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ChemMedMain 2026 • 8th Apr - Shift 2

Match Mass of Substance to Total Number of Atoms

Match List - I with List - II.

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

Choose the correct answer from the options given below :

Options

A

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

B

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

C

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

Correct
D

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

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometry

Step-by-Step Solution

To determine the correct matching between List - I and List - II, we calculate the total number of atoms for each given mass of the substance.

The total number of atoms in a given sample is calculated using the formula: Total number of atoms=Moles of the substance×Atomicity×NA\text{Total number of atoms} = \text{Moles of the substance} \times \text{Atomicity} \times N_A where Moles=Mass in gramsMolar mass in g/mol\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}}, and Atomicity\text{Atomicity} is the number of atoms per molecule or formula unit.

Calculation for A:

  • Substance: Water (H2O\text{H}_2\text{O})
  • Mass: 1.8 mg=1.8×10−3 g1.8 \text{ mg} = 1.8 \times 10^{-3} \text{ g}
  • Molar Mass: 18 g/mol18 \text{ g/mol}
  • Atomicity: 2+1=32 + 1 = 3 (2 Hydrogen atoms and 1 Oxygen atom)

Moles of H2O=1.8×10−3 g18 g/mol=10−4 mol\text{Moles of } \text{H}_2\text{O} = \frac{1.8 \times 10^{-3} \text{ g}}{18 \text{ g/mol}} = 10^{-4} \text{ mol}

Total number of atoms=10−4×3×NA=3×10−4×NA\text{Total number of atoms} = 10^{-4} \times 3 \times N_A = 3 \times 10^{-4} \times N_A

Thus, A matches with III.

Calculation for B:

  • Substance: Sulphuric acid (H2SO4\text{H}_2\text{SO}_4)
  • Mass: 9.8 mg=9.8×10−3 g9.8 \text{ mg} = 9.8 \times 10^{-3} \text{ g}
  • Molar Mass: 2(1)+32+4(16)=98 g/mol2(1) + 32 + 4(16) = 98 \text{ g/mol}
  • Atomicity: 2+1+4=72 + 1 + 4 = 7

Moles of H2SO4=9.8×10−3 g98 g/mol=10−4 mol\text{Moles of } \text{H}_2\text{SO}_4 = \frac{9.8 \times 10^{-3} \text{ g}}{98 \text{ g/mol}} = 10^{-4} \text{ mol}

Total number of atoms=10−4×7×NA=7×10−4×NA\text{Total number of atoms} = 10^{-4} \times 7 \times N_A = 7 \times 10^{-4} \times N_A

Thus, B matches with IV.

Calculation for C:

  • Substance: Carbon (C\text{C})
  • Mass: 1.8 mg=1.8×10−3 g1.8 \text{ mg} = 1.8 \times 10^{-3} \text{ g}
  • Molar Mass: 12 g/mol12 \text{ g/mol}
  • Atomicity: 11

Moles of C=1.8×10−3 g12 g/mol=0.15×10−3 mol=1.5×10−4 mol\text{Moles of } \text{C} = \frac{1.8 \times 10^{-3} \text{ g}}{12 \text{ g/mol}} = 0.15 \times 10^{-3} \text{ mol} = 1.5 \times 10^{-4} \text{ mol}

Total number of atoms=1.5×10−4×1×NA=1.5×10−4×NA\text{Total number of atoms} = 1.5 \times 10^{-4} \times 1 \times N_A = 1.5 \times 10^{-4} \times N_A

Thus, C matches with II.

Calculation for D:

  • Substance: Salt (NaCl\text{NaCl})
  • Mass: 5.85 mg=5.85×10−3 g5.85 \text{ mg} = 5.85 \times 10^{-3} \text{ g}
  • Molar Mass: 23+35.5=58.5 g/mol23 + 35.5 = 58.5 \text{ g/mol}
  • Atomicity: 1+1=21 + 1 = 2 (1 Na1\text{ Na} and 1 Cl1\text{ Cl})

Moles of NaCl=5.85×10−3 g58.5 g/mol=10−4 mol\text{Moles of } \text{NaCl} = \frac{5.85 \times 10^{-3} \text{ g}}{58.5 \text{ g/mol}} = 10^{-4} \text{ mol}

Total number of atoms=10−4×2×NA=2×10−4×NA\text{Total number of atoms} = 10^{-4} \times 2 \times N_A = 2 \times 10^{-4} \times N_A

Thus, D matches with I.

Conclusion:

The correct matching is: A-III, B-IV, C-II, D-I\text{A-III, B-IV, C-II, D-I}

This corresponds to Option C.

Match Mass of Substance to Total Number of Atoms | Chemistry PYQ Solution - JEE Challenger