Question

A $25.0 \,mm \times 40.0 \,mm$ piece of gold foil is $0.25\, mm$ thick. The density of gold is $19.32\, g / cm ^{3}$. How many gold atoms are in the sheet? (Atomic weight : $Au =197.0$ )

• A. $7.7 \times 10^{23}$
• B. $1.5 \times 10^{23}$
• C. $4.3 \times 10^{21}$
• D. $1.47 \times 10^{22}$

Answer 1

Answer: (D)

Volume of gold foil $=25 \times 40 \times 0.25\, mm ^{3}$
$=250 \times 10^{-3} \, cm ^{3}$
Mass of gold foil $=19.32 \times 250 \times 10^{-3} g =4.83\, g$
No. of gold atoms $=\frac{4.83}{197} \times N_{A}=1.47 \times 10^{22}$