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JEE Main 2026 · 8 Apr · Shift 2, Chemistry question 51

Match List - I with List - II.

Choose the correct answer from the options given below :

List - I: Mass of substance

  • A
    1.8 mg water
  • B
    9.8 mg sulphuric acid
  • C
    1.8 mg carbon
  • D
    5.85 mg salt (NaCl)

List - II: Number of atoms

  • I
    2×10−4×NA2\times10^{-4}\times N_A
  • II
    1.5×10−4×NA1.5\times10^{-4}\times N_A
  • III
    3×10−4×NA3\times10^{-4}\times N_A
  • IV
    7×10−4×NA7\times10^{-4}\times N_A
Choose one option
Answer and worked solution

Official answer: C

Atoms in each sample.

  • A: 1.8 mg water is 10−410^{-4} mol, 3 atoms each: 3×10−4NA3\times10^{-4}N_A (III).
  • B: 9.8 mg HX2SOX4\ce{H2SO4} is 10−410^{-4} mol, 7 atoms each: 7×10−4NA7\times10^{-4}N_A (IV).
  • C: 1.8 mg carbon is 1.5×10−41.5\times10^{-4} mol (II).
  • D: 5.85 mg NaCl is 10−410^{-4} mol, 2 atoms each: 2×10−4NA2\times10^{-4}N_A (I).

Match: A-III, B-IV, C-II, D-I.

(C)

Topics:Mole concept
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