Question

Plateau Creek carries 5.0 m^3 /s of water with a selenium (Se) concentration of 0.0015 mg/L. A farmer withdraws water at a certain flowrate (m3 /s) from the creek for irrigation. After using the water, half of the water withdrawn returns to the creek and contains 1.00 mg/L of Se. Fish in the creek are sensitive to Se levels over 0.04 mg/L.
Required:

How much water (m3 /s) can the farmer withdraw from the stream to maintain Se at 0.04 mg/L after the contaminated water is returned to the creek?

Answer 1

Answer:

The correct answer is "4.8137 m³". The further explanation is given below.

Explanation:

Firstly we have to calculate the concentration of Se:

$C = 0.0015 \ mg/L\times \frac{1g}{1000 mg}\times \frac{1 \ mol}{79 \ g}$

   $=1.9\times 10^{-8} \ mol/L$

Concentration the fish can take:

$=0.04 \ mg/L\times \frac{1 \ g}{1000mg}\times \frac{1 \ mol}{79 \ g}$

According to the general dilution principle will be:

⇒  $M_1V_1 = M_2V_2$

The volume that can take the farmer will be:

$V_2 = 1.9\times 10^{-8} M\times \frac{5\times 10^3 \ L}{5.1\times 10-7 M}$

    $=186.27 \ L$

On converting this into m³, we get

= $0.18627 \ m^3$

Finally the volume the farmer can remove would be:

$V = 5-0.18627$

   $= 4.8137 \ m^3$