10.545\mathrm{~g}$
- $(4)\ 5.272\mathrm{~g}$
### Related Formula
Molarity conversion relation matrix:
$$\text{Moles} = \text{Molarity (M)} \times \text{Volume (L)}$$
$$\text{Mass} = \text{Moles} \times \text{Molar Mass}$$
### Core Logic
Let's perform molar quantities verification row-by-row:
* Molar mass properties: $\mathrm{CaCO_3} = 100\mathrm{~g/mol}$, $\mathrm{CaCl_2} = 40 + (35.5 \times 2) = 111\mathrm{~g/mol}$.
* Initial chemical moles calculated:
$$\text{Moles of } \mathrm{CaCO_3} = \frac{1000}{100} = 10\mathrm{~mol}$$
$$\text{Moles of } \mathrm{HCl} = 0.76 \times \frac{250}{1000} = 0.19\mathrm{~mol}$$
* Determine the limiting reactant via stoichiometric ratios: $\mathrm{HCl}$ acts as the Limiting Reagent (L.R.) because its proportional structural requirement is much smaller.
* Moles of product $\mathrm{CaCl_2}$ formed based on L.R. configuration:
$$\text{Moles of } \mathrm{CaCl_2} = \frac{0.19}{2} = 0.095\mathrm{~mol}$$
$$\text{Mass of } \mathrm{CaCl_2} = 0.095 \times 111 = 10.545\mathrm{~g}$$
### Pattern Recognition
Always compare the available moles divided by the respective stoichiometric coefficients to quickly find the Limiting Reagent: $10/1 \gg 0.19/2$. This trick saves execution seconds during complex numeric problems.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q27 (2025)
$10\mathrm{~mL}$ of $2\mathrm{~M~NaOH}$ solution is added to $20\mathrm{~mL}$ of $1\mathrm{~M~HCl}$ solution kept in a beaker. Now, $10\mathrm{~mL}$ of this mixture is poured into a volumetric flask of $100\mathrm{~mL}$ containing 2 moles of $\mathrm{HCl}$ and made the volume upto the mark with distilled water. The solution in this flask is :
- $0.2\mathrm{~M~NaCl}$ solution
- $20\mathrm{~M~HCl}$ solution
- $10\mathrm{~M~HCl}$ solution
- Neutral solution
### Related Formula
Number of millimoles ($n$) is given by:
$$n = M \times V_{\mathrm{mL}}$$
Molarity ($M$) of a diluted mixture:
$$M = \frac{\text{Total moles}}{\text{Total Volume in Liters}}$$
### Core Logic
Evaluate the first mixing step to determine the net acid-base state:
- Millimoles of $\mathrm{NaOH} = 10\mathrm{~mL} \times 2\mathrm{~M} = 20\mathrm{~mmol}$
- Millimoles of $\mathrm{HCl} = 20\mathrm{~mL} \times 1\mathrm{~M} = 20\mathrm{~mmol}$
Since millimoles are equal, $\mathrm{HCl}$ and $\mathrm{NaOH}$ completely neutralize each other, producing a neutral aqueous salt solution.
### Step 1: Analyze transfer to volumetric flask
Taking $10\mathrm{~mL}$ of this neutralized solution provides no excess $\mathrm{H}^+$ or $\mathrm{OH}^-$ ions.
It is added to a volumetric flask containing $2\mathrm{~mol}$ of pure $\mathrm{HCl}$.
### Step 2: Calculate final molarity of $\mathrm{HCl}$
The volume of the flask is made up to $100\mathrm{~mL} = 0.1\mathrm{~L}$:
$$M = \frac{2\mathrm{~mol}}{0.1\mathrm{~L}} = 20\mathrm{~M}$$
Hence, the resulting solution is $20\mathrm{~M~HCl}$.
### Pattern Recognition
Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence ($MV_{\text{acid}} = MV_{\text{base}}$) is reached, any sub-aliquot of that solution remains completely neutral.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Class 11 Chemistry: Ionic Equilibrium
Q37 (2025)
Mass of magnesium required to produce $220\mathrm{~mL}$ of hydrogen gas at STP on reaction with excess of dil. HCl is :
Given: Molar mass of Mg is $24\mathrm{~g~mol}^{-1}$ .
- 235.7 g
- 0.24 mg
- 236 mg
- 2.444 g
### Related Formula
The balanced chemical equation for the displacement reaction is:
$$\mathrm{Mg(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl}_2\mathrm{(aq)} + \mathrm{H}_2\mathrm{(g)}$$
At STP, 1 mole of any ideal gas occupies a volume of $22.4\mathrm{~L} = 22400\mathrm{~mL}$.
### Core Logic
From the stoichiometry of the reaction:
- 1 mole of $\mathrm{Mg}$ ($24\mathrm{~g}$) produces 1 mole of $\mathrm{H}_2$ ($22400\mathrm{~mL}$ at STP).
