Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
Show Hint
- Po
- Pr
- Pb
- Pt
The Correct Option is B
Solution and Explanation
$\text{No. of atoms} = \frac{\text{Mass in g}}{\text{Molar Mass (g/mol)}} \times N_A$
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms.
$M_{Pb} = 209$
$M_{Pr} = 141$
$M_{Po} = 207$
$M_{Pt} = 195$
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