# 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 **moles** of sodium carbonate in

#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.