To determine the ammonia concentration in a sample of lake water, three samples are prepared. in sample a, 10.0 ml of lake water is mixed with 5 ml of phenol solution and 2 ml of sodium hypochlorite solution, and diluted to 25.0 ml in a volumetric flask. in sample b, 10.0 ml of lake water is mixed with 5 ml of phenol solution, 2 ml of sodium hypochlorite solution, and 2.50 ml of a 5.50×10−4 m ammonia solution, and diluted to 25.0 ml. sample c is a reagent blank. it contains 10.0 ml of distilled water, 5 ml of phenol solution, and 2 ml of sodium hypochlorite solution, diluted to 25.0 ml. the absorbance of the three samples is then measured at 625 nm in a 1.00 cm cuvet. the results are shown in the table. sample absorbance (625 nm)a 0.419b 0.666c 0.045what is the molar absorptivity of the indophenol =m−1cm−1what is the concentration of ammonia in the lake water? [nh3]lake water=m

Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

they vary. it used to be meat, grain, bread, candy, beverages, donuts, and fruits/vegetables. now they added exercise in.