Question #58587

1 Answer
Nathan L.
Jun 16, 2017

#81.6# #"g soln"#

Explanation:

This is essentially a problem involving dilution, but without the volumes or molarities. The moles (and thus grams) of the solute (#"H"_2"SO"_4#) before and after the dilution must be the same.

We can therefore set up this question by asking,

"#87.0%# of what #=# #35.5%# of #200#?"

The number of grams present is

#(0.335)(200# #"g") = 71.0# #"g"#

So therefore, #71.0# #"g"# is #87%# of the original mass of the solution:

#"Original mass" = 71"g"((100%)/(87%)) = color(red)(81.6# #color(red)("g soln"#

One way we can check this answer is by seeing if, in both cases, the percentage multiplied by the total solution mass is the same number, #71.0# #"g"#:

#"mass"_1 = (0.870)(color(red)(81.6"g")) = 71.0# #"g"#

#"mass"_2 = (0.335)(200"g") = 71.0# #"g"#

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