# Question 027a8

Feb 10, 2017

See explanation.

#### Explanation:

Well, you got the first one right, but that's about it. The answer to the second question is a bit off.

You can solve the first one by using the percent concentration of the solution as a conversion factor.

A $\text{0.9% w/w}$ sodium chloride solution will contain $\text{0.9 g}$ of sodium chloride, the solute, for every $\text{100 g}$ of solution. This means that $\text{55 g}$ of solution will contain

55 color(red)(cancel(color(black)("g solution"))) * overbrace("0.9 g NaCl"/(100color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 0.9% w/w")) = "0.495 g NaCl"

Rounded to one significant figure, the number of sig figs you have for the percent concentration of the solution, the answer should be

$\text{mass of NaCl = 0.5 g}$

Now, here's how to tackle the second one. For starters, a $\text{76% w/w/}$ glycerol solution will contain $\text{76 g}$ of glycerol, the solute, for every $\text{100 g}$ of solution.

This is what the weight by weight, or mass by mass, percent concentration of a solution tells you -- how many grams of solute you get in $\text{100 g}$ of solution.

Molarity is defined as the number of moles of solute present in ${\text{1 dm}}^{3}$ of solution. Let's pick a ${\text{1 dm}}^{3}$ sample of this solution to work with.

You should know what

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 dm"^3 = 10^3"cm"^3)))" " sqrt(color(darkgreen)())" " " " color(red)(cancel(color(black)("1 cm"^3 = 10^3"dm}}^{3}}}}$

so this sample will have a volume of

1 color(red)(cancel(color(black)("dm"^3))) * (10^3"cm"^3)/(1color(red)(cancel(color(black)("dm"^3)))) = 10^3"cm"^3

Use the density of the solution to find its mass

10^3 color(red)(cancel(color(black)("cm"^3))) * overbrace("1.2 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("= 1.2 g cm"^(-3))) = "1200 g"

Since every $\text{100 g}$ of this solution contain $\text{76 g}$ of glycerol, this sample will contain

1200 color(red)(cancel(color(black)("g solution"))) * overbrace("76 g glycerol"/(100color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 76% w/w")) = "912 g glycerol"

Use the molar mass of glycerol to convert this to moles

912color(red)(cancel(color(black)("g"))) * "1 mole glycerol"/(92color(red)(cancel(color(black)("g")))) = "9.913 moles glycerol"#

Since this is how many moles of solute you have in ${\text{1 dm}}^{3}$ of solution, the molarity of the solution will be

${\text{molarity = 9.9 mol dm}}^{- 3}$

The answer is rounded to two sig figs.