\(PTHH:NaX+AgNO_3\rightarrow AgX\downarrow+NaNO_3\\ Từ.PT:\left(23+M_X\right).....\left(108+M_X\right)\left(g\right)\\ Theo.đề:10,29.....................18,79\left(g\right)\\ \Rightarrow18,79.\left(23+M_X\right)=10,29.\left(108+M_X\right)\\ \Leftrightarrow M_X=79,9\left(\dfrac{g}{mol}\right)\)
=> NTK trung bình của X là 79,9(đ.v.C)
\(b,\\ 2.đồng.vị:X_1,X_2\\ \%X_1=a\Rightarrow\%X_2=100\%-\left(a+10\%\right)a=110\%-a\\ \overline{NTK}_X=79,9\\ \Leftrightarrow\dfrac{A_{X_1}.a+A_{X_2}.\left(110\%-a\right)}{100\%}=79,9\left(1\right)\\ Mà:A_{X_2}-A_{X_1}=2\left(2\right)\left(Vì:N_{A_1}-N_{A_2}=2\right)\\ \left(1\right),\left(2\right)\Rightarrow A_{X_1}=79\left(đ.v.C\right)\\ A_{X_2}=81\left(đ.v.C\right)\)
