Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Some Basic Concepts of Chemistry question

2023 · 24 Jan · Shift 1 · Q15
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Some Basic Concepts of Chemistry
  5. /2023 · 24 Jan · Shift 1 · Q15

Some Basic Concepts of Chemistry question

2023 · 24 Jan · Shift 1 · Q15

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
When Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is heated in presence of oxygen, it converts to Fe2O3\mathrm{Fe_2O_3}Fe2​O3​. The number of correct statement/s from the following is ‾\underline{\hspace{2cm}}​ A. The equivalent weight of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is Molecular weight0.79{{\mathrm{Molecular\,weight}} \over {0.79}}0.79Molecularweight​ B. The number of moles of Fe 2+^{2+}2+ and Fe 3+^{3+}3+ in 1 mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is 0.79 and 0.14 respectively C. Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is metal deficient with lattice comprising of cubic closed packed arrangement of O 2−^{2-}2− ions D. The % composition of Fe 2+^{2+}2+ and Fe 3+^{3+}3+ in Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is 85% and 15% respectively
Numerical answer
View written solutionFree

Correct answer: 4

  1. Find the oxidation-state distribution in Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O

Let in 111 mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O:

  • moles of Fe2+=x\mathrm{Fe^{2+}} = xFe2+=x
  • moles of Fe3+=y\mathrm{Fe^{3+}} = yFe3+=y

Since total Fe per formula unit is 0.930.930.93, x+y=0.93x+y=0.93x+y=0.93

For electrical neutrality with one O2−\mathrm{O^{2-}}O2−: 2x+3y=22x+3y=22x+3y=2

Solving: From x=0.93−yx=0.93-yx=0.93−y, 2(0.93−y)+3y=22(0.93-y)+3y=22(0.93−y)+3y=2 1.86−2y+3y=21.86-2y+3y=21.86−2y+3y=2 y=0.14y=0.14y=0.14 Hence, x=0.93−0.14=0.79x=0.93-0.14=0.79x=0.93−0.14=0.79

So in 111 mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O:

  • Fe2+=0.79\mathrm{Fe^{2+}} = 0.79Fe2+=0.79 mol
  • Fe3+=0.14\mathrm{Fe^{3+}} = 0.14Fe3+=0.14 mol

  1. Check statement B

Statement B says the number of moles of Fe2+\mathrm{Fe^{2+}}Fe2+ and Fe3+\mathrm{Fe^{3+}}Fe3+ in 111 mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O are 0.790.790.79 and 0.140.140.14 respectively.

This matches our calculation.

✅ B is correct


  1. Check statement D

Percentage composition among total Fe ions: % Fe2+=0.790.93×100≈84.95%≈85%\%\,\mathrm{Fe^{2+}}=\frac{0.79}{0.93}\times 100\approx 84.95\%\approx 85\%%Fe2+=0.930.79​×100≈84.95%≈85% % Fe3+=0.140.93×100≈15.05%≈15%\%\,\mathrm{Fe^{3+}}=\frac{0.14}{0.93}\times 100\approx 15.05\%\approx 15\%%Fe3+=0.930.14​×100≈15.05%≈15%

So statement D is correct.

✅ D is correct


  1. Check statement C

Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O has fewer Fe atoms than stoichiometric FeO\mathrm{FeO}FeO, so it is metal deficient.

Also, wustite-type structure is based on a cubic close packed arrangement of O2−\mathrm{O^{2-}}O2− ions with Fe occupying octahedral holes, and some Fe vacancies are present.

✅ C is correct


  1. Check statement A (equivalent weight)

Equivalent weight in this oxidation process is: Equivalent weight=molar massn-factor\text{Equivalent weight} = \frac{\text{molar mass}}{n\text{-factor}}Equivalent weight=n-factormolar mass​

Here Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O is oxidized to Fe2O3\mathrm{Fe_2O_3}Fe2​O3​, where all iron becomes Fe3+\mathrm{Fe^{3+}}Fe3+.

Initially in 111 mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O:

  • 0.790.790.79 mol of Fe2+\mathrm{Fe^{2+}}Fe2+ are oxidized to Fe3+\mathrm{Fe^{3+}}Fe3+
  • each such Fe loses 111 electron

So total electrons lost per mole of Fe0.93O\mathrm{Fe_{0.93}O}Fe0.93​O: n=0.79n = 0.79n=0.79

Therefore, Equivalent weight=molar mass of Fe0.93O0.79\text{Equivalent weight} = \frac{\text{molar mass of }\mathrm{Fe_{0.93}O}}{0.79}Equivalent weight=0.79molar mass of Fe0.93​O​

✅ A is correct


  1. Total number of correct statements

Correct statements are: A, B, C, DA,\ B,\ C,\ DA, B, C, D

Hence, the number of correct statements is 4\boxed{4}4​


  1. Comparison with stored answer

Stored correct answer: 444

Our derived answer: 444

They match.

PreviousNext

More from Some Basic Concepts of Chemistry

  • 5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? ​ Given : Molar Mass of Na, O and H is…2023 · Numerical
  • The number of units, which are used to express concentration of solutions from the following is ​ Mass percent, Mole, Mole fraction, Molarity, ppm, Molality2023 · Numerical
  • In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ​ (Nearest Integer) (Given : Atomic mass Ba: 137 u, S: 32 u, O: 16 u)2023 · Numerical
  • '25 volume' hydrogen peroxide means2023 · MCQ
  • Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is ​ (Given : Molar mass of A = 84 g mol −1)2023 · Numerical
  • What is the mass ratio of ethylene glycol (C2​H6​O2​, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?2023 · MCQ
  • The volume of HCl, containing 73 g L −1, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is ​ mL. (Nearest Integer) (Given : molar masses of Na, Cl, O, H, are…2023 · Numerical
  • When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO 2​ was produced. The molar mass of compound is ​ g mol −1 (Nearest integer).2023 · Numerical