Among 10^{-9} g (each) of the following elem | vedclass

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(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$.Since the mass is

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$Pr$

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(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$.Since the mass is constant $(10^{-9} \ g)$ for all elements,the number of atoms is inversely proportional to the molar mass.Therefore,the element with the lowest molar mass will have the highest number of atoms.The molar masses are:$\bullet M_{Po} = 209 \ g/mol$$\bullet M_{Pr} = 141 \ g/mol$$\bullet M_{Pb} = 207 \ g/mol$$\bullet M_{Pt} = 195 \ g/mol$Comparing these values,$Pr$ has the lowest molar mass $(141 \ g/mol)$.Thus,$Pr$ will have the highest number of atoms.

Isotope	$\%$ Natural abundance	Molar mass
$^1H$	$99.985$	$1$
$^2H$	$0.015$	$2$

Among $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?
Element: $Pb, Po, Pr$ and $Pt$

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Among $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?Element: $Pb, Po, Pr$ and $Pt$