If #1000000# silver atoms weigh #1.79 xx 10^(-16) "g"#, what is the atomic mass of silver?

1 Answer

Answer:

The atomic mass of silver is #"107.87 amu"# or #"g/mol"#.

Explanation:

Actually, atomic mass is a relative mass of an atom of an element and is represented by the ratio of actual mass of an atom of the element to the mass of# 1/12#th part of a #""_6^12 "C"# atom.

So atomic mass is a "unitless" quantity. But actual mass of an atom is not unit-less quantity.

The mass of# 1/12#th of a #""_6^12 "C"# atom is taken as a unit of mass and is known as 1 amu or 1 u .

#"1 u" = 1/(N_A) g=1.66xx10^-24 g#,

where #N_A# is Avogadro's Number.

Example

When we write atomic mass of #"Na"#, it is #"22.989 amu"# for one atom, or #"22.989 g/mol"# in general.

When we write mass of one atom of #"Na"#, it is:

#"22.989 amu" xx (1.66xx10^-24 "g")/"1 amu" = 3.82xx10^-24 "g"# for one atom of #"Na"#.

Now the answer of the given question

Let the atomic mass of silver be #M_"Ag"#.

So one atom of silver weighs:

#M_"Ag" "amu" xx (1.66xx10^(-24) "g")/"1 amu" = 1.66xx10^(-24)M_"Ag"# #"g"#

#10^6# atoms of silver weighs:

#= 1.66xx10^-24M_"Ag"# #"g"# #xx 10^6 = 1.66xx10^-18 M_"Ag"# #"g"#

Equating this with the given value we can write:

#1.66xx10^-18 M_"Ag"# #"g"# #= 1.79xx10^-16 "g"#

#:. M_"Ag" = 1.79/1.66xx10^2~~"107.87 amu" or "g/mol"#

You can see this conversion work:

#"107.87 amu" = 107.87/(6.022xx10^(23)) "g"#, as stated at the top.

Therefore...

#107.87/(cancel(6.022xx10^(23))) "g" xx (cancel(6.022xx10^(23)))/"mol"#

#= "107.87 g/mol"#

So, the atomic mass of Silver is #"107.87 amu"# or #"g/mol"#.

Please inform if not clear.