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

Apr 16, 2016

The atomic mass of silver is $\text{107.87 amu}$ or $\text{g/mol}$.

#### 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$\frac{1}{12}$th part of a $\text{_6^12 "C}$ atom.

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

The mass of$\frac{1}{12}$th of a $\text{_6^12 "C}$ atom is taken as a unit of mass and is known as 1 amu or 1 u .

$\text{1 u} = \frac{1}{{N}_{A}} g = 1.66 \times {10}^{-} 24 g$,

where ${N}_{A}$ is Avogadro's Number.

Example

When we write atomic mass of $\text{Na}$, it is $\text{22.989 amu}$ for one atom, or $\text{22.989 g/mol}$ in general.

When we write mass of one atom of $\text{Na}$, it is:

$\text{22.989 amu" xx (1.66xx10^-24 "g")/"1 amu" = 3.82xx10^-24 "g}$ for one atom of $\text{Na}$.

Now the answer of the given question

Let the atomic mass of silver be ${M}_{\text{Ag}}$.

So one atom of silver weighs:

${M}_{\text{Ag" "amu" xx (1.66xx10^(-24) "g")/"1 amu" = 1.66xx10^(-24)M_"Ag}}$ $\text{g}$

${10}^{6}$ atoms of silver weighs:

$= 1.66 \times {10}^{-} 24 {M}_{\text{Ag}}$ $\text{g}$ $\times {10}^{6} = 1.66 \times {10}^{-} 18 {M}_{\text{Ag}}$ $\text{g}$

Equating this with the given value we can write:

$1.66 \times {10}^{-} 18 {M}_{\text{Ag}}$ $\text{g}$ $= 1.79 \times {10}^{-} 16 \text{g}$

$\therefore {M}_{\text{Ag" = 1.79/1.66xx10^2~~"107.87 amu" or "g/mol}}$

You can see this conversion work:

$\text{107.87 amu" = 107.87/(6.022xx10^(23)) "g}$, as stated at the top.

Therefore...

$\frac{107.87}{\cancel{6.022 \times {10}^{23}}} \text{g" xx (cancel(6.022xx10^(23)))/"mol}$

$= \text{107.87 g/mol}$

So, the atomic mass of Silver is $\text{107.87 amu}$ or $\text{g/mol}$.