What is the mass of #7.83 * 10^23# atoms Cl?

1 Answer
anor277
Nov 29, 2015

Slightly over 35.5 grams.

Explanation:

If I have Avogadro's number of chlorine atoms, I necessarily have 35.45 g chlorine.

Here we have a number that is in excess of Avogadro's number, #6.022xx10^(23)# #=# #N_A#.

Since I know that if I have #6.022xx10^(23)# atoms of chlorine I have a mass of #35.45# #g# of chlorine, then all I have to do is give the product, #35.45*g# #xx(7.83xx10^(23))/(6.022xx10^(23))# #=# #??# #g#.

Related questions
See all questions in The Mole
Impact of this question
3090 views around the world
You can reuse this answer
Creative Commons License