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Topic: %concrentraion-grams  (Read 4753 times)

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Offline FouRRaW

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%concrentraion-grams
« on: February 01, 2007, 01:35:28 PM »
How man grams of KCl are needed to prepare 0.8L of a 0.15% (m/v) KCL solution?

I get 0.0012 grams because of the following formula, but appareently the answer is 1.2g? What did i do wrong, wrong formula?

mass/volume x 100 = %concentrtion  (this formula may not be totally accurate but i'm told it's close enough, cause i think really it should be m/m not m/v)
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Offline Borek

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Re: %concrentraion-grams
« Reply #1 on: February 01, 2007, 01:38:14 PM »
Volume unit?
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Offline FouRRaW

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Re: %concrentraion-grams
« Reply #2 on: February 01, 2007, 01:40:33 PM »
sorry,but i'm not sure what you mean
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Offline Dan

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Re: %concrentraion-grams
« Reply #3 on: February 01, 2007, 01:52:15 PM »
He's asking what volume units you are using. ie. is this 0.15% g/mL, g/L, g/pint etc.?

Given the factor you are out by, it looks like you have used the wrong volume units.
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Offline english

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Re: %concrentraion-grams
« Reply #4 on: February 01, 2007, 02:15:35 PM »
"m" is kind of ambiguous.
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Offline Yggdrasil

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Re: %concrentraion-grams
« Reply #5 on: February 01, 2007, 06:55:04 PM »
Quote from: FouRRaW on February 01, 2007, 01:35:28 PM
How man grams of KCl are needed to prepare 0.8L of a 0.15% (m/v) KCL solution?


For %(w/v), 1g/mL = 100%(w/v) (since for water 1mL has a mass of 1g).  You can also think of %(w/v) as being grams per 100mL.
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