Question

A golg ring weighs 1.97 grams.The number of atoms it contains?

Answer 1

Answer:Atomic weight of the Gold (Au) is = 196.97 (197g apprx.)

The Avogadro's number in one mole are = 6.022X10^23

That means 6.022X10^23  atoms weigh 196.97 g.

                   

Now

        1.0 g/ 196.97 g  = 0.0051 mol

So the number of atoms in one gram of gold are:

0.0051 x 6.022X10^23   = 3.057318373 x 10^21

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