Question #df0c5

1 Answer
Nov 17, 2015

#"1.7 M"#


Once again, your go-to strategy for such problems is to pick a volume sample of the solution and work with the given density to find its mass.

Once you know its mass, use the given percent concentration by mass to determine how many grams of sodium chloride you have.

Once you know the mass of the solute, you can use its molar mass to determine how many moles you have.

I'll pick a #"1.00-L"# sample of this #10%"w/w"# #"NaCl"# solution. Keep in mind that this is not a sample that you have to pick every time, but it's indeed very useful for molarity calculations.

So, use the solution's density to determine its mass

#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.997755 g"/(1color(red)(cancel(color(black)("mL")))) = "997.755 g"#

A #"10% w/w"# solution means that you get #"10 g"# of solute for every #"100 g"# of solution. As a result, the amount of sodium chloride you have in this much solution will be

#997.755color(red)(cancel(color(black)("g solution"))) * "10 g NaCl"/(100color(red)(cancel(color(black)("g solution")))) = "99.7755 g NaCl"#

Now use sodium chloride's molar mass to find the number of moles you have in this many grams

#99.7755color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.707 moles NaCl"#

Since molarity is defined as moles of solute per liters of solution, you can use the volume of the sample we picked to get the solution's molarity

#c = n/V#

#c = "1.707 moles"/"1.00 L" = color(green)("1.7 M")#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the percent concentration of the solution.

SIDE NOTE I don't like the value given to you for the density of the solution at #25^@"C"#. As far as I know, #10%# sodium chloride solutions have a density of #"1.070.7 g/mL"# at that temperature.

You can check that out here

Anyway, the strategy is exactly the same regardless of what values are given to you. I recommend using different sample volumes to double-check the result.

Remember, the molarity must come out to be the same regardless of the initial sample you pick.