# Question 2e7d7

May 16, 2016

$\text{1.6 g}$

#### Explanation:

As you know, a solution's molality tells you how many moles of solute you get per kilogram of solvent.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molality" = "moles of solute"/"kilogram of solvent} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, sodium carbonate, ${\text{Na"_2"CO}}_{3}$, is the solute and water is the solvent.

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 $\text{100 mL}$ of water to get a $\text{0.15 molal}$ solution.

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 ${\text{1 g mL}}^{- 1}$.

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to get

100color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.1 kg"

Now, a $\text{0.15-molal}$ solution will have $0.15$ moles of sodium carbonate in $\text{1 kg}$ of solvent, which means that your solution will have

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

$0.015 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Na"_2"CO"_3))) * "106 g"/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = color(green)(|bar(ul(color(white)(a/a)"1.6 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two **sig figs.