Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# An ammonia solution is 6.00 M and the density is 0.950 g/mL. What is the mass/mass percent concentration of ##"NH"_3## (17.04 g/mol)? Thank you

##10.8%##

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 ##"1 L"## sample of solution. As you know, is defined as moles of per **liter of solution**.

In this case, ##"1 L"## of ##"6.00 M"## ammonia solution will contain ##6.00## **moles** of ammonia.

Now, you know that this solution has a of ##"0.950 g mL"^(-1)##. Use it to find the **mass** of ##"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 **, ##"% m/m"##, tells you how many grams of solute you have in ##"100 g"## of **solution**.

##color(blue)(|bar(ul(color(white)(a/a)"% m/m" = "grams of solute / 100 g solution"color(white)(a/a)|)))##

In this case, you know that ##"950 g"## of solution contain ##"102.24 g"## of solute, which means that ##"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 ##"100 g"## of solution, you can say that the solution's ##"%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 .