Question

What volume of a .200 M solution of K2SO4 solution contains 85.6 g of K2SO4?

Answer 1

Hey @johnybgood199 and welcome to \(\LARGE \sf \bbox[#40B9E9]{\color{white}{Open}}\bbox[#A8CE91]{\color{white}{Study!}}\)

In order to solve this problem we need primarily two equations:\[\Large C=\frac{ n }{ V }\]and
\[\Large n=\frac{ m }{ M }\]
Where:
\(C\) is the concentration [M]
\(V\) is the volume [L]
\(n\) is the amount of substance [mol]
\(m\) is the mass [g]
\(M\) is the molar mass [g/mol]

You know how to obtain the molar mass for\(\ \sf K_{2}SO_{4}\)?

Answer 2

Yes 174.259 g/mol

Answer 3

2.456L

Answer 4

Perfect, then you just put into the equations and solve for the volume \(V\)

Answer 5

85.6g * 1mol / 174.259g = 0.491 mol

V = 0.491mol/0.2mol

Answer 6

V = 2.456 L

Answer 7

is that right?

Answer 8

It sure is, but in your last calculations change the mol to M. Also I suggest you give the final answer with only 3 significant figures as there are only given 3 in the question. Else well done!

Answer 9

So it become mol/M

Answer 10

And ones again welcome to OpenStudy!

Answer 11

Thank you for your help

Answer 12

Can I ask you another question?