# Question #027a8

##### 1 Answer

#### Answer:

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 **for every**

#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

#"mass of NaCl = 0.5 g"#

Now, here's how to tackle the second one. For starters, a **for every**

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

**Molarity** is defined as the number of moles of solute present in

You should know what

#color(blue)(ul(color(black)("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**

#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

#"molarity = 9.9 mol dm"^(-3)#

The answer is rounded to two **sig figs**.