Q22 / 75 · JEE Main 2025 (3 Apr Shift 1) · Chemistry
Among $10^{-9} \mathrm{~g}$ (each) of the following elements, which one will have the highest number of atoms? Element : $\mathrm{Pb}, \mathrm{Po}, \mathrm{Pr}$ and Pt
- APo
- Correct answerBPr
- CPb
- DPt
Solution
$\text { No. of atoms }=\frac{\text { Massin } g}{\operatorname{Molar} \operatorname{Mas}(\mathrm{~g} / \mathrm{mol})} \times \mathrm{N}_{\mathrm{A}}$
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms.
- $\mathrm{M}_{\mathrm{Po}}=209$
- $\mathrm{M}_{\mathrm{Pr}}=141$
- $\mathrm{M}_{\mathrm{Pb}}=207$
- $\mathrm{M}_{\mathrm{Pt}}=195$
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms.
- $\mathrm{M}_{\mathrm{Po}}=209$
- $\mathrm{M}_{\mathrm{Pr}}=141$
- $\mathrm{M}_{\mathrm{Pb}}=207$
- $\mathrm{M}_{\mathrm{Pt}}=195$
Hence, the correct answer is option: 'B'.