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 5.
Q62 (2024)
A sample of $CaCO_3$ and $MgCO_3$ weighed $2.21\text{ g}$ is ignited to constant weight of $1.152\text{ g}$. The composition of mixture is:
(Given molar mass in $\mathrm{g\,mol^{-1}}$
$\mathrm{CaCO_3:100, MgCO_3:84}$)
- $1.187\mathrm{~g}\mathrm{~CaCO}_3 + 1.023\mathrm{~g}\mathrm{~MgCO}_3$
- $1.023\mathrm{~g}\mathrm{~CaCO}_3 + 1.023\mathrm{~g}\mathrm{~MgCO}_3$
- $1.187\mathrm{~g}\mathrm{~CaCO}_3 + 1.187\mathrm{~g}\mathrm{~MgCO}_3$
- $1.023\mathrm{~g}\mathrm{~CaCO}_3 + 1.187\mathrm{~g}\mathrm{~MgCO}_3$
### Related Formula
$$\mathrm{CaCO_3(s)} \xrightarrow{\Delta} \mathrm{CaO(s)} + \mathrm{CO_2(g)}$$
$$\mathrm{MgCO_3(s)} \xrightarrow{\Delta} \mathrm{MgO(s)} + \mathrm{CO_2(g)}$$
### Core Logic
Let the weight of $\mathrm{CaCO}_3$ be $x\text{ g}$.
Then, the weight of $\mathrm{MgCO}_3 = (2.21 - x)\text{ g}$.
Moles of $\mathrm{CaCO}_3$ decomposed = Moles of $\mathrm{CaO}$ formed.
$\frac{x}{100} = \text{Moles of CaO formed}$
$\text{Weight of CaO formed} = \frac{x}{100} \times 56$
Moles of $\mathrm{MgCO}_3$ decomposed = Moles of $\mathrm{MgO}$ formed.
$\frac{(2.21 - x)}{84} = \text{Moles of MgO formed}$
$\text{Weight of MgO formed} = \frac{2.21 - x}{84} \times 40$
### Step 1: Setting up the Equation
The total weight of the residue ($\mathrm{CaO}$ + $\mathrm{MgO}$) is given as $1.152\text{ g}$.
$$\frac{2.21 - x}{84} \times 40 + \frac{x}{100} \times 56 = 1.152$$
### Step 2: Solving for x
$$\frac{88.4 - 40x}{84} + 0.56x = 1.152$$
$$1.0523 - 0.4761x + 0.56x = 1.152$$
$$0.0839x = 0.0997$$
$$x = 1.188\text{ g}$$
So, weight of $\mathrm{CaCO}_3 \approx 1.187\text{ g}$ (accounting for rounding)
Weight of $\mathrm{MgCO}_3 = 2.21 - 1.188 = 1.022\text{ g} \approx 1.023\text{ g}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q83 (2024)
Number of moles of methane required to produce 22g $CO_{2(g)}$ after combustion is $x \times 10^{-2}$ moles. The value of x is
### Step 1: Stoichiometric Equation
$$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$$
1 mole of $CH_4$ produces 1 mole of $CO_2$.
### Step 2: Moles Calculation
Molar mass of $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$
$$n_{CO_2} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{22}{44} = 0.5 \text{ moles}$$
Since 1 mole of $CH_4$ produces 1 mole of $CO_2$, the moles of $CH_4$ required is $0.5$ moles.
### Step 3: Finding x
$$0.5 \text{ moles} = 50 \times 10^{-2} \text{ moles}$$
$x = 50$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q56 (2026)
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
- $\mathrm{C}_2\mathrm{H}_4$
- $\mathrm{C}_4\mathrm{H}_{10}$
- $\mathrm{C}_2\mathrm{H}_2$
- $\mathrm{C}_2\mathrm{H}_6$
### Related Formula
$$\mathrm{C}_\mathrm{x}\mathrm{H}_\mathrm{y(g)} + \left(\mathrm{x} + \frac{\mathrm{y}}{4}\right)\mathrm{O}_{2(g)} \longrightarrow \mathrm{xCO}_{2(g)} + \frac{\mathrm{y}}{2}\mathrm{H}_2\mathrm{O}_{(\ell)}$$
### Core Logic
Let the volume of hydrocarbon be $V = 80$ mL.
Initial volume of $O_2 = 264$ mL.
At 273 K, $H_2O$ is liquid, so its volume is neglected.
