# If one has #4.50 xx 10^22# atoms of silver, what mass is available?

##### 1 Answer

#"8.06 g Ag"# ,

i.e. the mass of a handful of silver.

This is easiest in steps:

#"atoms" stackrel(N_A" ")(->) "mols" stackrel(M" ")(->) "mass"# where

#N_A = 6.0221413 xx 10^(23) "mol"^(-1)# isAvogadro's numberand#M# is themolar massin#"g/mol"# of the thing we are looking at.

So, to get to mols, we utilize the number of whatchamacallits in

#4.50 xx 10^(22) cancel"atoms" xx ("1 mol")/(6.0221413 xx 10^(23) cancel"whatevers")#

#=# #"0.0747 mols of any atom"#

To be specific to silver, we then need to use its **molar mass**

#0.0747 cancel"mols unknown element" xx ("107.8682 g Ag")/(cancel"1 mol Ag")#

#=# #color(blue)("8.06 g Ag")#