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 2.
Q47 (2025)
Quantitative analysis of an organic compound (X) shows following % composition.
$$\mathrm{C}:14.5\% \quad \mathrm{Cl}:64.46\% \quad \mathrm{H}:1.8\%$$
The empirical formula mass of the compound (X) is $\mathrm{x} \times 10^{-1}$. The value of $\mathrm{x}$ is:
(Given molar mass in $\mathrm{g\,mol^{-1}}$ of C: 12, H: 1, O: 16, Cl: 35.5)
### Step 1: Determine Oxygen Percentage
The total percentage must equal 100%. The remaining composition corresponds to Oxygen:
$$\%\mathrm{O} = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\%$$
### Step 2: Calculate Molar Ratios
Divide each mass percentage by its respective atomic weight:
- $\mathrm{C}: \frac{14.5}{12} = 1.208$
- $\mathrm{Cl}: \frac{64.46}{35.5} = 1.815$
- $\mathrm{H}: \frac{1.8}{1} = 1.800$
- $\mathrm{O}: \frac{19.24}{16} = 1.202$
### Step 3: Find Simple Integer Ratio
Divide by the lowest ratio value ($1.202$):
- $\mathrm{C}: \frac{1.208}{1.202} \approx 1 \rightarrow \times 2 = 2$
- $\mathrm{Cl}: \frac{1.815}{1.202} \approx 1.5 \rightarrow \times 2 = 3$
- $\mathrm{H}: \frac{1.800}{1.202} \approx 1.5 \rightarrow \times 2 = 3$
- $\mathrm{O}: \frac{1.202}{1.202} = 1 \rightarrow \times 2 = 2$
Thus, the empirical formula is $\mathrm{C}_2\mathrm{H}_3\mathrm{Cl}_3\mathrm{O}_2$.
### Step 4: Compute Mass
Empirical formula mass calculation:
$$\text{Mass} = (2 \times 12) + (3 \times 1) + (3 \times 35.5) + (2 \times 16)$$
$$\text{Mass} = 24 + 3 + 106.5 + 32 = 165.5\,\mathrm{g\,mol}^{-1}$$
Expressing in the requested format:
$$165.5 = 1655 \times 10^{-1} \Rightarrow x = 1655$$
### Pattern Recognition
Sees: Multi-element empirical calculation.
Trap: Forgetting to compute Oxygen by missing that the percentages do not sum to 100% initial value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q48 (2025)
The molarity of a $70\%$ (mass/mass) aqueous solution of a monobasic acid (X) is $\_\_\_\_\_\mathrm{M}$ (Nearest integer)
[Given : Density of aqueous solution of (X) is $1.25\mathrm{g}\,\mathrm{mL}^{-1}$
Molar mass of the acid is $70\mathrm{g\,mol}^{-1}$]
### Related Formula
Molarity formula based on mass percentage ($w/w$) and density ($d$):
$$\text{Molarity} = \frac{\%(w/w) \times d \times 10}{\text{Molar Mass of solute}}$$
### Step 1: Substitute Values
Given values: $\% = 70$, $d = 1.25\,\mathrm{g\,mL}^{-1}$, $\text{Molar Mass} = 70\,\mathrm{g\,mol}^{-1}$.
$$\text{Molarity} = \frac{70 \times 1.25 \times 10}{70} = 1.25 \times 10 = 12.5\,\mathrm{M}$$
Rounding to the nearest integer gives 13 (or 12.5 as written in standard templates; let us provide 13 matching nearest integer constraints).
### Pattern Recognition
Sees: Conversion of mass percentage to molarity tracking.
Shortcut: Using the classic shortcut formula $\frac{\% \times d \times 10}{M}$ simplifies the arithmetic immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q33 (2025)
Among $10^{-9}\mathrm{~g}$ (each) of the following elements, which one will have the highest number of atoms?
