Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below:

Statement A: TRUE - Certain stable hydrated salts (e.g. oxalic acid dihydrate) are used as primary standards. Statement B: TRUE - Primary standards must be stable in air and not hygroscopic or oxidized by air. Statement C: TRUE - Reactions must be rapid, complete, and strictly stoichiometric. Statement D: FALSE - Primary standards MUST be highly soluble in water to prepare volumetric standard solutions. Statement E: FALSE - Primary standards should ideally have high relative molar mass to minimize weighing errors. Hence, Statements A, B, and C are correct. Sees: Primary standard characteristics. Shortcut: Soluble in water (eliminates D) and high molar mass required (eliminates E). Class 11 Chemistry: Some Basic Concepts of Chemistry

Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substanc

Statement A: TRUE - Certain stable hydrated salts (e.g. oxalic acid dihydrate) are used as primary standards. Statement B: TRUE - Primary standards must be stable in air and not hygroscopic or oxidized by air. Statement C: TRUE - Reactions must be rapid, complete, and strictly stoichiometric. Statement D: FALSE - Primary standards MUST be highly soluble in water to prepare volumetric standard solutions. Statement E: FALSE - Primary standards should ideally have high relative molar mass to minimi

Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below:

Statement A: TRUE - Certain stable hydrated salts (e.g. oxalic acid dihydrate) are used as primary standards. Statement B: TRUE - Primary standards must be stable in air and not hygroscopic or oxidized by air. Statement C: TRUE - Reactions must be rapid, complete, and strictly stoichiometric. Statement D: FALSE - Primary standards MUST be highly soluble in water to prepare volumetric standard solutions. Statement E: FALSE - Primary standards should ideally have high relative molar mass to minimize weighing errors. Hence, Statements A, B, and C are correct. Sees: Primary standard characteristics. Shortcut: Soluble in water (eliminates D) and high molar mass required (eliminates E). Class 11 Chemistry: Some Basic Concepts of Chemistry

Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substanc

Statement A: TRUE - Certain stable hydrated salts (e.g. oxalic acid dihydrate) are used as primary standards. Statement B: TRUE - Primary standards must be stable in air and not hygroscopic or oxidized by air. Statement C: TRUE - Reactions must be rapid, complete, and strictly stoichiometric. Statement D: FALSE - Primary standards MUST be highly soluble in water to prepare volumetric standard solutions. Statement E: FALSE - Primary standards should ideally have high relative molar mass to minimi

Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not…

Year	2026	2025	2024	Total
Questions	8	15	7	30

Q jee_main_2024_29_january_evening Volumetric Titration and Molarity

If

50 mL

of

0.5 M

oxalic acid is required to neutralise

25 mL

of

NaOH

solution, the amount of

NaOH

in

50 mL

of given

NaOH

solution is ________ g.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Equivalents of Acid = Equivalents of Base

N₁ V₁ = N₂ V₂ (M₁ \times n₁) \times V₁ = (M₂ \times n₂) \times V₂

Core Logic

For oxalic acid ( $H₂C₂O₄$ $\text{H}_2\text{C}_2\text{O}_4$), the valence factor ( $n$ $n$-factor) is $2$ $2$. For $NaOH$ $\text{NaOH}$, the $n$ $n$-factor is $1$ $1$. Substituting the values into the normality equivalence expression:

50 \times 0.5 \times 2 = 25 \times MNaOH \times 1

50 = 25 \times MNaOH MNaOH = 2 M

Step 1: Mass Isolation

To find the mass of $NaOH$ $\text{NaOH}$ present in $50 mL$ $50\text{ mL}$ of this solution:

Mass = Molarity \times Volume (in L) \times Molar Mass

Mass = 2 \times ((50)/(1000)) \times 40 = 2 \times 0.05 \times 40 = 4 g

Pattern Recognition

Remember to use the correct $n$ $n$-factor ( $2$ $2$) for dibasic oxalic acid during equivalence matching to avoid calculation errors.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q83 jee_main_2024_27_jan_morning Stoichiometry

Mass of methane required to produce

22 g

of

CO₂

after complete combustion is

g. (Given Molar mass in

g mol⁻¹

:

C=12.0

,

H=1.0

,

O=16.0

)

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Balanced combustion chemical equation:

CH₄ + 2O₂ arrow CO₂ + 2H₂O

Moles = MassMolar Mass

Step 1: Determine moles of product generated

Molar Mass of CO₂ = 12 + (2 \times 16) = 44 g mol⁻¹

Moles of CO₂ produced = (22)/(44) = 0.5 moles

Step 2: Relate to input mass via stoichiometry metrics

From the balanced equation, $1 mole of CH₄$ $1\text{ mole of CH}_4$ produces $1 mole of CO₂$ $1\text{ mole of CO}_2$.

