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?
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.
.
+36916 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
+128475 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
