mình mất gốc giải chi tiết, dễ hiểu mình vote 5 sao

\(\begin{array}{l}
a)\\
\text{Đặt }\%_1^{1}H=x\%\to \%_1^{2}H=(100-x)\%\\
\to M_H=\dfrac{x+2(100-x)}{100}=1,008\\
\to x=99,2\\
\to \%_1^{1}H=99,2\%;\%_1^{2}H=0,8\%\\
b)\\
m_{H_2O}=5.1=5(g)\\
\to n_{H_2O}=\dfrac{5}{18}(mol)\\
\to n_H=2n_{H_2O}=\dfrac{5}{9}(mol)\\
\to n_{_1^{1}H}=\dfrac{5}{9}.99,2\%=\dfrac{31}{225}(mol)\\
\to \text{Số nguyên tử }_1^{1}H=\dfrac{31}{225}.6,022.10^{23}=8,3.10^{22}
\end{array}\)