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

1 Answer
Truong-Son N.
Aug 2, 2017

#"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)# is Avogadro's number and #M# is the molar mass in #"g/mol"# of the thing we are looking at.

So, to get to mols, we utilize the number of whatchamacallits in #"1 mol"#:

#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 #M# to identify the atom as silver. Look up the molar mass on a periodic table to find #M = "107.8682 g/mol"#. Thus:

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

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

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