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

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Aicharya · Accepted Answer

$	ext { No. of atoms }=\frac{	ext { Massin } g}{\operatorname{Molar} \operatorname{Mas}(\mathrm{~g} / \mathrm{mol})} 	imes \mathrm{N}_{\mathrm{A}}$ Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms. - $\mathrm{M}_{\mathrm{Po}}=209$ - $\mathrm{M}_{\mathrm{Pr}}=141$ - $\mathrm{M}_{\mathrm{Pb}}=207$ - $\mathrm{M}_{\mathrm{Pt}}=195$

Among $10^{-9} \mathrm{~g}$ (each) of the following elements, which one will have the highest number of…

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