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JEE MAIN 2026 · Some Basic Concepts in Chemistry

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 51

Updated
Single correct+4 / -1Answer matches NTA’s final key

Match List - I with List - II.

List - I (Mass of substance) List - II (Number of atoms)
A. 1.8 mg water I. 2×10−4×NA2 \times 10^{-4} \times N_A
B. 9.8 mg sulphuric acid II. 1.5×10−4×NA1.5 \times 10^{-4} \times N_A
C. 1.8 mg carbon III. 3×10−4×NA3 \times 10^{-4} \times N_A
D. 5.85 mg salt (NaCl) IV. 7×10−4×NA7 \times 10^{-4} \times N_A

Choose the correct answer from the options given below:

Choose one

Solution

Solution

Moles × atoms per formula unit, for each:

Substance Moles Atoms per unit Atoms
H2_2O, 1.8 mg 1.8×10−318=10−4\frac{1.8 \times 10^{-3}}{18} = 10^{-4} 3 3×10−4NA3 \times 10^{-4} N_A (III)
H2_2SO4_4, 9.8 mg 9.8×10−398=10−4\frac{9.8 \times 10^{-3}}{98} = 10^{-4} 7 7×10−4NA7 \times 10^{-4} N_A (IV)
C, 1.8 mg 1.8×10−312=1.5×10−4\frac{1.8 \times 10^{-3}}{12} = 1.5 \times 10^{-4} 1 1.5×10−4NA1.5 \times 10^{-4} N_A (II)
NaCl, 5.85 mg 5.85×10−358.5=10−4\frac{5.85 \times 10^{-3}}{58.5} = 10^{-4} 2 2×10−4NA2 \times 10^{-4} N_A (I)

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

Answer checked against the official answer key. The solution is our own working.

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