A silver sol is prepared by dispersing $2.16 g$ of silver (density $10.5 g c m^{- 3}$ ) into identical spherical particles of radius $20 nm$ ( $1 nm = 1 0^{- 7} cm$ ). Find the total surface area of all the particles in $m^{2}$ , to one decimal.

Numerical answer type, answer in m^2, tolerance 0.05.

Answer: 30.9 m^2

For N spheres of radius r, total volume =

m / ρ

and total area =

N \cdot 4 π r^{2} = 3 (m / ρ) / r

(the factor

π

cancels). Total volume =

2.16/10.5 = 0.2057 c m^{3}

. Area =

3 \times 0.2057/ (2.0 \times 1 0^{- 6}) = 3.086 \times 1 0^{5} c m^{2}

. In

m^{2}

this is 30.86, i.e. 30.9.