Brine is a solution of salt and water. if a tub contains 70 pounds of a 5% solution of brine, how much water (to the nearest tenth lb.) must evaporate to change it to an 7.5% solution? lbs. of water must evaporate.

Answer is: 23,33 ib.
mr - mass of solution. mr = mrs + m(water)
mrs - mass of solute.
ω - mass percent of solution, mass percent = the grams of solute divided by the grams of solution. ω = mrs÷mr
mr₁ = 70 ib.
ω₁ = 5% = 0,05.
mrs₁ = 0,05 · 70 ib = 3,5 ib.
mrs₁ = mrs₂, ω₂ = 7,5% = 0,075
mr₂ = mrs₂ ÷ ω₂ = 3,5 ib ÷ 0,075 = 46,66 ib.
m(water) = mr₁ - mr₂ = 70 - 46,66 =23,33 ib.

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Larus Larus · Answer 2 · 2019-05-31T03:11:36+02:00

Answer:

The correct answer is 23.3 lbs of water.

Explanation:

Let us first model the equation, which expresses the problem. Let x be the amount of water evaporated.

0.05(70) = (0.075)(70-x)

It is to be the noted that the expression is being modeled by keeping in mind the amount of salt (solute) will always be equivalent, and it is the water (solvent), which will modify thus, influencing the percentage of the amount of the solution.

Therefore, solving for x we get:

70-x = 0.05(70)/(0.075)

x = 23.3 lbs of water