A 25.0 mm x 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm3.How many gold atoms are in the sheet ? (Atomic weight : Au=197.0)

A 25.0 mm x 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm³.
How many gold atoms are in the sheet ? (Atomic weight : Au=197.0)

A
7.7 x 10²³
B
1.5 x 10²³
C
4.3 x 10²¹
D
1.47 x 10²²

Solution:

Volume of gold foil = 25 x 40 x 0.25 mm³
= 250 x10^-3cm³
Mass of gold foil = 19.32 x 250 x10^-3 g
= 4.83 g
No. of gold atoms = $= \frac{4 .83}{197} \times N_{A}$
=1.47 x 10²²

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