Question #027a8

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 #"0.9% w/w"# sodium chloride solution will contain #"0.9 g"# of sodium chloride, the solute, for every #"100 g"# of solution. This means that #"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

#"mass of NaCl = 0.5 g"#

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

Molarity is defined as the number of moles of solute present in #"1 dm"^3# of solution. Let's pick a #"1 dm"^3# sample of this solution to work with.

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 #"100 g"# of this solution contain #"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 #"1 dm"^3# of solution, the molarity of the solution will be

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

The answer is rounded to two sig figs.