## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The mass of 1 mole of gold is $196.967~g$. We can find the volume of this mass of gold as: $V = \frac{m}{\rho}$ $V = \frac{0.196967~kg}{19.3\times 10^3~kg/m^3}$ $V = 1.02\times 10^{-5}~m^3$ We can find the radius of the sphere as: $\frac{4}{3}\pi~R^3 = V$ $R^3 = \frac{3V}{4\pi}$ $R = (\frac{3V}{4\pi})^{(1/3)}$ $R = [\frac{(3)(1.02\times 10^{-5}~m^3)}{4\pi}]^{(1/3)}$ $R = 0.0135~m = 1.35~cm$ Since the diameter is $2R$, the diameter of the sphere is 2.7 cm.