95% ethanol??? (1 Viewer)

[tellmewhy]

hello everybody,

can anyone help me with this recipe:

Ethanol (C2H3OH - 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.



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?

thank you so much!

 

[BlackJack]

...assume the other 5% has the same molemular weight as ethanol does. You calculate how much pure ethanol you need: the 391.6g you have there. then to obtain the same amount when you weight 95% ethanol, you divide this by 0.95.

Observe 391.6 is 95% of 412.2...