Question 4393b

Jul 23, 2016

10.8%

Explanation:

Your strategy here will be to

• pick a sample of this ammonia solution
• use the solution's density to find the mass of the sample
• use the molar mass of ammonia to find the mass of solute

To make the calculations easier, pick a $\text{1 L}$ sample of solution. As you know, molarity is defined as moles of solute per liter of solution.

In this case, $\text{1 L}$ of $\text{6.00 M}$ ammonia solution will contain $6.00$ moles of ammonia.

Now, you know that this solution has a density of ${\text{0.950 g mL}}^{- 1}$. Use it to find the mass of $\text{1 L}$ of solution

1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("0.950 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("the given density")) = "950 g"

You know that this sample contains $6.00$ moles of ammonia, your solute. Use its molar mass to convert this to grams of solute

6.00 color(red)(cancel(color(black)("moles NH"_3))) * overbrace("17.04 g"/(1color(red)(cancel(color(black)("mole NH"_3)))))^(color(purple)("the given molar mass")) = "102.24 g"

Now, a solution's mass by mass percent concentration, $\text{% m/m}$, tells you how many grams of solute you have in $\text{100 g}$ of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% m/m" = "grams of solute / 100 g solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you know that $\text{950 g}$ of solution contain $\text{102.24 g}$ of solute, which means that $\text{100 g}$ of solution will contain

100 color(red)(cancel(color(black)("g solution"))) * ("102.24 g NH"_3)/(950color(red)(cancel(color(black)("g solution")))) = "10.8 g NH"_3

Since this is how many grams of ammonia you get per $\text{100 g}$ of solution, you can say that the solution's $\text{%m/m}$ will be equal to

"% m/m" = color(green)(|bar(ul(color(white)(a/a)color(black)(10.8%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.