What is the number of atoms present in $2.24 \ dm^3$ of $NH_3$ $(g)$ at $STP$?
- A$6.022 \times 10^{22}$
- B$2.4088 \times 10^{23}$
- C$1.8066 \times 10^{22}$
- D$6.022 \times 10^{23}$
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Assertion : The number of $O$ atoms in $16 \ g$ of oxygen and $16 \ g$ of ozone is same.
Reason : Each of the species represents $1 \ g$ molecule of oxygen.
Reason : Each of the species represents $1 \ g$ molecule of oxygen.
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Match List-$I$ with List-$II$. 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) $IV$. $7 \times 10^{-4} \times N_{A}$
| 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) | $IV$. $7 \times 10^{-4} \times N_{A}$ |
DifficultJEE MAIN 2026
View SolutionNumber of molecules in $100 \ mL$ of each of $O_2$,$NH_3$,and $CO_2$ at $STP$ are
Medium
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