How many atoms exist in a $23*g$ mass of sodium metal?

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Answer 1 · 2016-09-05T17:22:59

Yes (and sort of no)

1 mole of Na = $6.022140857(74)×10^"23"$ atoms .

Note the use of the word atoms .

1 mole of $Na^+$ would be $6.022140857(74)×10^"23"$ ions .

I may seem to be overly pedantic in this, but science is all about accuracy.

The number - $6.022140857(74)×10^"23"$ - is called the Avogadro constant and you can read more about it on Wikipedia.

Answer 2 · 2016-09-05T17:23:00

$1*"mole"$ of sodium metal $-=$ $6.022xx10^23$ atoms NOT ions.

$1*"mole"$ of $Na^+$ $-=$ $6.022xx10^23$ ions.

Of course we can make ions from atoms and atoms from ions and vice versa.

$Na(g) rarr Na^(+)(g) + e^(-)(g)$ .

Given $22.99*g$ of metal, there are molar quantities of atoms, ions, and electrons. The given stoichiometry demands this.

How many atoms exist in a 23*g23*g mass of sodium metal?