Question

A water company had its water analysed for Iron by AAS at 248.3 nm. The absorbance of the water, after a 1 in 5 fold dilution was 0.646 at 248.3 nm. A standard solution was prepared by dissolving 0.01483g of iron wire in acid and diluting to 250 mL.

10 mL of the standard was added to 90 mL of the 5 fold diluted test water and the absorbance at 248.3 nm was 0.813. Calculate the ppm in the water sample. A. 24.8ppm ​​B. 212.0ppm C.9.33ppm D. 2.42ppm

Answer 1

Given,

Water analysis is done for Iron by AAS .

Absorbance of water= 0.646

Path length= 248.3 nm

Volume of solution= 250 ml= 0.25 L

Weight of iron= 0.01483 g.

According to Beer's law

A= e L C

Where, A is absorbance.(no unit)

e is molar absorptivity (unit : L /mol cm)

L is path length. (in cm)

C is the concentration of solution.(in mol/ L)

Calculating e from above mentioned data,

0.646= 2.483× 10^-5 × e ×( 0.01483/56× 0.25) (molarity= no. Of moles of solute/ volume of solution (in L)).

10.49 × 10⁵/ 0.03682 = e

e= 284.89 × 10⁵ L/ mol cm.

Now,

New volume of solution= 10+ 90 ml= 100 ml= 0.1 L.

Absorbance= 0.813

e= 284.89 × 10⁵ L/ mol cm (from previous data as it's a constant value)

Path length= 248.3 nm= 2.483× 10^-5 cm.

Let mass of water be x

Again, calculating by beer's law:

0.813= 2.483 × 10^-5× 248.89× 10⁵ × x/(18×0.1).

1.4634/618 = x

x=0.0024679 g

Now, as 1 ppm= 1 mg per litre.

Therefore, 0.0024679 g = 24.8 ppm

Therefore option (A) is correct.