Question

What is the number of cations is present in 1.17 g of sodium chloride? Given Na=6*10power 23 mol

Answer 1

Answer:

$1.21x10^{22}cationsNa^+$

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:

$1.17gNaCl*\frac{1molNaCl}{58.45gNaCl}*\frac{1molNa^+}{1molNaCl}* \frac{6.022x10^{23}cationsNa^+}{1molNa^+} =1.21x10^{22}cationsNa^+$

Best regards.

Answer 2

1.17×6×1023 Wich equals 7.02×1023 Your final answer is 7.02×1023