Đáp án đúng: A
1,02 gam < mNa+K < 1,729 gam
1 lít nước khoáng đó có $\displaystyle \left\{ \begin{array}{l}1,3\,gam\,C{{l}^{-}}\\0,4\,gam\,HCO_{3}^{-}\\0,3\,gam\,SO_{4}^{2-}\\0,06\,gam\,C{{a}^{2+}}\\0,025\,gam\,M{{g}^{{}}}\\N{{a}^{+}};\,{{K}^{+}}\end{array} \right.\xrightarrow{{}}\left\{ \begin{array}{l}0,0366\,mol\,C{{l}^{-}}\\0,0066\,mol\,HCO_{3}^{-}\\{{3,125.10}^{-3}}\,mol\,SO_{4}^{2-}\\{{1,5.10}^{-3}}\,mol\,C{{a}^{2+}}\\{{1,04.10}^{-3}}\,mol\,M{{g}^{2+}}\\N{{a}^{+}},\,{{K}^{+}}\end{array} \right.$
Bảo toàn điện tích có${{n}_{N{{a}^{+}}\,+\,{{K}^{+}}}}\,=\,0,04437\,mol\,\Rightarrow \,1,02\,<\,{{m}_{N{{a}^{+}}\,+\,{{K}^{+}}}}\,<\,1,73$