AP EAMCET · Chemistry · Some Basic Concepts of Chemistry
The number of molecules of \(\mathrm{CO}_2\) liberated by the complete combustion of \(0.1 \mathrm{~g}\) atom of graphite in air is
- A \(3.01 \times 10^{22}\)
- B \(6.02 \times 10^{23}\)
- C \(6.02 \times 10^{22}\)
- D \(3.01 \times 10^{23}\)
Answer & Solution
Correct Answer
(C) \(6.02 \times 10^{22}\)
Step-by-step Solution
Detailed explanation
\(\underset{1\mathrm{~mol}}{\mathrm{C}(s)}+\mathrm{O}_2(g) \longrightarrow \underset{1\mathrm{~mol}}{\mathrm{CO}_2(g)}\) \(=6.023 \times 10^{23}\) \(\because 1\) mole of graphite on complete combustion gives \(\mathrm{CO}_2\) \(=6.023 \times 10^{23} \text { molecules }\)…
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