Question

How many sodium ions are present in 325 ml of 0.850 m na2so4?

Answer 1

Answer:

$Na^+_{ions}=3.33x10^{23}Na^+_{ions}$

Explanation:

Hello,

In this case, for the given molarity and volume of such solution, the moles of sodium sulfate are computed below:

$n_{Na_2SO_4}=0.850\frac{mol}{L}*0.325L=0.276molNa_2SO_4$

Now, by using the Avogadro's number, the ions result:

$Na^+_{ions}=0.276molNa_2SO_4*\frac{2molNa^+}{1molNa_2SO_4} *\frac{6.022x10^{23}Na^+_{ions}}{1molNa^+} \\\\Na^+_{ions}=3.33x10^{23}Na^+_{ions}$

Best regards.