Answer:
[tex]1.21x10^{22}cationsNa^+[/tex]
Explanation:
Hello,
In this case, one knows that sodium chloride's formula is NaCl, therefore, the amount of cations is computed via the following stoichiometric relationship, considering the Avogadro's number to relate moles of sodium cations with their actual amount:
[tex]1.17gNaCl*\frac{1molNaCl}{58.45gNaCl}*\frac{1molNa^+}{1molNaCl}* \frac{6.022x10^{23}cationsNa^+}{1molNa^+} =1.21x10^{22}cationsNa^+[/tex]
Best regards.
