1 đvC nặng số (g) là:
\(\dfrac{1,9926.10^{-23}}{12}\simeq1,66.10^{-24}\)
Theo bài ra, ta có:
\(\left\{{}\begin{matrix}NTK_{Ca}=40\left(đvC\right)\\1\left(đvC\right)=1,66.10^{-24}g\end{matrix}\right.\)
\(\rightarrow m_{5Ca}=1,66.10^{-24}.5.40=332.10^{-24}\left(g\right)\)
Khối lượng 5 nguyên tử Ca là:
\(1,9926\times10^{-23}\times5=9,963\times10^{-23}\left(g\right)\)