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(C) The number of molecules is calculated as: $	ext{Number of molecules} = 	ext{moles} 	imes N_A = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A

Vedclass · Accepted Answer

$8 \ g$ $H_2$

Vedclass · Answer

(C) The number of molecules is calculated as: $	ext{Number of molecules} = 	ext{moles} 	imes N_A = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A$. For $A$: $44 \ g$ $CO_2$ (molar mass $= 44 \ g/mol$): $	ext{moles} = 44/44 = 1 \ mol$. For $B$: $48 \ g$ $O_2$ (molar mass $= 32 \ g/mol$): $	ext{moles} = 48/32 = 1.5 \ mol$. For $C$: $8 \ g$ $H_2$ (molar mass $= 2 \ g/mol$): $	ext{moles} = 8/2 = 4 \ mol$. For $D$: $64 \ g$ $SO_2$ (molar mass $= 64 \ g/mol$): $	ext{moles} = 64/64 = 1 \ mol$. Since $C$ has the highest number of moles $(4 \ mol)$,it contains the maximum number of molecules.

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)	$IV$. $7 \times 10^{-4} \times N_{A}$

Maximum number of molecules will be in $:-$

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