Question #474d7

Question

1554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-25T16:50:55

$1"g"$ helium contains the most particles.

The relative atomic mass in grams of an element contains the number of particles equal to the Avogadro Constant $=6.02xx10^(23)"mol"^(-1)$

So $m/A_r$ $xx6.02xx10^(23)$ will give you the number of particles, in this case atoms, which are present.

Using approx. $A_r$ values:

For $Al$ $n=(1)/(27)xx6.02xx10^(23)=2.23xx10^(22)$

For $C$ $n=1/12xx6.02xx10^(23)=5xx10^(22)$

For $He$ $n=1/4xx6.02xx10^(23)=15xx10^(22)$

For $Fe$ $n=1/56xx6.02xx10^(23)=1.07xx10^(22)$

So you can see from these data that 1g of He contains the most atoms.