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)

1. A

7.7 x 1023

2. B

1.5 x 1023

3. C

4.3 x 1021

4. D

1.47 x 1022

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### 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 = $=\frac{4.83}{197}×{\mathrm{N}}_{\mathrm{A}}$
=1.47 x 1022

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