The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atom?

Options

A. 4g He
B. 46g Na
C. 0.40 g Ca
D. 12g He

Answer

D

Explanation

`"4 g He"=(46)/(23)"mole = 1 mole"=N_(A)" atoms"`
`"46 g Na"=(46)/(23)"mole = 0.5 moles"=(N_(A))/(2)"atoms"`
`"0.40 g Ca"=(0.40)/(40)"mole"=10^(-2)" mole"=10^(-2)N_(A)" pr "(N_(A))/(100)" atoms"`
`"12 g He"=(12)/(4)" mole = 3 mole "=3N_(A)" atoms"`
Thus, 12 g He contains maximum number of atoms.