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Density of ethanoic acid question

Hi guys,

Doing a bit of AS-level chemistry for revision and I came across this question. I'm not sure whether it's a typo or I'm just looking at it from the wrong angle.

The question says:

Calculate the Mr of ethanoic acid and use it to find the concentration of ethanoic acid in g dm^-3.

So I got the Mr to be approximately 60 and then I used Mr*moles to get the mass so I did 60 * 2.257*10^-3 which gave me 0.13542g in 0.025dm^-3

0.025 needs to be multiplied by 40 to get 1dm^3 so 0.13542*40=5.4168g dm^-3

The next part then says:

The density of pure ethanoic acid is 1.049g cm^-3 Use this value to calculate the concentration of ethanoic acid in vinegar as a volume percentage.

So I worked out that 1.049g cm^-3 = 1049g dm^-3 but this already doesn't seem right. Can anyone see where I've gone wrong? @JN17 ?

Thanks
(edited 1 year ago)
Reply 1
Avatar for JN17
JN17
10
Could you send the full question please, some numbers from the first part don't make much sense without the question:tongue:
What you've done for the second part is correct (the conversion) but again the question doesn't make too much sense
Reply 2
Avatar for Dav 1506
Dav 1506
Hi,
I've come across this question and that is not the correct conversion.

To get from cm^3 to dm^3 you have to divide by 1000

So I spoke with another teacher and it's to do with the idea of density!

The value in the question given is the density of pure ethanoic acid- 1.049 gcm^3. Which is not not mixed with water so you have no concentration. Where as the value we have is derived from a solution of ethanoic acid which Is x1000 more than the value given.

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