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Topic: Dilutions  (Read 3614 times)

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Offline kadaj

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Dilutions
« on: July 26, 2007, 04:34:05 AM »
Hi everybody,

If a solution A contains 100mg of quanine sulfate and it dissolved in 100ml of 0.5M of sulfuric acid and the second solution, solution B was made up by diluting 10ml of solution A using 0.5M sulfuric acid and made up 100ml, calculate the new concentration.

Using this value, solution C was made up by diluting 10ml of solution B and made it up to 25ml, calculate the new concentration for solution C.

Thank you
Disobedience is man's original virtue.It's through disobedience and rebellion that progress has been made.

Offline sdekivit

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Re: Dilutions
« Reply #1 on: July 26, 2007, 04:43:24 AM »
first: 100 mg in 100 mL solution (= solution A): concentration = ?

--> then: solution B is 10 mL A to 100 mL solution: dilution factor = ? and use that to calculate the new concentration.

--> then: solution C = 10 mL B to 25 mL solution: dilution factor = ? --> concentration = ?

Offline kadaj

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Re: Dilutions
« Reply #2 on: July 26, 2007, 04:57:11 AM »
Sorry, but I meant using mol/L units

I tried this using C1V1=C2V2 formula and I ended up with 5M for concentration in solution A, 0.5M for solution B and 0.2M for solution C, please correct me if im wrong

Thank you
Disobedience is man's original virtue.It's through disobedience and rebellion that progress has been made.

Offline AWK

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Re: Dilutions
« Reply #3 on: July 26, 2007, 05:05:30 AM »
solution B 100/10=10 times diluted
solution C 25/10 = 2.5 times
hence solution C is 25 times diluted with regards to solution A (you can use any units of concentration for solution A)
AWK

Offline kadaj

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Re: Dilutions
« Reply #4 on: July 26, 2007, 05:10:39 AM »
Cheers mate  ;D that helps
Disobedience is man's original virtue.It's through disobedience and rebellion that progress has been made.

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