It would be a ehtanol plant
Volume of the solution =
= 2 L solution x 
Volume of solute = 7.5 mL
Volume of water (solvent) = 2000 mL - 7.5 mL = 1992.5 mL water
Explanation:
The given data is as follows.
= 211 nm, 
l = 1 cm, 7% < Transmittance < 85%
Suppose the aqueous solution follows Lambert-Beer's law. Therefore,
Absorbance = 
Hence, for 7% transmittance the value of absorbance will be as follows.
Absorbance = 
= 1.155
For 85% transmittance the value of absorbance will be as follows.
Absorbance = 
= 0.07058
According to Lambert-Beer's law.
A = 
where, A = absorbance
= molar extinction coefficient
C = concentration
Therefore, concentration for 7% absorbance is as follows.

= 
= 
= 0.187 mmol/L
Concentration for 85% absorbance is as follows.

= 
= 
= 0.01144 mmol/L
Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.
Ecell = E°cell - RT/vF * lnQ
R is the gas constant: 8.3145 J/Kmol
T is the temperature in kelvin: 273.15K = 0°C, 25°C = 298.15K
v is the amount of electrons, which in your example seems to be six (I'm not totally sure)
F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
Q is the concentration of products divided by the concentration of reactants, in which we ignore pure solids and liquids: [Mg2+]³ / [Fe3+]²
Standard conditions is 1 mol, at 298.15K and 1 atm
To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:
E°cell = cathode - anode
Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above.
When you have found E°cell, you can just solve with the numbers I gave you.
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