How many moles of Ag contain #4.49 * 10^23# atoms Ag?

1 Answer
Al E.
May 13, 2018

Recall Avogadro's constant,

#1"mol" = 6.02*10^23"units"#

where a unit can be anything. It's analogous to a dozen, for instance.

Hence,

#4.49*10^23"atoms" * "mol"/(6.02*10^23"atoms") approx 0.746"mol"#

of silver will contain that many atoms.

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