Question

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

Answer 1

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.