Question #2e7d7
1 Answer
Explanation:
As you know, a solution's molality tells you how many moles of solute you get per kilogram of solvent.
#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#
In your case, sodium carbonate,
Your strategy here will be to use the molality of the solution as a conversion factor to help you find how many moles of sodium carbonate must be added to
Notice that the problem provides you with the volume of solvent, but that molality requires mass of solvent. Since no mention of the density of water was made, you can assume it to be equal to
This means that your sample of water will have a mass of
#100 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "100 g"#
To get this to kilograms, use the conversion factor
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#
to get
#100color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.1 kg"#
Now, a
#0.1color(red)(cancel(color(black)("kg solvent"))) * ("0.15 moles Na"_2"CO"_3)/(1color(red)(cancel(color(black)("kg solvent")))) = "0.015 moles Na"_2"CO"_3#
Finally, use sodium carbonate's molar mass to go from moles to grams
#0.015color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * "106 g"/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = color(green)(|bar(ul(color(white)(a/a)"1.6 g"color(white)(a/a)|)))#
I'll leave the answer rounded to two **sig figs.