A gold foil has area 2.0 cm^2 and thickness 1.0x10^-4 cm. The density of gold is 19.7 g cm^-3 (Au = 197, N_A = 6.02x10^23 mol^-1). The number of gold atoms in the foil is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc012

A gold foil has area $2.0 c m^{2}$ and thickness $1.0 \times 1 0^{- 4} cm$ . The density of gold is $19.7 g c m^{- 3}$ (Au = 197, $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ ). The number of gold atoms in the foil is:

Volume

= 2.0 \times 1.0 \times 1 0^{- 4} = 2.0 \times 1 0^{- 4} c m^{3}

; mass

= 19.7 \times 2.0 \times 1 0^{- 4} = 3.94 \times 1 0^{- 3} g

. Moles

= 3.94 \times 1 0^{- 3} /197 = 2.0 \times 1 0^{- 5}

. Atoms

= 2.0 \times 1 0^{- 5} \times 6.02 \times 1 0^{23} = 1.204 \times 1 0^{19}