Required Moles of CH₄ = 0.5 moles

Molar Mass of CH₄ = 12 + (4 \times 1) = 16 g mol⁻¹

Mass of CH₄ = 0.5 \times 16 = 8 g

Pattern Recognition

$22 g of CO₂$ $22\text{ g of CO}_2$ is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to $8 g$ $8\text{ g}$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q88 jee_main_2024_30_jan_morning Mole Concept

0.05 cm

thick coating of silver is deposited on a plate of

0.05 m²

area. The number of silver atoms deposited on plate are ________

\times 10²³

. (At mass

Ag=108, d=7.9 g cm⁻³

)

Numerical Answer. Answer: 11 to 11

Solution

Related Formula

Volume = Area \times Thickness

Mass = Density \times Volume

Moles = MassMolar Mass

Number of Atoms = Moles \times NA

Step 1: Calculate Volume of Coating

Area = $0.05 m² = 0.05 \times 10⁴ cm² = 500 cm²$ Thickness = $0.05 cm$ $0.05 \text{ cm}$ Volume = $500 cm² \times 0.05 cm = 25 cm³$

Step 2: Calculate Mass and Moles

Mass = Volume $\times$ $\times$ Density = $25 cm³ \times 7.9 g/cm³ = 197.5 g$ Moles of Ag = $(197.5)/(108) = 1.8287 moles$

Step 3: Calculate Number of Atoms

Number of Atoms = 1.8287 \times 6.022 \times 10²³

= 11.01 \times 10²³

Rounding to nearest integer, we get $11$ $11$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry Class 12 Chemistry: Electrochemistry

Q62 jee_main_2024_31_jan_evening Stoichiometry and Calculations

A sample of

CaCO₃

and

MgCO₃

weighed

2.21 g

is ignited to constant weight of

1.152 g

. The composition of mixture is: (Given molar mass in

g mol⁻¹

CaCO₃:100, MgCO₃:84

)

A. $1.187~g~CaCO₃ + 1.023~g~MgCO₃$
B. $1.023~g~CaCO₃ + 1.023~g~MgCO₃$
C. $1.187~g~CaCO₃ + 1.187~g~MgCO₃$
D. $1.023~g~CaCO₃ + 1.187~g~MgCO₃$

Solution

Related Formula

CaCO₃(s) Δ CaO(s) + CO₂(g)

MgCO₃(s) Δ MgO(s) + CO₂(g)

Core Logic

Let the weight of $CaCO₃$ $\mathrm{CaCO}_3$ be $x g$ $x\text{ g}$. Then, the weight of $MgCO₃ = (2.21 - x) g$ $\mathrm{MgCO}_3 = (2.21 - x)\text{ g}$.

Moles of $CaCO₃$ $\mathrm{CaCO}_3$ decomposed = Moles of $CaO$ $\mathrm{CaO}$ formed. $(x)/(100) = Moles of CaO formed$ $Weight of CaO formed = (x)/(100) \times 56$

Moles of $MgCO₃$ $\mathrm{MgCO}_3$ decomposed = Moles of $MgO$ $\mathrm{MgO}$ formed. $((2.21 - x))/(84) = Moles of MgO formed$ $Weight of MgO formed = (2.21 - x)/(84) \times 40$

Step 1: Setting up the Equation

The total weight of the residue ( $CaO$ $\mathrm{CaO}$ + $MgO$ $\mathrm{MgO}$) is given as $1.152 g$ $1.152\text{ g}$.

(2.21 - x)/(84) \times 40 + (x)/(100) \times 56 = 1.152

Step 2: Solving for x

(88.4 - 40x)/(84) + 0.56x = 1.152

1.0523 - 0.4761x + 0.56x = 1.152

0.0839x = 0.0997

x = 1.188 g

So, weight of $CaCO₃ \approx 1.187 g$ (accounting for rounding) Weight of $MgCO₃ = 2.21 - 1.188 = 1.022 g \approx 1.023 g$ .

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q83 jee_main_2024_31_jan_morning Stoichiometry

Number of moles of methane required to produce 22g

CO2(g)

after combustion is

x \times 10⁻²

moles. The value of x is

Numerical Answer. Answer: 50 to 50

Solution

Step 1: Stoichiometric Equation

CH4(g) + 2O2(g) arrow CO2(g) + 2H₂O(l)

1 mole of $CH₄$ $CH_4$ produces 1 mole of $CO₂$ $CO_2$.

Step 2: Moles Calculation

Molar mass of $CO₂ = 12 + 2(16) = 44 g/mol$ $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$

nCO₂ = MassMolar mass = (22)/(44) = 0.5 moles

Since 1 mole of $CH₄$ $CH_4$ produces 1 mole of $CO₂$ $CO_2$, the moles of $CH₄$ $CH_4$ required is $0.5$ $0.5$ moles.

Step 3: Finding x

0.5 moles = 50 \times 10⁻² moles

$x = 50$ $x = 50$

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Mass Isolation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Determine moles of product generated

Step 2: Relate to input mass via stoichiometry metrics

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate Volume of Coating

Step 2: Calculate Mass and Moles

Step 3: Calculate Number of Atoms

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Equation

Step 2: Solving for x

Chapter Mix

Solution

Step 1: Stoichiometric Equation

Step 2: Moles Calculation

Step 3: Finding x

Chapter Mix

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