Volume of $CO_2$ formed = $80x$ mL.
Volume of $O_2$ used = $80\left(x + \frac{y}{4}\right)$ mL.
Unreacted $O_2$ = $264 - 80\left(x + \frac{y}{4}\right)$ mL.
Total residual volume = $V_{CO_2} + V_{unreacted \ O_2} = 224$ mL.
$$80x + 264 - 80\left(x + \frac{y}{4}\right) = 224$$
$$264 - \frac{80y}{4} = 224$$
$$40 = 20y \implies y = 2$$
After treatment with KOH, $CO_2$ is absorbed. The remaining volume is unreacted $O_2$, which is 64 mL.
$$264 - 80\left(x + \frac{y}{4}\right) = 64$$
Substitute $y = 2$:
$$264 - 80\left(x + \frac{1}{2}\right) = 64$$
$$264 - 80x - 40 = 64$$
$224 - 80x = 64$
$$80x = 160 \implies x = 2$$
The hydrocarbon is $\mathrm{C}_2\mathrm{H}_2$.
### Pattern Recognition
Volume decrease by KOH indicates the volume of $CO_2$ produced.
$V_{CO_2} = 224 - 64 = 160$ mL.
$V_{HC} = 80$ mL. So $x = \frac{160}{80} = 2$.
Total volume reduction = 264 - 64 = 200 mL ($O_2$ consumed).
$O_2$ consumed = $80(x + y/4) = 200 \implies 2 + y/4 = 2.5 \implies y/4 = 0.5 \implies y = 2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q57 (2026)
14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K.
Which of the following statements is incorrect?
[Given : Molar mass in $\mathrm{g \ mol^{-1}}$ of Ca–40, Cl–35.5, H–1]
- $\text{0.35 mol of }H_{2}\text{ gas is evolved.}$
- $7.84 \mathrm{~L}\text{ of }\mathrm{H}_{2}\text{ gas is evolved.}$
- $\text{33.3 g of }\mathrm{CaCl}_2\text{ is produced.}$
- $\text{The limiting reagent is calcium metal.}$
### Related Formula
$$\mathrm{Ca}_{(\mathrm{s})} + 2\mathrm{HCl}_{(\mathrm{g})} \longrightarrow \mathrm{CaCl}_{2(\mathrm{s})} + \mathrm{H}_{2(\mathrm{g})}$$
### Core Logic
Number of moles of Calcium ($n_{Ca}$) = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{14.0}{40} = 0.35$ mol.
Since HCl is in excess, Calcium is the limiting reagent.
From stoichiometry, 1 mole of Ca produces 1 mole of $H_2$ gas and 1 mole of $CaCl_2$.
Moles of $H_2$ evolved = 0.35 mol.
Volume of $H_2$ at STP (1 atm, 273 K) = $0.35 \times 22.4$ L = $7.84$ L.
Mass of $CaCl_2$ produced = $0.35 \times \text{Molar mass of } CaCl_2$ = $0.35 \times (40 + 71) = 0.35 \times 111 = 38.85$ g.
Option (3) states 33.3 g of $CaCl_2$ is produced, which is incorrect.
### Step 1: Final Conclusion
Statement (3) is incorrect.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q55 (2026)
By usual analysis, 1.00g of compound (X) gave 1.79g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer)
(Given, molar mass in $\text{g mol}^{-1}: \text{O} = 16, \text{Mg} = 24, \text{P} = 31$)
- $(1) \ 50$
- $(2) \ 30$
- $(3) \ 20$
- $(4) \ 40$
### Related Formula
$$\% \text{ of } \text{P} = \frac{\text{Moles of } \text{Mg}_2\text{P}_2\text{O}_7 \times 2 \times 31}{\text{Mass of compound}} \times 100$$
### Core Logic
Molar mass of $\text{Mg}_2\text{P}_2\text{O}_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 \text{ g/mol}$.
$$\text{Percentage of P} = \frac{\left(\frac{1.79}{222} \times 2 \times 31\right)}{1} \times 100 = 49.99\% \approx 50\%$$
### Step 1: Final Calculation
Rounding to the nearest integer gives 50.
### Pattern Recognition
Sees: Gravimetric analysis involving magnesium pyrophosphate.
Trap: Forgetting the factor of 2 for phosphorus atoms in $\text{Mg}_2\text{P}_2\text{O}_7$.
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
