Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: socomplicated on August 18, 2004, 01:43:30 PM
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hello everybody,
can anyone help me with this recipe:
Ethanol; MW 46.07
8.5 M - The ethanol can be weighed [391.6 grams of 100% ethanol or 412.2 grams of 95% ethanol (v/v)]. Weigh the appropriate amount and dilute to a final volume of 1 liter with water.
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i don't understand how they get 412.2g of 95% EtOH. i have calculated over and over again, but can't figure out how they get 412.2g. do any of you know how to do it?
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391.6 g eth * (100 gsolution/95 g eth) = 412.2 g solution