....................

 Ta có $\rm D=0,97g/cm^3$ 

$\rm M_{Na}=23g/mol$ 

Thể tích nguyên tử $\rm Na$ 

$\rm V_{nguyên~tử}=\dfrac{\dfrac{M_{Na}}{D}.68\%}{6,023.10^{23}}=2,677.10^{-23}(cm^3)$ 

Bán kính nguyên tử $\rm Na$ 

$\rm R=\sqrt[3]{\dfrac{3.V_{nguyên~tử}}{4}}=\sqrt[3]{\dfrac{3.2,677.10^{-23}}{4}}=2,718.10^{-8}(cm)=2,718 \mathop{A}\limits^{o}$