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# Quick concentration question. watch

1. Just trying to work out the concentration of HCl, acetone and iodine in a 25ml reaction volume.

4M Acetone - 5ml
1M HCL - 5ml
0.01M iodine - 5ml

starch solution - 1ml
water - 9ml

C= n/v.

So for example to get this concentration of Acetone. I've been doing 25/1000 to get 0.025 for answer in mol L-1.

So 4/0.025 = 160 mol L-1. Is this the answer? Or because there is 5ml of this in the 25ml total volume I would then have to divide by 5 again to get 32 mol L-1?
Just trying to work out the concentration of HCl, acetone and iodine in a 25ml reaction volume.

4M Acetone - 5ml
1M HCL - 5ml
0.01M iodine - 5ml

starch solution - 1ml
water - 9ml

C= n/v.

So for example to get this concentration of Acetone. I've been doing 25/1000 to get 0.025 for answer in mol L-1.

So 4/0.025 = 160 mol L-1. Is this the answer? Or because there is 5ml of this in the 25ml total volume I would then have to divide by 5 again to get 32 mol L-1?
1ml = 1dm3
1cm3 = 1000dm3

Number of moles of acetone, HCl and iodine used:

Moles of acetone = volume(dm3) x concentration(mol dm-3)
= (5/1000)dm3 x 4mol dm-3
= 0.02 mol

Moles of HCl = volume(dm3) x concentration(mol dm-3)
= (5/1000)dm3 x 1mol dm-3
= 5x10-3 mol

Moles of iodine = volume(dm3) x concentration(mol dm-3)
= (5/1000)dm3 x 0.01mol dm-3
= 5x10-5 mol

Concentration of acetone, HCl and iodine in the 25ml solution:

Concentration of acetone = (moles x 1000) / volume
= (0.02 x 1000) / 25cm3
= 0.8mol dm-3

Concentration of HCl = (moles x 1000) / volume
= (5x10-3 x 1000) / 25cm3
= 0.2mol dm-3

Concentration of iodine = (moles x 1000) / volume
= (5x10-5 x 1000) / 25cm3
= 2x10-3mol dm-3

I could be entirely wrong though...

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