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# The initial volumes of the two solvents are the same. Hence deduce the concentration

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1. Need help with part (ii)
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2. The answer should be 3.84 the same as the concentration of the aqueous ammonia solution since the number of moles are the same and so is the volume.
C=n/V
3. (Original post by Senal17)
The answer should be 3.84 the same as the concentration of the aqueous ammonia solution since the number of moles are the same and so is the volume.
C=n/V
The answer is 0.16 mol dm-3?!
Need help with part (ii)
You know the volume and concentration of the initial ammonia solution so you can work out the mumber of moles there.

You then work out the number of moles in the aqueous layer after shaking.

Thus the number of moles in the other layer = mol(initially) - moles(aqueous)

So if you know the volume (which you do) you can work out the concentration...
5. (Original post by charco)
You know the volume and concentration of the initial ammonia solution so you can work out the mumber of moles there.

You then work out the number of moles in the aqueous layer after shaking.

Thus the number of moles in the other layer = mol(initially) - moles(aqueous)

So if you know the volume (which you do) you can work out the concentration...
Moles initial= 75/1000 x 4= 0.3
Moles (aqueous)= 0.096

0.3-0.096 = 0.204 mol
mole/vol = conc
0.204/3.84 = 0.053 moldm-3
but in the markscheme it says 0.16 moldm-3
Moles initial= 75/1000 x 4= 0.3

Moles (aqueous)= 0.096
This is moles in 25ml BUT there is 75 ml altogether

Hence moles = 3 x 0.096 = 0.288 mol

0.3-0.096 = 0.204 mol
So this should be 0.3 - 0.288 = 0.012 mol

mole/vol = conc
0.204/3.84 = 0.053 moldm-3
but in the markscheme it says 0.16 moldm-3
and the concentration = 0.012/0.075 = 0.16 mol dm-3
7. (Original post by charco)

This is moles in 25ml BUT there is 75 ml altogether

Hence moles = 3 x 0.096 = 0.288 mol

So this should be 0.3 - 0.288 = 0.012 mol

and the concentration = 0.012/0.075 = 0.16 mol dm-3
Thanks

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