[Spacenova]

I've prepared standard solution according to: 28 mg standard/100 ml. From this take out 1,0 ml and diluted to 100 ml
Test solution: 0,1 g test material/50 ml. Take out from this 10,00 ml and diluted to 50 ml.

To 10 ml standard and test material add 5,00–5,00 ml 8,5 m/V% NaOH. After this add 10,00-10,00 ml Sulfanilic acid and 10,00-10,00 ml 2 M HCl. and 10,00-10,00 ml naphtyl etilen diammin solution.

This was UV measuring method

Standard specific absorbance calc= absorbance*500/ weight in mg
Test specific absorbance calc= absorbance*12500 / weight in mg

Is this calculation OK?

[mjc123]

How did you get these values? They look wrong to me.
Your standard solution is much more diluted than your test solution, so the specific absorption of the standard should be higher than the test sample. (Check units. Have you gone a factor of 1000 wrong somewhere?)
(And what do you mean by "5.00-5.00 ml" etc? Have you incorrectly copied a range?)

[Spacenova]

Test solution
Prepare a solution containing 1.0 mg of
test material 250.0 ml of methanol R.
Reference solution
Dissolve70.0mg of ref material R
in methanol R and diluteto250.0ml with the same solvent.
Dilute 5.0 ml of the solution to 500.0 ml with
methanol R .
Into three 50 ml volumetric flasks introduce 10.0 ml of
the test solution, 10.0 ml of the reference solution and
10 ml of methanol R as a blank. To each flask add 5 ml
of dilute sodium hydroxide solution R
, close the flask, mix and allow to stand at room temperature for 30 min.
Add 10 ml of sulphanilic acid solution R
and 10 ml of dilute hydrochloric acid R
and mix. After exactly 4 min,
add 10 ml of naphthylethylenediamine dihydrochloride
solution R ,dilutetovolumewith water R
and mix.

This is the correct method. How can calculate specific absorbance of test method and reference method?

[mjc123]

You have not answered my question. How did you arrive at the answers you gave? If we can't see your working, we can't tell where you might have gone wrong.
As to the sample prep,

Reference solution
Dissolve70.0mg of ref material R
in methanol R and diluteto250.0ml with the same solvent.
Dilute 5.0 ml of the solution to 500.0 ml with
methanol R

OK, this is equivalent to dissolving 28 mg in 100 ml, and diluting 1 ml to 100 ml. (What are all the Rs for?) But

Test solution
Prepare a solution containing 1.0 mg of
test material 250.0 ml of methanol R.

This is not equivalent to 0.1 g in 50 ml, or to the diluted solution (≡ 0.1 g in 250 ml). What is right? What did you do?

dilutetovolumewith water R and mix.

Ah, you didn't mention that. Dilute to what volume? If it's 50 ml, then I think your expression for specific absorbance of test sample is right - for the amounts you said you used, not what you have just quoted. But the expression for the standard is still out by a factor of 1000.

[Spacenova]

So Referencia solution calculation: absorbance*500/weight in mg/1000
Test material:absorbance*12500/weight in g

500:5X100
12500: 500X5X5

Is it OK?

[mjc123]

No! I said test sample right, standard wrong!
You still haven't answered my question. Show us how you worked out those numbers!

[Spacenova]

So Referencia solution calculation: absorbance*500/weight in mg/1000
Test material:absorbance*12500/weight in g

500:5X100
12500: 500X5X5

[mjc123]

That still doesn't help. Where do those numbers come from. 5 what * 100 what? Why 2 numbers in one case but 3 in the other?