A (25.0 mm × 40.0 mm) piece of gold foil is 0.25 mm thick. The density of gold is $19.70 g / cm^{3}$ .
How many gold atoms are removed if the thickness of the sheet is reduced to 0.1 mm
(Atomic mass of Au = 197, $N_{A} = 6 \times 1 0^{23}$ )

Question

A (25.0 mm × 40.0 mm) piece of gold foil is 0.25 mm thick. The density of gold is $19.70 g / cm^{3}$ .

How many gold atoms are removed if the thickness of the sheet is reduced to 0.1 mm

(Atomic mass of Au = 197, $N_{A} = 6 \times 1 0^{23}$ )

TestHub · Accepted Answer

The change in thickness is Δ t = ( 0.25 − 0.10 ) mm = 0.15 mm = 0.015 c m . The volume removed is V = ( 2.5 × 4.0 ) × 0.015 = 0.15 c m 3 . The mass removed is m = 19.7 × 0.15 = 2.955 g . The number of moles removed is n = 197 2.955 ​ = 0.015 m o l . The number of atoms removed is 0.015 × 6 × 1 0 23 = 9.0 × 1 0 21 atoms. Answer: (C) 9.0 × 1 0 21 atoms

Chemistry - Mole Concept Question with Solution | TestHub

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