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# A *simple* concentration/volume question. watch

1. What volume of 0.200 mol dm^-3 potassium sulfate solution is required to make, by dilution with water, 1.00 dm^3 of a solution with potassium ion concentration of 0.100 mol dm^-3

A 100 cm^3
B 250 cm^3
C 400 cm^3
D 500 cm^3

Some explanation, would be appreciated as well.
What volume of 0.200 mol dm^-3 potassium sulfate solution is required to make, by dilution with water, 1.00 dm^3 of a solution with potassium ion concentration of 0.100 mol dm^-3

A 100 cm^3
B 250 cm^3
C 400 cm^3
D 500 cm^3

Some explanation, would be appreciated as well.
For this question, you have to work backwards and use the formula n=CV

If you have a K+ conc of 0.100 mol dm-3 and a volume of 1.00 dm3, that means you have 0.100 mol K+ ions (since n=CV).

K2SO4 ==> 2K+ + SO42-

One mole of potassium sulphate will dissociate in water to produce two moles of K+ ions and so you have to work backwards. If you have 0.100 mol of K+ ions, you must have had 0.0500 mol of K2SO4 to begin with.

Now, to find the volume you have to rearrange n=CV, so V=n/C.

Therefore the volume of K2SO4 = 0.0500/0.200 = 0.250 dm3 = 250cm3. Therefore the answer is B.

Hope this helps you out
3. (Original post by Jpw1097)
For this question, you have to work backwards and use the formula n=CV

If you have a K+ conc of 0.100 mol dm-3 and a volume of 1.00 dm3, that means you have 0.100 mol K+ ions (since n=CV).

K2SO4 ==> 2K+ + SO42-

One mole of potassium sulphate will dissociate in water to produce two moles of K+ ions and so you have to work backwards. If you have 0.100 mol of K+ ions, you must have had 0.0500 mol of K2SO4 to begin with.

Now, to find the volume you have to rearrange n=CV, so V=n/C.

Therefore the volume of K2SO4 = 0.0500/0.200 = 0.250 dm3 = 250cm3. Therefore the answer is B.

Hope this helps you out
Thanks!

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Updated: February 12, 2016
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