Match List-I with List-II. begin array c l c l hline List-I & Mass of substance & List-II & Number of atoms \\ hline A. & 1.8\ mg water & I. & 2 × 10 -4 × N A \\ hline B. & 9.8\ mg sulphuric acid & II. & 1.5 × 10 -4 × N A \\ hline C. & 1.8\ mg carbon & III. & 3 × 10 -4 × N A \\ hline D. & 5.85\ mg salt (ce NaCl ) & IV. & 7 × 10 -4 × N A \\ hline end array Choose the correct answer from the options given below :

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2026 — 8 April, Evening Shift — Question 43

Match List-I with List-II.

\begin{array}{|c|l|c|l|} \hline \text{List-I} & \text{Mass of substance} & \text{List-II} & \text{Number of atoms} \\ \hline \text{A.} & 1.8\ \text{mg water} & \text{I.} & 2 \times 10^{-4} \times N_A \\ \hline \text{B.} & 9.8\ \text{mg sulphuric acid} & \text{II.} & 1.5 \times 10^{-4} \times N_A \\ \hline \text{C.} & 1.8\ \text{mg carbon} & \text{III.} & 3 \times 10^{-4} \times N_A \\ \hline \text{D.} & 5.85\ \text{mg salt }(\ce{NaCl}) & \text{IV.} & 7 \times 10^{-4} \times N_A \\ \hline \end{array}

Choose the correct answer from the options given below :

(A) No. of atoms in $1.8\,\mathrm{mg}$ water $=\frac{1.8\times10^{-3}}{18}\times N_{\mathrm{A}}\times3 =3\times10^{-4}N_{\mathrm{A}}$

(B) No. of atoms in $9.8\,\mathrm{mg}\,\mathrm{H_2SO_4}$

(D) No. of atoms in $5.85\,\mathrm{mg}\,\mathrm{NaCl}$ $=\frac{5.85\times10^{-3}}{58.5}\times N_{\mathrm{A}}\times2 =2\times10^{-4}N_{\mathrm{A}}$

Correct match: $A-\mathrm{III},\ B-\mathrm{IV},\ C-\mathrm{II},\ D-\mathrm{I}$

Answer key and solution verified before publishing.

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