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 3.
Q48 (2025)
Butane reacts with oxygen to produce carbon dioxide and water following the equation given below: [cite: 461, 463]
$$\mathrm{C_4H{10}(g) + \frac{13}{2}O_2(g)
ightarrow 4CO_2(g) + 5H_2O(l)} $$
If $174.0\text{ kg}$ of butane is mixed with $320.0\text{ kg}$ of $\text{O}_2$, the volume of water formed in litres is $\dots$. (Nearest integer) [cite: 465, 466]
[Given: (a) Molar mass of $\text{C}$, $\text{H}$, $\text{O}$ are $12$, $1$, $16\text{ g mol}^{-1}$ respectively, (b) Density of water $= 1\text{ g mL}^{-1}$] [cite: 467, 468]
### Related Formula
$$\text{Moles } (n) = \frac{\text{Mass in grams}}{\text{Molar Mass}}$$
$$\text{Volume of water } (V) = \frac{\text{Mass of water}}{\text{Density of water}}$$
### Core Logic
First, calculate initial molar quantities for both reactants:
- Molar mass of $\text{C}_4\text{H}_{10} = 4(12) + 10(1) = 58\text{ g/mol}$
- Initial Moles of butane $= \frac{174.0 \times 10^3\text{ g}}{58\text{ g/mol}} = 3000\text{ mol} = 3 \times 10^3\text{ mol} $
- Molar mass of $\text{O}_2 = 32\text{ g/mol}$
- Initial Moles of oxygen $= \frac{320.0 \times 10^3\text{ g}}{32\text{ g/mol}} = 10000\text{ mol} = 10 \times 10^3\text{ mol} $
### Step 1: Identify Limiting Reagent
Let's test the stoichiometric requirements via calculation ratios:
- Ratio for $\text{C}_4\text{H}_{10} = \frac{3000}{1} = 3000$
- Ratio for $\text{O}_2 = \frac{10000}{13/2} = 1538.46$
Since the ratio for $\text{O}_2$ 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:
$$\text{Moles of } \text{H}2\text{O} = 5 \times \frac{\text{Moles of } \text{O}2}{13/2} = 5 \times \frac{2}{13} \times 10000 = \frac{100000}{13}\text{ mol} $$
Convert moles to mass ($M{\text{H}2\text{O}} = 18\text{ g/mol}$):
$$\text{Mass of water} = \frac{100000}{13} \times 18 = 138461.5\text{ g} \approx 138.46\text{ kg} $$
Since density $= 1\text{ g/mL} = 1\text{ kg/L}$, the net volume is exactly:
$$V{\text{water}} = 138.46\text{ Litres} \approx 138\text{ Litres} $$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q36 (2025)
The elemental composition of a compound is $54.2\%\text{ C}$, $9.2\%\text{ H}$ and $36.6\%\text{ O}$. If the molar mass of the compound is $132\text{ g mol}^{-1}$, the molecular formula of the compound is:
[Given : The relative atomic mass of $\mathrm{C} : \mathrm{H} : \mathrm{O} = 12 : 1 : 16$]
- $\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{O}_{3}$
- $\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}$
- $\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{3}$
- $\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}_{2}$
### Core Logic
Let's find the empirical formula by calculating the relative molar ratios of each element:
* Carbon (C):
$$\text{Moles} = \frac{54.2}{12} = 4.516$$
* Hydrogen (H):
$$\text{Moles} = \frac{9.2}{1} = 9.200$$
* Oxygen (O):
$$\text{Moles} = \frac{36.6}{16} = 2.287$$
Next, divide each value by the smallest number of moles ($2.287$) to get the simplest whole-number ratio:
* C: $\frac{4.516}{2.287} \approx 1.97 \approx 2$
* H: $\frac{9.200}{2.287} \approx 4.02 \approx 4$
* O: $\frac{2.287}{2.287} = 1$
Thus, the Empirical Formula is $\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}$.
### Step 2: Molecular Formula Determination
1. Calculate the empirical formula mass:
$$\text{Empirical Mass} = (2 \cdot 12) + (4 \cdot 1) + 16 = 24 + 4 + 16 = 44\text{ g/mol}$$
2. Find the multiplier ($n$):
$$n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{132}{44} = 3$$
3. Determine the molecular formula:
$$\text{Molecular Formula} = (\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O})_{3} = \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{3}$$
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 ($n$). Multiplication gives the definitive molecular formula.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q50 (2025)
Consider the following reaction occurring in the blast furnace.
