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A pot contains 6 L of brine at a concentration of 110 g/L. How much of the water should be boiled off to increase the concentration to 200 g/L?

 Feb 13, 2020
 #1
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+3

 

A pot contains 6 L of brine at a concentration of 110 g/L. How much of the water should be boiled off to increase the concentration to 200 g/L?

 

If there's 6 L of brine at 110 g/L, there has to be 660 g of salt in it. 

That way, the ratio obtained by division is (660 g) / (6 L)  = 110 g/L 

 

When you boil water off, the salt doesn't boil off, so the resulting stronger brine will still contain 660 g of salt. 

 

So what must you divide 660 by to get 200?        (660 g) / (x)  =  200 g/L 

 

Multiply both sides by (x)                                (x) • (660 g) / (x)  =  (x) • (200 g/L) 

Cancel out the x's on the left                           (x) • (660 g) / (x)  =  (x) • (200 g/L) 

Then you have                                                        (660 g)         =  (x) • (200 g/L) 

 

Divide both sides by 200 g/L                             (660 g) / (200 g/L)  =  x 

 

Swap sides for no other reason than 

because my personal preference is 

to have the unknown on the left                            x  =  (660 / 200) L  =  3.3 L 

 

That's how much water you need remaining 

in the brine so subtract that amount from the 

original 6 L to find the amount needed to be 

boiled off                                                                 6.0  – 3.3 L  =  2.7 L that you need to boil off 

 

Check back at the first to see if this would be 

reasonable.  You started with 6 L, so imagine 

boiling half of the water off.  That would double 

the concentration to 220, but that's too strong, 

so it's reasonable that you'd boil off slightly less 

than half of the 6 L.  2.7 looks like a good answer. 

.

 Feb 13, 2020
 #4
avatar+109740 
0

Thx, Guest......

 

Checking to see that a solution [ pardon the pun ]  is "reasonable" is always a good thing  !!!

 

 

 

cool cool cool

CPhill  Feb 13, 2020
 #2
avatar+22108 
+2

6000 l   x 110 g/l = 660 gm

 

 

660 gm / x liters = 200 g / liter       solve for 'x'      then amount to be boiled off  =    6 Liters - x

 Feb 13, 2020
 #3
avatar+109740 
+1

Note that  pure water  contains   0 concentration of brine....so....

 

1L = 1000 g

110g/L  =  110g/1000g =  11% concentraion =.11

200g /L  = 200g/1000g  =  20% concentration  = .20

 

So

 

6 ( .11)  - x* (0) =  (6 - x) (.20)      where x  is the amount of water to be  boiled off  in L

 

6 (.11)  = 1.2 - .2x

 

.66 = 1.2 - .2x      rearrange as

 

.2x  = 1.2 - .66

 

.2x = .54

 

x = 2.7  L 

 

 

cool cool cool

 Feb 13, 2020

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