### Step 1: Calculate moles of $\mathrm{H}_2$ gas produced
$$n_{\mathrm{H}_2} = \frac{220\mathrm{~mL}}{22400\mathrm{~mL/mol}} \approx 9.8214 \times 10^{-3}\mathrm{~mol}$$$
### Step 2: Calculate mass of Magnesium required
Since the molar ratio of $\mathrm{Mg}$ to $\mathrm{H}_2$ is $1:1$:
$$n_{\mathrm{Mg}} = 9.8214 \times 10^{-3}\mathrm{~mol}$$
$$\text{Mass of Mg} = 9.8214 \times 10^{-3}\mathrm{~mol} \times 24\mathrm{~g/mol}$$
$$\text{Mass of Mg} \approx 0.2357\mathrm{~g} = 235.7\mathrm{~mg} \approx 236\mathrm{~mg}$$
This matches Option (3).
### Pattern Recognition
Always keep a close eye on unit prefixes in options. A mass of $0.2357\mathrm{~g}$ corresponds to $235.7\mathrm{~mg}$, which rounds directly to $236\mathrm{~mg}$, whereas $235.7\mathrm{~g}$ is off by a factor of 1000.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q34 (2025)
At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively :
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given: Molar mass of $\mathrm{N}_2 = 28\text{ g mol}^{-1}$, $\mathrm{O}_2 = 32\text{ g mol}^{-1}$ and $\mathrm{Ar} = 40\text{ g mol}^{-1}$ respectively)
- 4.26, 19.3
- 2.59, 11.85
- 5.46, 17.8
- 2.96, 11.2
### Related Formula
$$P_i = X_i \cdot P_{\text{total}} = \frac{n_i}{n_{\text{total}}} \cdot P_{\text{total}}$$
Ratio of partial pressures:
$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$
### Core Logic
Assume a sample of dry air with total mass $= 100 \text{ g}$:
- Mass of $\mathrm{N}_2 = 70.0 \text{ g}$
- Mass of $\mathrm{O}_2 = 27.0 \text{ g}$
- Mass of $\mathrm{Ar} = 3.0 \text{ g}$
Now, convert masses to moles:
$$n_{\mathrm{N}_2} = \frac{70.0}{28} = 2.5 \text{ moles}$$
$$n_{\mathrm{O}_2} = \frac{27.0}{32} = 0.84375 \text{ moles}$$
$$n_{\mathrm{Ar}} = \frac{3.0}{40} = 0.075 \text{ moles}$$
Calculate ratios:
(i) Ratio of partial pressure of nitrogen to oxygen:
$$\frac{P_{\mathrm{N}_2}}{P_{\mathrm{O}_2}} = \frac{n_{\mathrm{N}_2}}{n_{\mathrm{O}_2}} = \frac{2.5}{0.84375} \approx 2.96$$
(ii) Ratio of partial pressure of oxygen to argon:
$$\frac{P_{\mathrm{O}_2}}{P_{\mathrm{Ar}}} = \frac{n_{\mathrm{O}_2}}{n_{\mathrm{Ar}}} = \frac{0.84375}{0.075} \approx 11.25 \approx 11.2$$
### Pattern Recognition
Since total pressure cancels out in a ratio of partial pressures, we only need to calculate the mole ratio directly from the given mass percentages divided by their respective molar masses.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Class 11 Physics: Kinetic Theory of Gases
Q48 (2025)
A $20 \text{ mL}$ sample of a sodium iodide solution yields $4.74 \text{ g}$ of silver iodide precipitate when treated with an excess of silver nitrate solution. The molarity of the initial sodium iodide solution is _________ M (as the nearest integer value).
Given molar masses:
$\text{Na} = 23, \, \text{I} = 127, \, \text{Ag} = 108, \, \text{N} = 14, \, \text{O} = 16 \text{ g mol}^{-1}$.
### Related Formula
Precipitation reaction stoichiometry:
$$\text{NaI(aq)} + \text{AgNO}_3\text{(aq)} \longrightarrow \text{AgI(s)} + \text{NaNO}_3\text{(aq)}$$
Molarity calculation formula:
$$M = \frac{\text{Moles of solute (NaI)}}{\text{Volume of solution in Liters (L)}}$$
### Execution
Step 1: Determine the molar mass of the Silver Iodide ($\text{AgI}$) precipitate:
$$\text{Molar Mass of AgI} = 108 + 127 = 235 \text{ g mol}^{-1}$$
Step 2: Calculate the moles of $\text{AgI}$ precipitated:
$$\text{Moles of AgI} = \frac{4.74 \text{ g}}{235 \text{ g mol}^{-1}} \approx 0.02017 \text{ mol}$$
Step 3: Apply the 1:1 reaction stoichiometry to find the moles of $\text{NaI}$:
$$\text{Moles of NaI} = \text{Moles of AgI} = 0.02017 \text{ mol}$$
Step 4: Compute the molarity of the solution, converting $20 \text{ mL}$ to $0.020 \text{ L}$:
$$\text{Molarity [NaI]} = \frac{0.02017 \text{ mol}}{0.020 \text{ L}} = 1.0085 \text{ M}$$
Rounding to the nearest integer value gives **1**.
### Pattern Recognition
Precipitation reactions involving silver halides follow a strict 1:1 mole ratio between the halide source and the silver precipitate. Converting mass into moles and dividing by the volume in liters quickly yields the molarity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
