Eva has a saltwater aquarium. The tank has 95 liters of water, and she puts 3.36 kg of salts in it to make it the right salinity for her fish.

Question

She just bought a new, larger tank. She put 130 liters of water and 5 kg of salt in the new tank.

A) Which is saltier? The old tank or the new tank? Show your work.

B) She wants her new tank to have the same salinity as her old tank. How can she adjust the amount of water or salt to achieve her goal? Describe your reasoning and show your work.

Answer 1

A) The goal of this problem is to determine which tank has the higher concentration of salt. You can think of the concentration as how much salt perunit of water. In this case the original tanke had 3.36 kg of salt in 95 L of water so 3.36 kg/95 L. If we do the math with a calculator we get 0.035 kg salt/ L of water. For the new tank we will do the same kind of math but swap out the numbers for those of the new tank which had 5 kg of salt in 130 liters. 5 kg/130 L=0.038 kg salt/ L of water. When we compare the two numbers we see that the second tank has a slightly higher concentration and is therefore more salty. B) For the second part of the problem a proportion is the easiest way to solve it. We want to make the second tank have the same concentration as the first tank. They tell us that we can either adjust the salt or the water of the second tank to achieve this. I will show how we can either one to make this work. First we will adjust the amount of salt in the second tank. The proportion will be set up to show that the concentration on one side of the equation is equal to the concentration on the other. The only thing that will be a variable is the amount of salt in the second tank which we will call X. So the first tank has 3.36 kg of salt in 95 gallons of water. Our left side of the equation will be 3.36/95. Our second tank has X kg of salt in 130 gallons of water. Our right side of the equation will be X/130. If we set the two sides equal we get 3.36/95=X/130. Next we cross multiply which means multiplying the numerator of one side of the equation with the denominator of the other. We end up with 3.36*130=90*X. If we simplify the left side we get 436.8=90*X. If we divide both sides by 90 we get 4.853=X. So we would need to add 4.853 kg of salt to the second 130 gallon tank to make the saltiness equal. The other way of doing this would be to vary the amount of water in the second tank and keep the amount of salt constant. The left side of the equation would still be 3.36/95 but the right side would now be our original amount of salt (5kg) divided by an unknown amount of water. We will represent the amount of water with the variable Y. We end up with 3.36/95=5/Y. We cross multiply again to get 3.36*Y=95*5. When we simplify we get 3.36*Y=475. When we divide both sides by 3.36 we get Y=141.37. SO the second tank would need to have 141.37 gallons of water with 5 kg of salt to be as salty as the first tank.

Answer 2

A) This question, in asking which tank is saltier is asking which tank tank has a higher concentration of salt in it. To find concentration you simply have to divide the mass of salt in each tank by the volume of each respective tank.

1. Original Tank:
2. concentration(kg/L) = 3.36 kg / 95 L = .03536 kg/L
3. New Tank:
4. concentration(kg/L) = 5 kg / 130 L = .03846 kg/L

As shown above, the new tank is saltier as it has a higher concentration of salt in it.

B) Question B is simply a missing variable problem where you are trying to see how adjusting the amount of salt or the amount of water in the new tank can lead to the same salinity(salt concentration) as in the old tank.

1. Changing the amount of salt:
2. known variables: volume(130 L), concentration(.03536 kg/L)
3. missing variables: mass of salt needed(? kg)
4. Step 1: set up the equation
5. concentration(kg/L) = mass(kg)/volume(L)
6. Step 2: rearrange the variables in order to solve for mass
7. mass(kg) = concentration(kg/L) * volume(L)
8. Step 3: plug in your known variables and solve
9. mass(kg) = 130 L ( .03536 kg/L)

= 4.5968 kg

1. Changing the amount of water:
2. known variables: concentration(.03536 kg), mass of salt(5 kg)
3. missing variables: volume(? L)
4. Step 1: set up the equation
5. concentration(L) = mass(kg)/volume(L)
6. Step 2: rearrange the variables to solve for volume
7. volume(L) = mass(kg)/concentration(kg/L)
8. Step 3: plug in your know variables and solve
9. volume(L) = 5 kg /.03536 kg/L

= 141.4027 L