What is number of atoms present in $2.24 \mathrm{dm}^3 \mathrm{NH}_{3(\mathrm{~g})}$ at STP?
What is number of atoms present in $2.24 \mathrm{dm}^3 \mathrm{NH}_{3(\mathrm{~g})}$ at STP?
- $6.022 \times 10^{22}$
- $2.4088 \times 10^{23}$
- $1.8066 \times 10^{22}$
- $6.022 \times 10^{23}$
Solution
$\begin{aligned} \text { For } \mathrm{NH}_3 & \\ 22.4 \mathrm{dm}^3 & =1 \mathrm{~mol} \\ & =6.022 \times 10^{22} \text { molecules } \\ & =6.022 \times 10^{22} \times 4 \text { atoms } \\ \therefore \quad 2.24 \mathrm{dm}^3 & =0.1 \mathrm{~mol} \\ & =0.6022 \times 10^{22} \text { molecules } \\ & =0.6022 \times 10^{22} \times 4 \text { atoms } \\ & =2.4088 \times 10^{23} \text { atoms }\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
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