MHT CET · Chemistry · Some Basic Concepts of Chemistry
What is number of atoms present in \(2.24 \mathrm{dm}^3 \mathrm{NH}_{3(\mathrm{~g})}\) at STP?
- A \(6.022 \times 10^{22}\)
- B \(2.4088 \times 10^{23}\)
- C \(1.8066 \times 10^{22}\)
- D \(6.022 \times 10^{23}\)
Answer & Solution
Correct Answer
(B) \(2.4088 \times 10^{23}\)
Step-by-step Solution
Detailed explanation
\(\text {For } \mathrm{NH}_3 \)
\( 22.4 \mathrm{dm}^3 =1 \mathrm{~mol} \)
\( =6.022 \times 10^{22} \text { molecules } \)
\( =6.022 \times 10^{22} \times 4 \text { atoms } \)
\( \therefore 2.24 \mathrm{dm}^3 =0.1 \mathrm{~mol} \)
\( =0.6022 \times 10^{22} \text { molecules } \)
\( =0.6022 \times 10^{22} \times 4 \text { atoms } \)
\( =2.4088 \times 10^{23} \text { atoms }\)
\( 22.4 \mathrm{dm}^3 =1 \mathrm{~mol} \)
\( =6.022 \times 10^{22} \text { molecules } \)
\( =6.022 \times 10^{22} \times 4 \text { atoms } \)
\( \therefore 2.24 \mathrm{dm}^3 =0.1 \mathrm{~mol} \)
\( =0.6022 \times 10^{22} \text { molecules } \)
\( =0.6022 \times 10^{22} \times 4 \text { atoms } \)
\( =2.4088 \times 10^{23} \text { atoms }\)
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