Question #f9eac

1 Answer
Sep 8, 2016

Answer:

#1.363 * 10^(24)" Na"^(+)"cations"#

Explanation:

If by amount of sodium ions you mean how many of them you have in that sample of sodium carbonate, #"Na"_2"CO"_3#, then all you have to do here is use two conversion factors

  • one to take you from grams to moles
  • one to take you from moles to number of ions

Now, before moving forward, notice that one mole of sodium carbonate contains

  • two moles of sodium cations, #2 xx "Na"^(+)#
  • one mole of carbonate anions, # 1 xx "CO"_3^(2-)#

So, use the molar mass of sodium carbonate to calculate the number of moles present in that sample

#120.0 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"CO"_3)/(106.0color(red)(cancel(color(black)("g")))) = "1.132 moles Na"_2"CO"_3#

This means that your sample will contain

#1.132 color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * ("2 moles Na"^(+))/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3))))#

# = "2.264 moles Na"^(+)#

Now all you have to do is use Avogadro's number

#color(purple)(bar(ul(|color(white)(a/a)color(black)("1 mole" = 6.022 * 10^(23)"particles")color(white)(a/a)|)))#

to calculate the number of sodium ions present in that many moles

#2.264 color(red)(cancel(color(black)("moles Na"^(+)))) * (6.022 * 10^(23)" Na"^(+)"cations")/(1color(red)(cancel(color(black)("mole Na"^(+)))))#

# = color(green)(bar(ul(|color(white)(a/a)color(black)(1.363 * 10^(24)" Na"^(+)"cations")color(white)(a/a)|)))#

The answer is rounded to four sig figs.