Gold of density $19.3 g c m^{- 3}$ is dispersed as spherical particles of radius $15 nm$ . Find the number of particles in $1.00 g$ of gold, in units of $1 0^{15}$ , to the nearest integer. Use $π = 3.14$ and $1 nm = 1 0^{- 7} cm$ .

Numerical answer type, answer in x 10^15, tolerance 0.5.

Answer: 4 x 10^15

Radius =

1.5 \times 1 0^{- 6}

cm, so volume of one particle =

\frac{4}{3} (3.14) (1.5 \times 1 0^{- 6})^{3} = 1.413 \times 1 0^{- 17} c m^{3}

. Its mass =

19.3 \times 1.413 \times 1 0^{- 17} = 2.727 \times 1 0^{- 16}

g. Number in 1.00 g =

1/ (2.727 \times 1 0^{- 16}) = 3.67 \times 1 0^{15}

, which rounds to 4 in units of

1 0^{15}