KCET · Chemistry · Some Basic Concepts of Chemistry
The number of oxygen atoms in \( 4.4 \mathrm{gm} \) of \( \mathrm{CO}_{2} \) is,
- A \( 1.2 \times 10^{23} \)
- B \( 6 \times 10^{22} \)
- C \( 6 \times 10^{23} \)
- D \( 12 \times 10^{23} \)
Answer & Solution
Correct Answer
(A) \( 1.2 \times 10^{23} \)
Step-by-step Solution
Detailed explanation
\( 1 \) molecule of \( \mathrm{CO}_{2} \) contains \( 2 \) oxygen atoms
Therefore, \( 1 \) mole of \( \mathrm{CO}_{2} \) contains \( 6.022 \times 10^{23} \times 2 \) oxygen atoms.
\( \Rightarrow 44 g \) of \( C O_{2}=6.022 \times 10^{23} \times 2 \) oxygen atoms \( \Rightarrow 4.4 g \) of \( C O_{2}=\frac{6.022 \times 10^{23} \times 2 \times 4.4}{44}=12.044 \times 10^{22} \) oxygenatoms.
Hence \( 4.4 \mathrm{~g} \) of \( \mathrm{CO}_{2} \) will contain \( 1.2044 \times 10^{23} \) oxygen atoms.
Therefore, \( 1 \) mole of \( \mathrm{CO}_{2} \) contains \( 6.022 \times 10^{23} \times 2 \) oxygen atoms.
\( \Rightarrow 44 g \) of \( C O_{2}=6.022 \times 10^{23} \times 2 \) oxygen atoms \( \Rightarrow 4.4 g \) of \( C O_{2}=\frac{6.022 \times 10^{23} \times 2 \times 4.4}{44}=12.044 \times 10^{22} \) oxygenatoms.
Hence \( 4.4 \mathrm{~g} \) of \( \mathrm{CO}_{2} \) will contain \( 1.2044 \times 10^{23} \) oxygen atoms.
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