Question

Question #4393b

Answer 1

`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 `"1 L"` sample of solution. As you know, molarity is defined as moles of solute 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 density 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 percent concentration, `"% 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 sig figs.