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# Density of ethanoic acid question

1. 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
Blake
2. Could you send the full question please, some numbers from the first part don't make much sense without the question
What you've done for the second part is correct (the conversion) but again the question doesn't make too much sense
3. 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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