If $1000000$ $1000000$ silver atoms weigh $1.79 \times 10^{- 16} g$ $1.79 xx 10^(-16) "g"$ , what is the atomic mass of silver?

Question

If $1000000$ $1000000$ silver atoms weigh $1.79 \times 10^{- 16} g$ $1.79 xx 10^(-16) "g"$ , what is the atomic mass of silver?

1 Answer

Impact of this question

5352 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-16T02:46:05

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

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 $\frac{1}{12}$ $1/12$ th part of a $_{6}^{12} C$ $""_6^12 "C"$ atom.

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

The mass of $\frac{1}{12}$ $1/12$ th of a $_{6}^{12} C$ $""_6^12 "C"$ atom is taken as a unit of mass and is known as 1 amu or 1 u .

$1 u = \frac{1}{N_{A}} g = 1.66 \times 10^{- 24} g$ $"1 u" = 1/(N_A) g=1.66xx10^-24 g$ ,

where $N_{A}$ $N_A$ is Avogadro's Number.

Example

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

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

$22.989 amu \times \frac{1.66 \times 10^{- 24} g}{1 amu} = 3.82 \times 10^{- 24} g$ $"22.989 amu" xx (1.66xx10^-24 "g")/"1 amu" = 3.82xx10^-24 "g"$ for one atom of $Na$ $"Na"$ .

Now the answer of the given question

Let the atomic mass of silver be $M_{Ag}$ $M_"Ag"$ .

So one atom of silver weighs:

$M_{Ag} amu \times \frac{1.66 \times 10^{- 24} g}{1 amu} = 1.66 \times 10^{- 24} M_{Ag}$ $M_"Ag" "amu" xx (1.66xx10^(-24) "g")/"1 amu" = 1.66xx10^(-24)M_"Ag"$ $g$ $"g"$

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

$= 1.66 \times 10^{- 24} M_{Ag}$ $= 1.66xx10^-24M_"Ag"$ $g$ $"g"$ $\times 10^{6} = 1.66 \times 10^{- 18} M_{Ag}$ $xx 10^6 = 1.66xx10^-18 M_"Ag"$ $g$ $"g"$

Equating this with the given value we can write:

$1.66 \times 10^{- 18} M_{Ag}$ $1.66xx10^-18 M_"Ag"$ $g$ $"g"$ $= 1.79 \times 10^{- 16} g$ $= 1.79xx10^-16 "g"$

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

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.

Science

Math

Humanities

... and beyond

If $1000000$ $1000000$ silver atoms weigh $1.79 \times 10^{- 16} g$ $1.79 xx 10^(-16) "g"$ , what is the atomic mass of silver?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If 100000010000001000000 silver atoms weigh 1.79 xx 10^(-16) "g"1.79×10−16g1.79 xx 10^(-16) "g", what is the atomic mass of silver?

1 Answer

Explanation:

Related questions

Impact of this question

If $1000000$ $1000000$ silver atoms weigh $1.79 \times 10^{- 16} g$ $1.79 xx 10^(-16) "g"$ , what is the atomic mass of silver?