0,125nm=1,25.10^{-8}cm
Giả sử n_{Cr}=1(mol)
->m_{Cr}=1.52=52(g)
->V_{Cr}={52}/{7,2}=7,22(cm^3)
Lại có V_{1\ ng tu\ Cr}=4/{3}.\pi.(1,25.10^{-8})^3\approx 8,18.10^{-24}(cm^3)
->V_{Cr\ t t}=8,18.10^{-24}.6,022.10^{23}\approx 4,926(cm^3)
->x={4,926}/{7,22}.100\%\approx 68,23(\%)