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…

Q56 jee_main_2026_21_jan_morning Eudiometry

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 :

A. $mathrmC_2mathrmH_4$ $\mathrm{C}_2\mathrm{H}_4$
B. $mathrmC_4mathrmH_10$ $\mathrm{C}_4\mathrm{H}_{10}$
C. $mathrmC_2mathrmH_2$ $\mathrm{C}_2\mathrm{H}_2$
D. $mathrmC_2mathrmH_6$ $\mathrm{C}_2\mathrm{H}_6$

Solution

### Related Formula

mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(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 =

80left(x + fracy4right)

mL. Unreacted

O_2

=

264 - 80left(x + fracy4right)

mL. Total residual volume =

V_CO_2 + V_unreacted \ O_2 = 224

mL.

80x + 264 - 80left(x + fracy4right) = 224

264 - frac80y4 = 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 - 80left(x + fracy4right) = 64

Substitute

y = 2

:

264 - 80left(x + frac12right) = 64

264 - 80x - 40 = 64

224 - 80x = 64

80x = 160 implies x = 2

The hydrocarbon is

mathrmC_2mathrmH_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 = frac16080 = 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 jee_main_2026_21_jan_morning Stoichiometry and Limiting Reagent

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

mathrmg \ mol^-1

of Ca–40, Cl–35.5, H–1]

A. $text0.35 mol of H_2text gas is evolved.$
B. $7.84 mathrm~Ltext of mathrmH_2text gas is evolved.$
C. $text33.3 g of mathrmCaCl_2text is produced.$
D. $textThe limiting reagent is calcium metal.$

Solution

### Related Formula

mathrmCa_(mathrms) + 2mathrmHCl_(mathrmg) longrightarrow mathrmCaCl_2(mathrms) + mathrmH_2(mathrmg)

### Core Logic Number of moles of Calcium (

n_Ca

) =

fractextGiven masstextMolar mass = frac14.040 = 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 textMolar 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 jee_main_2026_21_jan_evening Percentage Composition and Stoichiometry

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

textg mol^-1: textO = 16, textMg = 24, textP = 31

)

A. $(1) \ 50$ $(1) \ 50$
B. $(2) \ 30$ $(2) \ 30$
C. $(3) \ 20$ $(3) \ 20$
D. $(4) \ 40$ $(4) \ 40$

Solution

### Related Formula

\% text of textP = fractextMoles of textMg_2textP_2textO_7 times 2 times 31textMass of compound times 100

### Core Logic Molar mass of

textMg_2textP_2textO_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 text g/mol

.

textPercentage of P = fracleft(frac1.79222 times 2 times 31right)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

textMg_2textP_2textO_7

. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

Q56 jee_main_2026_22_january_morning Stoichiometry and Molar Volume

In the reaction,

2Al(s) + 6HCl(aq) rightarrow 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g)

A. $text11.2 L H_2(g) text at STP is produced for every mole of HCl consumed .$
B. $text67.2 L H_2(g) text at STP is produced for every mole of Al that reacts.$
C. $text12 L HCl(aq) is consumed for every 6L H_2(g) text produced.$
D. $text33.6 L H_2(g) text is produced regardless of temperature and pressure for every mole of Al that reacts.$

Solution

### Related Formula

textVolume of gas at STP = textMoles times 22.4 text L

### Core Logic From the balanced chemical equation:

2Al(s) + 6HCl(aq) rightarrow 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g)

6 moles of HCl produce 3 moles of

H_2

. Therefore, 1 mole of HCl produces

frac36 = 0.5

moles of

H_2

. Volume of

H_2

produced at STP for 1 mole of HCl:

V = 0.5 times 22.4text L = 11.2text L

### Step 1: Check other options Option (2): 2 moles of Al produce 3 moles of

H_2

. 1 mole Al produces 1.5 moles

H_2 = 1.5 times 22.4 = 33.6text L

(Incorrect). Option (4) is incorrect because volume depends on temperature and pressure. ### Pattern Recognition Standard stoichiometry. Directly map the mole ratio from the balanced equation to molar volume at standard conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

Q53 jee_main_2026_22_january_evening Limiting Reagent and Stoichiometry

textA + 2textB rightarrow textAB_2

36.0text g

of 'A' (Molar mass:

60text g mol^-1

) and

56.0text g

of 'B' (Molar mass:

80text g mol^-1

) are allowed to react. Which of the following statements are correct? (A) 'A' is the limiting reagent (B)

77.0text g

of

textAB_2

is formed (C) Molar mass of

textAB_2

is

140text g mol^-1

(D)

15.0text g

of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below:

A. C and D only
B. A and C only
C. B and D only
D. A and B only

Solution

### Related Formula

textMoles (n) = fractextGiven MasstextMolar Mass

textMolar Mass of textAB_2 = M_A + 2 M_B

### Core Logic Step 1: Calculate initial moles:

n_A = frac3660 = 0.6text mol

n_B = frac5680 = 0.7text mol

Step 2: Identify Limiting Reagent (LR):

textRatio for A = frac0.61 = 0.6, quad textRatio for B = frac0.72 = 0.35

Since ratio of B is smaller, B is the limiting reagent. Step 3: Evaluate product formed and remaining reactant:

textMolar mass of textAB_2 = 60 + 2(80) = 220text g mol^-1

textMoles of textAB_2 text formed = 0.35text mol

textMass of textAB_2 text formed = 0.35 times 220 = 77.0text g quad text(Statement B is correct)

textMoles of A reacted = 0.35text mol

textMoles of A remaining = 0.6 - 0.35 = 0.25text mol

textMass of A remaining = 0.25 times 60 = 15.0text g quad text(Statement D is correct)

### Pattern Recognition Sees: Initial masses of reactants in stoichiometric equation. Shortcut: Compare

n/textcoefficient

to find LR, then compute formed product mass and unreacted mass directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

Solution & Explanation

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions

Solution

Solution

Solution

Solution

Solution

More Some Basic Concepts of Chemistry Questions — jee_main_2026_22_january_evening

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Solution & Explanation

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions

Solution

Solution

Solution

Solution

Solution

More Some Basic Concepts of Chemistry Questions — jee_main_2026_22_january_evening

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back