A gold leaf of area $1.0 c m^{2}$ and thickness $1.0 \times 1 0^{- 5} cm$ is made from gold of density $19.7 g c m^{- 3}$ ( $Au = 197$ ). The number of gold atoms in the leaf is (take $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ )

Answer: D

Volume

= 1.0 \times 1.0 \times 1 0^{- 5} = 1.0 \times 1 0^{- 5} c m^{3}

. Mass

= 19.7 \times 1 0^{- 5} = 1.970 \times 1 0^{- 4} g

. Moles

= 1.970 \times 1 0^{- 4} /197 = 1.0 \times 1 0^{- 6}

mol. Atoms

= 1.0 \times 1 0^{- 6} \times 6.02 \times 1 0^{23} = 6.02 \times 1 0^{17}