Match List - I with List - II. 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 x 10^-4 x NA II. 1.5 x 10^-4 x NA III. 3 x 10^-4 x NA IV. 7 x 10^-4 x NA Choose the correct answer from the options given below :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q51 – JEE-MAIN 2026

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Home Chemistry Basic ChemistryQuestion #47

Match List - I with List - II.

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

Choose the correct answer from the options given below :

Clue 1 of 3: Mole Concept

To find the number of atoms, you need to first calculate the number of moles of the substance and then multiply by Avogadro's number and the number of atoms per molecule/formula unit.

🧠 Full Solution Path

Step 1: Calculate Moles and Atoms per Molecule for each substance✦ Active

First, convert the given mass from mg to g and calculate the number of moles for each substance using their respective molar masses. Then, identify the number of atoms present in one molecule or formula unit of each substance.

\begin{array}{ll} A. Water (H_{2} O) : & Molar mass = 18 g/mol \\ Moles = \frac{1.8 \times 10^{- 3} g}{18 g/mol} = 1 \times 10^{- 4} mol \\ Atoms per molecule = 3 (2H + 1O) \\ B. Sulphuric acid (H_{2} S O_{4}) : & Molar mass = 98 g/mol \\ Moles = \frac{9.8 \times 10^{- 3} g}{98 g/mol} = 1 \times 10^{- 4} mol \\ Atoms per molecule = 7 (2H + 1S + 4O) \\ C. Carbon (C) : & Molar mass = 12 g/mol \\ Moles = \frac{1.8 \times 10^{- 3} g}{12 g/mol} = 1.5 \times 10^{- 4} mol \\ Atoms per molecule = 1 (monatomic) \\ D. Salt (N a C l) : & Molar mass = 58.5 g/mol \\ Moles = \frac{5.85 \times 10^{- 3} g}{58.5 g/mol} = 1 \times 10^{- 4} mol \\ Atoms per formula unit = 2 (1Na + 1Cl) \end{array}

Step 2: Calculate Total Number of Atoms for each substance○ Expand

Multiply the moles by Avogadro's number ( $N_{A}$ ) and the number of atoms per molecule/formula unit to find the total number of atoms.

\begin{array}{ll} A. Water: & (1 \times 10^{- 4}) \times N_{A} \times 3 = 3 \times 10^{- 4} N_{A} \\ B. Sulphuric acid: & (1 \times 10^{- 4}) \times N_{A} \times 7 = 7 \times 10^{- 4} N_{A} \\ C. Carbon: & (1.5 \times 10^{- 4}) \times N_{A} \times 1 = 1.5 \times 10^{- 4} N_{A} \\ D. Salt (NaCl): & (1 \times 10^{- 4}) \times N_{A} \times 2 = 2 \times 10^{- 4} N_{A} \end{array}

Step 3: Match List-I with List-II and identify the correct option○ Expand

Based on the calculations, the matches are: A-III, B-IV, C-II, D-I. Comparing this with the given options, Option 3 is the correct match.

\begin{array}{ll} A. 3 \times 10^{- 4} N_{A} & \to III \\ B. 7 \times 10^{- 4} N_{A} & \to IV \\ C. 1.5 \times 10^{- 4} N_{A} & \to II \\ D. 2 \times 10^{- 4} N_{A} & \to I \end{array}

💡 Teacher's Secret Hint

Double-check unit conversions (mg to g) and the count of atoms in each molecule/formula unit.

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Chemistry Question 51 – JEE-MAIN 2026

Clue 1 of 3: Mole Concept

🧠 Full Solution Path