Match List-I with List-II. List-I (Mass of su | vedclass

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(C) To find the number of atoms, we calculate the moles of each substance and multiply by the number of atoms per molecule/formula unit and Avogadro's

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

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(C) To find the number of atoms, we calculate the moles of each substance and multiply by the number of atoms per molecule/formula unit and Avogadro's number $(N_{A})$.$A$. $1.8 	ext{ mg}$ water $(H_{2}O, M=18 	ext{ g/mol})$: Moles = $1.8 	imes 10^{-3} 	ext{ g} / 18 	ext{ g/mol} = 10^{-4} 	ext{ mol}$. Each molecule has $3$ atoms $(2H + 1O)$. Total atoms = $10^{-4} 	imes 3 	imes N_{A} = 3 	imes 10^{-4} 	imes N_{A}$ $(III)$.$B$. $9.8 	ext{ mg}$ sulphuric acid $(H_{2}SO_{4}, M=98 	ext{ g/mol})$: Moles = $9.8 	imes 10^{-3} 	ext{ g} / 98 	ext{ g/mol} = 10^{-4} 	ext{ mol}$. Each molecule has $7$ atoms $(2H + 1S + 4O)$. Total atoms = $10^{-4} 	imes 7 	imes N_{A} = 7 	imes 10^{-4} 	imes N_{A}$ $(IV)$.$C$. $1.8 	ext{ mg}$ carbon $(C, M=12 	ext{ g/mol})$: Moles = $1.8 	imes 10^{-3} 	ext{ g} / 12 	ext{ g/mol} = 1.5 	imes 10^{-4} 	ext{ mol}$. Total atoms = $1.5 	imes 10^{-4} 	imes N_{A}$ $(II)$.$D$. $5.85 	ext{ mg}$ salt (NaCl, $M=58.5 	ext{ g/mol}$): Moles = $5.85 	imes 10^{-3} 	ext{ g} / 58.5 	ext{ g/mol} = 10^{-4} 	ext{ mol}$. Each formula unit has $2$ atoms $(1Na + 1Cl)$. Total atoms = $10^{-4} 	imes 2 	imes N_{A} = 2 	imes 10^{-4} 	imes N_{A}$ $(I)$.Thus, the correct match is $A-III, B-IV, C-II, D-I$.

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

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