Answer
$N_{Pb} = 6.24 \times 10^{18}$
Work Step by Step
The mass of a single $^{206}Pb$ atom is,
$m_{Pb} = 206 u \times 1.661 \times 10^{-24} g/u $
$m_{Pb} = 3.42 \times 10^{-22} g$
so the number of $^{206}Pb$ atom in $2.135 \times 10^{–3} g$ is
$N_{Pb} = \frac{2.135 \times 10^{–3} g}{3.42 \times 10^{-22} g}$
$N_{Pb} = 6.24 \times 10^{18}$
