MHT CET · Chemistry · Some Basic Concepts of Chemistry
Which among the following elements has highest number of atoms in \(1 \mathrm{~g}\) each ? (at. no. \(\mathrm{Au}=197, \mathrm{Na}=23, \mathrm{Cu}=63 \cdot 5, \mathrm{Fe}=56\) )
- A \(\mathrm{Cu}_{(\mathrm{s})}\)
- B \(\mathrm{Na}_{(\mathrm{s})}\)
- C \(\mathrm{Au}_{(\mathrm{s})}\)
- D \(\mathrm{Fe}_{(\mathrm{s})}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{Na}_{(\mathrm{s})}\)
Step-by-step Solution
Detailed explanation
i. \(1 \mathrm{~g} \mathrm{Fe}=\frac{1}{56} \mathrm{~mol}=\frac{1}{56} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Fe}\)
ii. \(\lg \mathrm{Au}=\frac{1}{197} \mathrm{~mol}=\frac{1}{197} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Au}\)
iii. \(1 \mathrm{~g} \mathrm{Na}=\frac{1}{23} \mathrm{~mol}=\frac{1}{23} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Na}\)
iv. \(\lg \mathrm{Cu}=\frac{1}{63.5} \mathrm{~mol}=\frac{1}{63.5} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Cu}\)
Hence, \(1 \mathrm{~g}\) of Na has highest number of atoms.
ii. \(\lg \mathrm{Au}=\frac{1}{197} \mathrm{~mol}=\frac{1}{197} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Au}\)
iii. \(1 \mathrm{~g} \mathrm{Na}=\frac{1}{23} \mathrm{~mol}=\frac{1}{23} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Na}\)
iv. \(\lg \mathrm{Cu}=\frac{1}{63.5} \mathrm{~mol}=\frac{1}{63.5} \times 6.022 \times 10^{23}\) atoms of \(\mathrm{Cu}\)
Hence, \(1 \mathrm{~g}\) of Na has highest number of atoms.
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