Question

What is the Mass of 6.022 *10^23 atoms of Ag

Answer 1

6.023×10

23

atoms will be contained by

6.023×10

23

108( mass of 1 silver atom)

×6.023×10

23

=0.108g

Answer 2

Answer:

Atomic Mass Of Ag = 107/108

$1 \: mole = 6.022 \times 10 ^{23}$

Gram Molecular Mass of Ag = 107/108 g

Number Of Atoms Present in Gram Molecular Mass Of Element

=

1 Mole

=)107/108g of ag contains 6.022 x 10^24 atoms

Hope it helped