A #75mL# volume of #3.80molL^-1# #"iron(II) sulfate"# is diluted to #4.0L#. What is the new concentration?

Question

1401 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T18:35:24

#"Molarity"# #=# #"Moles of solute"/"Volume of solution"#

And thus we use the above equation to give us (i), the initial moles of solute:

#"Moles of"# #FeSO_4# #=# #"Volume"# #xx# #"Molarity"#

#75xx10^-3Lxx3.8*mol*L^-1# #=# #0.285*mol#

But this is diluted to a volume of #4.0*L#, and so to (ii), the new concentration:

#"Concentration"# #=# #(0.285*mol)/(4.0*L)=0.071*mol*L^-1#

A #75*mL# volume of #3.80*mol*L^-1# #"iron(II) sulfate"# is diluted to #4.0*L#. What is the new concentration?