Question 1188338
.
A 700-gram brine solution contains 20% salt by mass. If 300 grams of water is evaporated
from it, what percent of the new brine solution is salt?
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Let me explain you, &nbsp;what will happen &nbsp;<U>IN &nbsp;REALITY</U> &nbsp;with the salt and with the brine solution.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;From Wikipedia, the free Encyclopedia  &nbsp;&nbsp;&nbsp;&nbsp;https://en.wikipedia.org/wiki/Brine


<pre>
    Brine is a high-concentration solution of salt (NaCl) in water (H2O). In diverse contexts, 
    brine may refer to the salt solutions ranging from about 3.5% (a typical concentration of seawater, 
    on the lower end of that of solutions used for brining foods) up to about 26% (a typical saturated solution, 
    depending on temperature). Brine forms naturally due to evaporation of ground saline water but 
    it is also generated in the mining of sodium chloride.[1] Brine is used for food processing and cooking 
    (pickling and brining), for de-icing of roads and other structures, and in a number of technological processes. 
    It is also a by-product of many industrial processes, such as desalination, so it requires wastewater treatment 
    for proper disposal or further utilization (fresh water recovery).
</pre>


In the course of water evaporating, &nbsp;the process will go smoothly until the salt concentration will reach 
the value of &nbsp;26%, &nbsp;approximately.



At this point, &nbsp;the brine solution becomes &nbsp;SATURATED &nbsp;and its concentration does not  change any more.


The exceed mass of the salt will go as a sediment &nbsp;( ! not a mixture ! ) &nbsp;at the bottom of the reservoir.


The concentration of the salt in the brine will remain &nbsp;26%.



See this Wikipedia article &nbsp;&nbsp;&nbsp;&nbsp;https://en.wikipedia.org/wiki/Brine



It is the process which students learn in the middle school in their standard &nbsp;Science class.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Nevertheless, &nbsp;although the process seems to be complicated and is non-linear, 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;it can be computed, &nbsp;in full, &nbsp;to the end. &nbsp;This computation is given below.



<pre>
700-grams brine solution contains 20% salt by mass.

    So, there are  0.2*700 = 140  grams of salt in the solution, at the beginning,
    and 700-140 = 560 grams of water, correspondingly.


After 300 grams of water evaporated, 560-300 = 260 grams of water remained.

This amount  of liquid contains 26% of salt, by mass


    {{{x/(260+x)}}} = 0.26,


which gives x = 91.35 grams of salt dissolved.


The rest of salt,  140 grams - 91.35 grams = 48.65 grams, went from the original brine solution to the sediment.


<U>ANSWER</U>.  After 300 grams of water evaporated, the remained brine contains 260 grams of water and 91.35 grams of salt.

         This brine is saturated at 26% concentration.  The amount of salt of 48.65 grams went from the original brine solution to the sediment.
</pre>

Solved, &nbsp;answered and explained.