If one has $4.50 \times 10^{22}$ $4.50 xx 10^22$ atoms of silver, what mass is available?

Question

If one has $4.50 \times 10^{22}$ $4.50 xx 10^22$ atoms of silver, what mass is available?

1 Answer

Impact of this question

2500 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-02T21:03:36

$8.06 g Ag$ $"8.06 g Ag"$ ,

i.e. the mass of a handful of silver.

This is easiest in steps:

$atoms N_{A} - - \to mols M - - \to mass$ $"atoms" stackrel(N_A" ")(->) "mols" stackrel(M" ")(->) "mass"$

where $N_{A} = 6.0221413 \times 10^{23} {mol}^{- 1}$ $N_A = 6.0221413 xx 10^(23) "mol"^(-1)$ is Avogadro's number and $M$ $M$ is the molar mass in $g/mol$ $"g/mol"$ of the thing we are looking at.

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

$4.50 xx 10^(22) cancel"atoms" xx ("1 mol")/(6.0221413 xx 10^(23) cancel"whatevers")$

$=$ $"0.0747 mols of any atom"$

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:

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

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

Science

Math

Humanities

... and beyond

If one has $4.50 \times 10^{22}$ $4.50 xx 10^22$ atoms of silver, what mass is available?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If one has 4.50 xx 10^224.50×10224.50 xx 10^22 atoms of silver, what mass is available?

1 Answer

Related questions

Impact of this question

If one has $4.50 \times 10^{22}$ $4.50 xx 10^22$ atoms of silver, what mass is available?