Calculate molality of a #"1-L"# solution of #93%# #"H"_2"SO"_4# mass by volume. The density of the solution is #"1.84 g/mL"# ?
1 Answer
Here's what I got.
Explanation:
For starters, you know that you're dealing with a
To make the calculations easier, pick a sample of this solution that has a volume of
#100 color(red)(cancel(color(black)("mL solution"))) * "1.84 g"/(1color(red)(cancel(color(black)("mL solution")))) = "184 g"#
Now, in order to find the molality of the solution, you need to figure out the number of moles of sulfuric acid present for every
This sample contains
#"184 g " - " 93 g" = "91 g"#
of water, which implies that a sample that contains
#10^3 color(red)(cancel(color(black)("g water"))) * ("93 g H"_2"SO"_4)/(91 color(red)(cancel(color(black)("g water")))) = "1021.98 g H"_2"SO"_4#
To convert the mass of sulfuric acid to moles, use the molar mass of the compound.
#1021.98 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.079 color(red)(cancel(color(black)("g")))) = "10.42 moles H"_2"SO"_4#
Since this represents the number of moles of sulfuric acid present for every
#color(darkgreen)(ul(color(black)("molality = 10.4 mol kg"^(-1))))#
I'll leave the answer rounded to three sig figs, but keep in mind that the value you have for the concentration of the solution justifies only two sig figs for the answer.
SIDE NOTE: Notice that the fact that the solution given to you has a volume of