Q1 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry

Match List-I with List-II.
List-I
Mass of substance		List-II
Number of atoms
A. $1.8$ mg waterI. $2\times 10^{-4}\times N_A$
B. $9.8$ mg sulphuric acidII. $1.5\times 10^{-4}\times N_A$
C. $1.8$ mg carbonIII. $3\times 10^{-4}\times N_A$
D. $5.85$ mg salt (NaCl)IV. $7\times 10^{-4}\times N_A$
Choose the correct answer from the options given below:

  • A
    A-IV, B-III, C-I, D-II
  • B
    A-III, B-II, C-IV, D-I
  • Correct answer
    C
    A-III, B-IV, C-II, D-I
  • D
    A-III, B-IV, C-I, D-II

Solution

For A: Molar mass of $H_2O = 18$ g/mol. Moles of $H_2O = \dfrac{1.8 \times 10^{-3}}{18} = 10^{-4}$. Since each $H_2O$ molecule has $3$ atoms, the number of atoms = $3 \times 10^{-4} \times N_A$.

For B: Molar mass of $H_2SO_4 = 98$ g/mol. Moles of $H_2SO_4 = \dfrac{9.8 \times 10^{-3}}{98} = 10^{-4}$. Since each $H_2SO_4$ molecule has $7$ atoms, the number of atoms = $7 \times 10^{-4} \times N_A$.

For C: Molar mass of $C = 12$ g/mol. Moles of $C = \dfrac{1.8 \times 10^{-3}}{12} = 1.5 \times 10^{-4}$. The number of atoms = $1.5 \times 10^{-4} \times N_A$.

For D: Molar mass of $NaCl = 58.5$ g/mol. Moles of $NaCl = \dfrac{5.85 \times 10^{-3}}{58.5} = 10^{-4}$. Since each $NaCl$ formula unit has $2$ atoms, the number of atoms = $2 \times 10^{-4} \times N_A$.

Thus, the correct matching is A-III, B-IV, C-II, D-I.

Answer: A-III, B-IV, C-II, D-I

Hence, the correct answer is option: 'C'.

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