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 4.
Q77 (2024)
Given below are two statements :
Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.
Statement (II) : In this titration phenolphthalein can be used as indicator.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both Statement I and Statement II are correct}$
- $\text{Statement I is correct but Statement II is incorrect}$
- $\text{Statement I is incorrect but Statement II is correct.}$
- $\text{Both Statement I and Statement II are incorrect.}$
### Core Logic
Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like $NaOH$. This is because it is highly pure, non-hygroscopic, stable, and has a relatively high molar mass, making its concentration reliable and stable over time.
Statement (II): KHP is a weak acid and $NaOH$ is a strong base. The titration of a weak acid with a strong base yields an equivalence point in the weakly basic range (pH > 7). Phenolphthalein changes colour in the pH range 8.3 to 10.0, making it the perfect indicator for this titration.
### Step 1: Evaluate Statements
Statement I is correct.
Statement II is correct.
### Pattern Recognition
Weak Acid vs Strong Base $\rightarrow$ Equivalence pH > 7 $\rightarrow$ Phenolphthalein is the indicator of choice.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Equilibrium
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q89 (2024)
Consider the following reaction:
$$3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl$$
If $72 \mathrm{~mmol}$ of $PbCl_2$ is mixed with $50 \mathrm{~mmol}$ of $(NH_4)_3PO_4$, then amount of $Pb_3(PO_4)_2$ formed is ... mmol. (nearest integer)
### Related Formula
$$ \text{Moles of Product} = \text{Moles of Limiting Reagent} \times \frac{\text{Stoichiometry of Product}}{\text{Stoichiometry of Limiting Reagent}} $$
### Core Logic
From the balanced chemical equation:
$3 \text{ moles of } PbCl_2 \text{ react with } 2 \text{ moles of } (NH_4)_3PO_4$.
Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients:
For $PbCl_2$: $\frac{72}{3} = 24$
For $(NH_4)_3PO_4$: $\frac{50}{2} = 25$
Since $24 < 25$, $PbCl_2$ is the limiting reagent and will completely consume.
### Step 1: Calculate Product Moles
Moles of $Pb_3(PO_4)_2$ formed depends entirely on $PbCl_2$.
3 mmol of $PbCl_2$ produces 1 mmol of $Pb_3(PO_4)_2$.
Therefore, 72 mmol of $PbCl_2$ will produce:
$\frac{1}{3} \times 72 = 24 \mathrm{~mmol}$ of $Pb_3(PO_4)_2$.
### Pattern Recognition
Always identify the Limiting Reagent by taking the ratio $n / \text{coefficient}$. The smallest ratio dictates the extent of the reaction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q1495 (2024)
If $50\text{ mL}$ of $0.5\text{ M}$ oxalic acid is required to neutralise $25\text{ mL}$ of $\mathrm{NaOH}$ solution, the amount of $\mathrm{NaOH}$ in $50\text{ mL}$ of given $\mathrm{NaOH}$ solution is ________ g.
### Related Formula
$$\text{Equivalents of Acid} = \text{Equivalents of Base}$$
$$N_1 V_1 = N_2 V_2 \implies (M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2$$
### Core Logic
For oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$), the valence factor ($n$-factor) is $2$. For $\text{NaOH}$, the $n$-factor is $1$.
Substituting the values into the normality equivalence expression:
$$50 \times 0.5 \times 2 = 25 \times M_{\text{NaOH}} \times 1$$
$$50 = 25 \times M_{\text{NaOH}} \implies M_{\text{NaOH}} = 2\text{ M}$$
### Step 1: Mass Isolation
To find the mass of $\text{NaOH}$ present in $50\text{ mL}$ of this solution:
$$\text{Mass} = \text{Molarity} \times \text{Volume (in L)} \times \text{Molar Mass}$$
$$\text{Mass} = 2 \times \left(\frac{50}{1000}\right) \times 40 = 2 \times 0.05 \times 40 = 4\text{ g}$$
### Pattern Recognition
Remember to use the correct $n$-factor ($2$) for dibasic oxalic acid during equivalence matching to avoid calculation errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q83 (2024)
Mass of methane required to produce $22\text{ g}$ of $\text{CO}_2$ after complete combustion is $\text{\quad\quad}$ g.
(Given Molar mass in $\text{g mol}^{-1}$: $\text{C}=12.0$, $\text{H}=1.0$, $\text{O}=16.0$)
### Related Formula
Balanced combustion chemical equation:
$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$
$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$
### Step 1: Determine moles of product generated
$$\text{Molar Mass of CO}_2 = 12 + (2 \times 16) = 44\text{ g mol}^{-1}$$
$$\text{Moles of CO}_2 \text{ produced} = \frac{22}{44} = 0.5\text{ moles}$$
### Step 2: Relate to input mass via stoichiometry metrics
From the balanced equation, $1\text{ mole of CH}_4$ produces $1\text{ mole of CO}_2$.
$$\text{Required Moles of CH}_4 = 0.5\text{ moles}$$
$$\text{Molar Mass of CH}_4 = 12 + (4 \times 1) = 16\text{ g mol}^{-1}$$
$$\text{Mass of CH}_4 = 0.5 \times 16 = 8\text{ g}$$
### Pattern Recognition
$22\text{ g of CO}_2$ is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to $8\text{ g}$.
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q88 (2024)
$0.05\text{ cm}$ thick coating of silver is deposited on a plate of $0.05\text{ m}^2$ area. The number of silver atoms deposited on plate are ________ $\times 10^{23}$.
(At mass $Ag=108, d=7.9\text{ g cm}^{-3}$)
### Related Formula
$$\text{Volume} = \text{Area} \times \text{Thickness}$$
$$\text{Mass} = \text{Density} \times \text{Volume}$$
$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$
$$\text{Number of Atoms} = \text{Moles} \times N_A$$
### Step 1: Calculate Volume of Coating
Area = $0.05 \text{ m}^2 = 0.05 \times 10^4 \text{ cm}^2 = 500 \text{ cm}^2$
Thickness = $0.05 \text{ cm}$
Volume = $500 \text{ cm}^2 \times 0.05 \text{ cm} = 25 \text{ cm}^3$
### Step 2: Calculate Mass and Moles
Mass = Volume $\times$ Density = $25 \text{ cm}^3 \times 7.9 \text{ g/cm}^3 = 197.5 \text{ g}$
Moles of Ag = $\frac{197.5}{108} = 1.8287 \text{ moles}$
### Step 3: Calculate Number of Atoms
$$\text{Number of Atoms} = 1.8287 \times 6.022 \times 10^{23}$$
$$= 11.01 \times 10^{23}$$
Rounding to nearest integer, we get $11$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Class 12 Chemistry: Electrochemistry
