How many moles of $Pb$ is in $9.34 * 10^15$ atoms of $Pb$ ?

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Answer 1 · 2016-02-25T18:21:36

Approx. $10^-8$ $"moles"$ .

$N_A$ $=$ $6.022xx10^23$

So, $(9.34xx10^15*"lead atoms")/(6.022xx10^23" lead atoms mole"^-1)$ $=$ $??"mole"$

How many moles of PbPb is in 9.34 * 10^159.34 * 10^15 atoms of PbPb?