1.

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

Answer»

4g He
46g Na
0.40 g Ca
12g He

Solution :`"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.


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