Element: $\mathrm{Pb}$, $\mathrm{Po}$, $\mathrm{Pr}$ and $\mathrm{Pt}$
- $\mathrm{Po}$
- $\mathrm{Pr}$
- $\mathrm{Pb}$
- $\mathrm{Pt}$
### Related Formula
$$\text{Number of atoms} = \frac{\text{Mass (in g)}}{\text{Molar Mass (g/mol)}} \times N_{\mathrm{A}}$$
### Core Logic
For equal mass ($10^{-9}\mathrm{~g}$), the number of atoms is inversely proportional to the molar mass ($M$):
$$\text{Number of atoms} \propto \frac{1}{M}$$
### Step 1: Comparison of Molar Masses
Molar masses of given elements:
- $M_{\mathrm{Po}} = 209\mathrm{~g/mol}$
- $M_{\mathrm{Pr}} = 141\mathrm{~g/mol}$
- $M_{\mathrm{Pb}} = 207\mathrm{~g/mol}$
- $M_{\mathrm{Pt}} = 195\mathrm{~g/mol}$
Praseodymium ($\mathrm{Pr}$) has the lowest molar mass ($141\mathrm{~g/mol}$), so it contains the maximum number of atoms.
### Pattern Recognition
Same mass $\rightarrow$ lowest molar mass has highest number of atoms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q47 (2025)
The amount of calcium oxide produced on heating $150\mathrm{~kg}$ limestone (75% pure) is _________________ kg. (Nearest integer)
Given : Molar mass (in g mol$^{-1}$ ) of Ca-40, O-16, C-12
### Related Formula
$$\mathrm{CaCO}_3 \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO}_2$$
$$\text{Pure mass} = \text{Total mass} \times \frac{\text{Purity \%}}{100}$$
### Core Logic
1. Find the pure mass of calcium carbonate ($CaCO_3$) present:
$$\text{Mass of } CaCO_3 = 150 \times \frac{75}{100} = 112.5 \mathrm{~kg} = 112500 \mathrm{~g}$$
2. Convert this mass into moles (Molar mass of $CaCO_3 = 40+12+48 = 100 \mathrm{~g\cdot mol^{-1}}$):
$$\text{moles of } CaCO_3 = \frac{112500}{100} = 1125 \text{ moles}$$
3. From the stoichiometry of the reaction, 1 mole of $CaCO_3$ yields 1 mole of $CaO$:
$$\text{moles of } CaO = 1125 \text{ moles}$$
$$\text{Mass of } CaO = 1125 \times 56 \mathrm{~g} = 63000 \mathrm{~g} = 63 \mathrm{~kg}$$
### Pattern Recognition
Always multiply by the purity fraction first before entering regular stoichiometric conversion chains. Since the formula weight of limestone is exactly $100$, tracking percentages directly mirrors mole factors seamlessly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q48 (2025)
Fortification of food with iron is done using $\mathrm{FeSO_4\cdot 7H_2O}$. The mass in grams of the $\mathrm{FeSO_4\cdot 7H_2O}$ required to achieve $12\mathrm{~ppm}$ of iron in $150\mathrm{~kg}$ of wheat is _______. (Nearest integer)
[Given: Molar mass of Fe, S and O respectively are 56, 32 and $16\mathrm{~g~mol^{-1}}$]
### Related Formula
$$\text{ppm} = \frac{\text{Mass of solute (g)}}{\text{Total mass of solution/mixture (g)}} \times 10^6$$
### Core Logic
Let the required mass of pure iron be $w\mathrm{~g}$. The total mass of the wheat mixture is $150\mathrm{~kg} = 150 \times 10^3\mathrm{~g}$.
Applying the parts-per-million concentration condition:
$$12 = \frac{w}{150 \times 10^3} \times 10^6 \implies 12 = w \times 6.666$$
$$w = \frac{12 \times 150 \times 10^3}{10^6} = 1.8\mathrm{~g} \text{ of Iron}$$
Now, determine the molar mass of the complete green vitriol salt crystal template, $\mathrm{FeSO_4 \cdot 7H_2O}$:
$$M = 56 + 32 + (4 \times 16) + (7 \times 18) = 56 + 32 + 64 + 126 = 278\mathrm{~g~mol^{-1}}$$
Set up a stoichiometric mass balance proportion to find the total salt mass $w_1$:
$$\text{Moles of Fe} = \frac{1.8}{56} = \frac{w_1}{278}$$
$$w_1 = \frac{1.8 \times 278}{56} = \frac{500.4}{56} \approx 8.935\mathrm{~g}$$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
