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

Answer 1

Wyatt Adkins

#"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#

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Answer 2

Cameron Andrews

The new concentration of the iron(II) sulfate solution is 0.0715 mol/L.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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