SOLUTION: In a tropical fish store, one tank contains 65 kilograms of seawater with 7% salt. a. How much pure water must be added to have 5% salt? b. How much water must be evaporated to c

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Question 1161827: In a tropical fish store, one tank contains 65 kilograms of seawater with 7% salt.
a. How much pure water must be added to have 5% salt?
b. How much water must be evaporated to contain 9% salt?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
you start with 65 kilograms of seawater that is 7% salt.
that means that 7% of the seawater is salt and 93% of the seawater is pure water.
that means you have 4.55 kilograms of salt and 60.45 kilograms of pure water.

if you want to get 5% seawater, you have to add pure water.
if you want to get 9% seawater, you have to add salt.

we'll use the variable called x to determine the amount of pure water to add to get 5% seawater and then we'll use the variable called x to determine the amount of salt to add to get 9% seawater.

if you want 5% seawater, than you have to have 95% pure water.
your first equation is therefore:
.95 * (65 + x) = 60.45 + x
solve this equation and you will get x = 26 kilograms.
60.45 + 26 = 86.45 kilograms of pure water.
add 4.55 kilograms of salt (you didn't add any more salt to the mixture, only pure water) to get 91 kilograms of seawater.
86.45 / 91 = .95 = 95% pure water.
that means 5% salt which is what you want.

if you want 9% salt, then your second equation becomes:
.09 * (65 + x) = 4.55 + x
solve this equation and you will get x = 1.428571429 kilograms.
4.55 + x = 5.978571q429 kilograms of salt.
add 60.45 kilograms of water (you didn't add any pure water to the mixture, only salt) to get 65.97857143 kilograms of seawater.
4.978571429 / 65.97857143 = .09 = 9% salt.

your solutions are therefore.
add 26 kilograms of pure water to 65 kilograms of seawater that is 7% salt to get 91 kilograms of seawater that is 5% salt.
add 1.428571429 kilograms of salt to 65 kilograms of seawater that is 7% salt to get 65.97857143 kilograms of seawater that is 9% salt.

HOWEVER, .....

you were not asked to add salt to get seawater that is 9% salt.
you were asked to evaporate pure water to get seawater that is 9% salt.
that's a different equation.

that equation is:
60.45 - x = .91 * (65 - x)
you are evaporating x kilograms of pure water from a mixture that is 93% pure water to get a mixture that is 91% pure water, making it 9% salt.
solve this equation and you will get x = 14.4444444.....
60.45 - x = 46.0055555... kilograms of pure water that has not been evaporated and is therefore left in the mixture.
65 - x = 50.555555..... kilograms of seawater that is left after the pure water has been evaporated.
46 .0055555..... / 50.55555.... = .91 = 91% pure water.
that makes the mixture 9% salt.

so your solutions become:
add 26 kilograms of pure water to 65 kilograms of seawater that is 7% salt to get 91 kilograms of seawater that is 5% salt.
evaporate 14.44444.... kilograms of pure water from seawater that is 7% salt to get 50.55555..... kilograms of seawater that is 9% salt.

in both cases, the amount of salt was not changed. the only changes was in the amount of pure water in the mixture.




Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.

(a)  The amount (the mass) of salt is  0.07*65 kilograms = 4.55 kilograms.


     To dilute it to 5% salt by mass, the amount (the mass) of water "m" should be added to get


         = 0.05.


    Solve this equation for "m".


         = 65 + m

        7*13 = 65 + m

        91   = 65 + m

        m    = 91-65 = 26 kilograms of pure water, or 26 liters.


    ANSWER.  26 liters (kilograms) of water should be added.


Part (a) is just solved. 


(b)  The mass "m" of water should be evaporated to satisfy this concentration eauation


         = 0.09.


    Solve it for "m"


         = 65 - m

        50.555      = 65 - m

        m           = 65 - 50.555 = 14.444  kilograms of water must be evaporated.


    ANSWER.  14.444 kilograms / (liters) of water must be evaporated.

Solved.

-------------------

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    - How much water must be evaporated
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