Gold $($molar mass $\left.=197.0\mathrm{g}{\mathrm{mol}}^{-1}\right)$ has a face-centred cubic unit cell. If its density is $19.7\mathrm{g}{\mathrm{cm}}^{-3}$, the atomic radius of gold will be about 

Since $\rho =N\left(M/N_{\mathrm{A}}\right)/a^3=N\left(M/N_{\mathrm{A}}\right)/(2\sqrt{2}r{)}^3$, We have

    $r=\frac{1}{2\sqrt{2}}{\left[\frac{N\left(M/N_{\mathrm{A}}\right)}{\rho }\right]}^{1/3}=\frac{1}{2\sqrt{2}}{\left[\frac{4\left(197.0\mathrm{g}{\mathrm{mol}}^{-1}\right)/\left(6.022\times {10}^{23}{\mathrm{mol}}^{-1}\right)}{19.7\mathrm{g}{\mathrm{cm}}^{-3}}\right]}^{1/3}$$=\frac{1}{2\sqrt{2}}\left(4.05\times {10}^{-8}\mathrm{cm}\right)=1.432\times {10}^{-8}\mathrm{cm}=143.2\mathrm{pm}$