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Some Basic Concepts of Chemistry · Question 7

JEE Main 2026 (Online) 4th April Evening ShiftMCQMedium📈 Topic in 19% papers+4−1
1

The correct order of total number of atoms present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP is :

  1. C > A > B
  2. C > B > A
  3. B > C > A
  4. B > A > C

Answer

(4) B > A > C

Solution

Number of atoms = moles of molecules × atoms per molecule (× NA, common to all)

(A) Cyclohexane, C6H12 (6 + 12 = 18 atoms per molecule)

⇒ atoms = 2 × 18 = 36 mol of atoms

(B) Sucrose, C12H22O11 (45 atoms per molecule), M = 342 g mol-1

⇒ n = 684342 = 2 mol ⇒ atoms = 2 × 45 = 90 mol of atoms

(C) Dihydrogen, H2; at STP (1 bar, 273.15 K) molar volume = 22.7 L

⇒ n = 90.822.7 = 4 mol ⇒ atoms = 4 × 2 = 8 mol of atoms

∴ B (90 NA) > A (36 NA) > C (8 NA)

Answer: (4)

JEE Main 2026 (Online) 4th April Evening Shift

Keep practising — Some Basic Concepts of Chemistry

2

The mass of iron converted into Fe3O4 by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol-1 respectively) Assume iron is present in excess :

  1. 2.1 g
  2. 4.2 g
  3. 21 g
  4. 42 g
JEE Main 2026 (Online) 2nd April Morning Shift
3

The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1. It has two carbon atoms. The weight (in g) of CO2(g) formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol-1 C : 12, H : 1, O : 16)

  1. 5.68
  2. 11.44
  3. 22.74
  4. 17.05
JEE Main 2026 (Online) 2nd April Evening Shift
4

For the given reaction; CaCO3 + 2HCl CaCl2 + H2O + CO2 If 90 g CaCO3 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL-1, then which of the following option is correct? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol-1 respectively.

  1. 64.97 g of HCl remains unreacted
  2. 60.32 g of HCl remains unreacted
  3. 32.85 g of CaCO3 remains unreacted
  4. 97.30 g of HCl reacted
JEE Main 2026 (Online) 28th January Evening Shift
5

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3 g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The % yield of this reaction is :-

  1. 59.5\%
  2. 72.5\%
  3. 97.5\%
  4. 81.5\%
JEE Main 2026 (Online) 24th January Evening Shift
6

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba(OH)2 solution is used when the HCl solution was titrated against it. The numerical value of x is 10-1. (Nearest integer) Given : Molar mass of HCl and Ba(OH)2 are 36.5 and 171.0 g mol-1 respectively.

JEE Main 2026 (Online) 23rd January Morning Shift
7

In the reaction, 2Al(s) + 6HCl(aq) 2Al3+(aq) + 6Cl-(aq) + 3H2(g)

  1. 12 L HCl(aq) is consumed for every 6 L H2(g) produced.
  2. 11.2 L H2(g) at STP is produced for every mole of HCl consumed.
  3. 33.6 L H2(g) is produced regardless of temperature and pressure for every mole of Al that reacts.
  4. 67.2 L H2(g) at STP is produced for every mole of Al that reacts.
JEE Main 2026 (Online) 22nd January Morning Shift
8

At T(K), 100 g of 98\% H2SO4 (w/w) aqueous solution is mixed with 100 g of 49\% H2SO4 (w/w) aqueous solution. What is the mole fraction of H2SO4 in the resultant solution? (Given : Atomic mass H = 1 u; S = 32 u; O = 16 u). (Assume that temperature after mixing remains constant)

  1. 0.9
  2. 0.663
  3. 0.1
  4. 0.337
JEE Main 2026 (Online) 22nd January Evening Shift
9

A + 2B AB2 36.0 g of 'A' (Molar mass : 60 g mol-1) and 56.0 g of 'B' (Molar mass : 80 g mol-1) are allowed to react. Which of the following statements are correct ? Choose the correct answer from the options given below :

  1. A and C Only
  2. A and B Only
  3. C and D Only
  4. B and D Only
JEE Main 2026 (Online) 22nd January Evening Shift
10

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 g mol-1 of Ca – 40, Cl – 35.5, H – 1]

  1. The limiting reagent is calcium metal.
  2. 33.3 g of CaCl2 is produced.
  3. 7.84 L of H2 gas is evolved.
  4. 0.35 mol of H2 gas is evolved.
JEE Main 2026 (Online) 21st January Morning Shift
11

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 :

  1. C2H2
  2. C2H4
  3. C4H10
  4. C2H6
JEE Main 2026 (Online) 21st January Morning Shift

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