SOLUTION: 1.How much water must be evaporated from 170 liters of a 75% solution of salt inoder to obtain an 85% brine solution. 2.There is 60 gallons of 50% solution of glycerin and water.H

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Question 193293: 1.How much water must be evaporated from 170 liters of a 75% solution of salt inoder to obtain an 85% brine solution.
2.There is 60 gallons of 50% solution of glycerin and water.How many gallons of water must be added to such solution inroder to reduve the glycerin concentration to 12%.
3.How many gallons of 40% salt solution and 80% salt should be mixed inorder to have 40 gallons of 50% salt solution?
4.How much water must be added to 100 liters of denatured alcohol 90% pure to dilute into 75% alcohol content.
THANK YOU~

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
How much water must be evaporated from 170 liters of a 75% solution of salt in
order to obtain an 85% brine solution.
:
Let x = amt of water to be evaporated
write an equation for the amt of salt in each solution
.75(170) = .85(170-x)
127.5 = 144.5 - .85x
.85x = 144.5 - 127.5
x =
x = 20 liters to be evaporated
:
You can check this, the amt of salt remains the same, (only the % changes)
.75(170) = .85(150)
:
:
2.There is 60 gallons of 50% solution of glycerin and water.
How many gallons of water must be added to such solution in order to reduce
the glycerin concentration to 12%.
Let x = amt of water to be added
.5(60) = .12(60 + x)
30 = 7.2 + .12x
30 - 7.2 = .12x
22.8 = .12x
x =
x = 190 gallons has to be added
:
Check solution:
.5(60) = .12(60+190)
:
:
3.How many gallons of 40% salt solution and 80% salt should be mixed in order
to have 40 gallons of 50% salt solution?
Let x = amt of 80% salt solution required
then. since the total amt is to be 40 gal:
(40-x) = amt of 40% solution required
.8x + .4(40-x) = .5(40)
See if you can solve and check this one out, email me if you can't
;
:
4.How much water must be added to 100 liters of denatured alcohol 90% pure to dilute into 75% alcohol content.
Let x = amt of water to be added
A amt of alcohol equation
.9(100) = .75(100+x)
If I have explained this well, you should be able to do these last two problems
Let me know if I have (or haven't)

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