A 25.0 mm × 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 )

Question

A 25.0 mm × 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 × 1023 (B) 1.5 × 1023 (C) 4.3 × 1021 (D) 1.47 × 1022

Tardigrade Admin · Accepted Answer

Volume of gold foil =25 × 40 × 0.25 mm 3 =250 × 10-3 cm 3 Mass of gold foil =19.32 × 250 × 10-3 g =4.83 g No. of gold atoms =(4.83/197) × NA=1.47 × 1022

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

$7.7 \times 1 0^{23}$

$1.5 \times 1 0^{23}$

$4.3 \times 1 0^{21}$

$1.47 \times 1 0^{22}$

Volume of gold foil $= 25 \times 40 \times 0.25 m m^{3}$
$= 250 \times 1 0^{- 3} c m^{3}$
Mass of gold foil $= 19.32 \times 250 \times 1 0^{- 3} g = 4.83 g$
No. of gold atoms $= \frac{4.83}{197} \times N_{A} = 1.47 \times 1 0^{22}$

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

7.7×1023

1.5×1023

4.3×1021

1.47×1022