The density of gold is 19.3 g/cc. how many milligrams of gold are required to coat a small ball bearing of diameter 1.00 mm? the thickness of the coating is to be 0.200 mm. report answer to 3 sig figs.

Solution: Density of gold is 19.3 g/cc or 19.3 g/cm^{3}. Diameter of small ball is 1.0 mm. Thus, radius will be 0.5 mm. Since, 1 mm =0.1 cm thus, 0.5 mm =0.05 cm.

First calculate the volume of small ball from formula of volume of sphere as follows:

V=\frac{4}{3}\pi r^{3}

V=\frac{4}{3}(3.14)(0.05 cm)^{3}=5.02\times 10^{-4}cm^{3}

Now, if a layer of gold covered the small ball with thickness 0.200 mm or 0.02 cm, a sphere with radius more than the small ball is formed. The radius of bigger sphere will be sum of radius of small ball and the thickness of gold layer. Thus, R=0.05+0.02=0.07 cm

Calculate volume of bigger sphere:

V_{1}=\frac{4}{3}\pi R^{3}

V_{1}=\frac{4}{3}(3.14)(0.07 cm)^{3}=1.43\times 10^{-3}cm^{3}

now, from the figure attached, volume of gold used can be calculated as follows:

V_{gold}=V_{1}-V

=1.43\times 10^{-3}cm^{3}-5.02\times 10^{-4}cm^{3}

=9.28\times 10^{-4}cm^{3}

Mass of gold can be calculated from volume and density as follows:

m=d×V=19.3 g/cm^{3}×9.28\times 10^{-4}cm^{3}=0.0179 g

Since, 1 g=10^{3}mg

Thus,

Mass of gold will be 17.9 mg. This is up to three significant figures.

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