In a textile mill, a double-effect evaporator system concentrates weak liquor containing 4% (by mass) caustic soda to produce lye contains 25% solids (by mass). What is the weight of water evaporated per 100 g feed in the evaporator?
(A) 125.0 g
(B) 50.0 g
(C) 84.0 g
(D) 16.0 g

Hint: To solve this question, we first need to understand what is lye. Lye can be defined as a metal hydroxide or caustic basic solution of a highly soluble strong alkali in water. Generally, lye refers to NaOH (sodium hydroxide) solutions.

Complete answer:
Caustic soda is the common name for the sodium hydroxide of NaOH.
Now, it is given to us that the weak liquor contains 4% (by mass) caustic soda.
So, we can say that 100 gm of the liquor feed contains 4 gm caustic soda or NaOH.
Let us assume that x gm of lye containing 25% (by mass) solids or NaOH is produced.
So, the weight of caustic soda in the lye produced = $\dfrac{25}{100}x$ gm.
Since the weight of caustic soda in the feed liquor and the product lye will be the same, so
\[\begin{align}
  & \dfrac{25}{100}x=4 \\
 & x=16\text{ gm} \\
\end{align}\]
Hence, 16 gm of lye is produced when 100 gm of liquor is concentrated in the double-effect evaporator system.
So, the weight of the water evaporated will be to reduce the feed liquor to lye product will be 100 - 16 = 84 gm

Hence the correct answer is option (C) 84.0g.

Note:
It should be noted that a metal hydroxide lye can also be produced by leaching wood ashes. The process of extracting solute from its carrier substance through a solvent is known as leaching.
Lye can be used during soap making, to cure foods, for cleaning, digest animal carcass tissues, identify funguses, etc.