$$\mathrm {F e} _ {3} \mathrm {O} _ {4 (\mathrm {s})} + 4 \mathrm {C O} _ {(\mathrm {g})} \rightarrow 3 \mathrm {F e} _ {(\mathrm {l})} + 4 \mathrm {C O} _ {2 (\mathrm {g})}$$
$x$ kg of iron is produced when $2.32\times 10^{3}\mathrm{kg}$ $\mathrm{Fe_3O_4}$ and $2.8\times 10^{2}\mathrm{kg}$ $\mathrm{CO}$ are brought together in the furnace. The value of $x$ is ______ (nearest integer)
{Given:
Molar mass of $\mathrm{Fe_3O_4} = 232\mathrm{g mol^{-1}}$
Molar mass of $\mathrm{CO} = 28\mathrm{g mol^{-1}}$
Molar mass of $\mathrm{Fe} = 56\mathrm{g mol^{-1}}$}
### Related Formula
$$\text{Moles } (n) = \frac{\text{Mass in grams}}{\text{Molar Mass}}$$
### Core Logic
First, calculate the input moles for each reactant:
- Moles of $$Fe_{3}O_{4} = \frac{2.32 \times 10^3 \times 10^3\text{ g}}{232\text{ g mol}^{-1}} = 10,000\text{ moles}$$
- Moles of $$CO = \frac{2.8 \times 10^2 \times 10^3\text{ g}}{28\text{ g mol}^{-1}} = 10,000\text{ moles}$$
Next, identify the limiting reagent by comparing the available moles to the stoichiometric coefficients:
- For $Fe_3O_4$: $\frac{10000}{1} = 10000$
- For $CO$: $\frac{10000}{4} = 2500$
Since $2500 < 10000$, **carbon monoxide ($CO$) is the limiting reagent**.
Now, determine the production yield of iron based on the limiting reagent ($CO$):
$$\text{Moles of } Fe \text{ produced} = \frac{3}{4} \times n(CO) = \frac{3}{4} \times 10000 = 7500\text{ moles}$$
Convert these moles into kilograms to find the final mass ($x$):
$$\text{Mass of } Fe = \frac{7500 \times 56\text{ g/mol}}{1000\text{ g/kg}} = 420\text{ kg}$$
### Pattern Recognition
Always identify the limiting reagent first by normalizing the mole quantities with their respective stoichiometric coefficients before calculating product yields.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q35 (2025)
Concentrated nitric acid is labelled as $75\%$ by mass. The volume in mL of the solution which contains $30\mathrm{\ g}$ of nitric acid is
Given: Density of nitric acid solution is $1.25\mathrm{\ g/mL}$
- 45
- 55
- 32
- 40
### Related Formula
Mass percentage definition:
$$\%\text{ w/w} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$
Density conversion equation:
$$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}}$$
### Core Logic
A value of $75\%\text{ w/w HNO}_3$ implies that $75\mathrm{\ g}$ of pure $\text{HNO}_3$ is present in $100\mathrm{\ g}$ of solution.
We need to find the volume that provides exactly $30\mathrm{\ g}$ of pure acid solute.
### Step 1: Calculate Solution Mass and Volume
Mass of solution needed for $30\mathrm{\ g}$ solute:
$$\text{Mass} = \frac{100}{75} \times 30 = 40\mathrm{\ g}$$
Converting mass to volume using solution density ($1.25\mathrm{\ g/mL}$):
$$\text{Volume} = \frac{40\mathrm{\ g}}{1.25\mathrm{\ g/mL}} = 32\mathrm{\ mL}$$
### Pattern Recognition
Break concentration steps down clearly: $\text{Mass of solute} \rightarrow \text{Mass of solution} \rightarrow \text{Volume of solution}$. Combining operations: $\text{Volume} = \frac{\text{Mass solute}}{\%} \times \frac{100}{\text{density}} = \frac{30}{75} \times \frac{100}{1.25} = 0.4 \times 80 = 32$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q1306 (2025)
Choose the correct statements.
(A) Weight of a substance is the amount of matter present in it.
(B) Mass is the force exerted by gravity on an object.
(C) Volume is the amount of space occupied by a substance.
(D) Temperatures below $0^{\circ}\mathrm{C}$ are possible in Celsius scale, but in Kelvin scale negative temperature is not possible.
(E) Precision refers to the closeness of various measurements for the same quantity.
- (B), (C) and (D) Only
- (A), (B) and (C) Only
- (A), (D) and (E) Only
- (C), (D) and (E) Only
### Related Formula
$$T_{\mathrm{K}} = T_{^{\circ}\mathrm{C}} + 273.15$$
Absolute zero ($0\text{ K}$) represents the lowest theoretical temperature limit.
### Core Logic
Analyzing each statement based on foundational definitions :
* (A) & (B) Incorrect: Mass is the actual matter present; weight is the gravitational force exerted on that mass. These definitions are reversed in the statements.
* (C) Correct: Volume correctly defines the space occupied by a substance .
* (D) Correct: Celsius values can be negative, whereas Kelvin scale strictly defaults to absolute zero ($0\text{ K}$) as minimum .
* (E) Correct: Precision measures how close experimental trials lie relative to each other .
Therefore, statements (C), (D), and (E) are correct.
### Pattern Recognition
Absolute temperature scale (Kelvin) can never possess real negative values because $0\text{ K}$ represents complete cessation of molecular motion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
