Frequently Asked Question

Questions asked by users might not always be phrased in the clearest way.

Diclofenac is a non-steroidal anti-inflammatory drug. if the water discharge of your diclofenac manufacturing facility contains 5.0 g/l diclofenac and you wish to remove 99.0% of the drug from 100 l wastewater using n-octane, how much n- octane do you need for a single liquid-liquid extraction? (ko=25.0) a) 257 L B) 396 l c) 425 l d) 437 l

Accepted Answer

Here's how to solve this problem:
Understanding the Concept
This problem involves liquid-liquid extraction, a technique used to separate components from a mixture by exploiting their different solubilities in two immiscible liquids (in this case, water and n-octane).
Key Information:

Diclofenac concentration in wastewater: 5.0 g/L
Desired removal percentage: 99.0%
Volume of wastewater: 100 L
Distribution coefficient (Ko): 25.0 (This represents the ratio of diclofenac concentration in n-octane to its concentration in water at equilibrium)
Calculation
1. Calculate the mass of diclofenac to be removed:

99% of 5.0 g/L
100 L = 495 g
2. Set up the mass balance equation:

Let 'x' be the volume of n-octane needed.

The mass of diclofenac in water after extraction: (5.0 g/L
100 L) - 495 g = 5 g

The mass of diclofenac in n-octane after extraction: 495 g
3. Apply the distribution coefficient:

Ko = (495 g / x) / (5 g / 100 L) = 25.0
4. Solve for x (volume of n-octane):

x = (495 g
100 L) / (5 g
25.0) = 396 L
Answer: You would need 396 L of n-octane for a single liquid-liquid extraction.
Therefore, the correct answer is (B).


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