Question

A sample of cooling liquid containing ethylene glycol and water is brought to your lab. The cooling tower contains 2,500 tons of this coolant to keep the chillers at an oil refinery working properly. It is critical that the concentration of the ethylene glycol be maintained at a concentration of 44.0±0.5%(w/w), to prevent freezing and transfer the heat properly. Since it’s late and an unexpected cold front has the plant manager worried, you decide to use the refractometer to run this test as fast as possible, yielding a result of 1.3687. Based upon published literature values, does the coolant meet its specification? If not, calculate how much water or ethylene glycol should be added to correct the situation. Would the final volume fill an Olympic-sized pool (660,000 gal)?

The RI indicates its only 36% so I know that part. I'm just not sure on the last two parts of the question.

Answer 1

Knowing that the current concentration of ethylene glycol is only 36%, you can use the following equation to determine the amount of it that is needed:

C₁m₁ = C₂m₂

Where C 1 and 2 are concentrations and m 1 and 2 are the corresponding masses.

Now, substitute with the values you already know: (36%)(2500tons) = (44%)(m₂)

Now simply solve for the desired mass, getting m₂= 3055.55tons. And remember to subtract the previous quantity, to know how much of it there is to add to the current mass, getting then: V= 3055.55 - 2500 = 555.55 tons of ethylene glycol.

From the initial concentration you can know how much water and ethylene glycol you had in the first place:
36% = x / 2500tons, x= 900tons of ethylene glycol and 1600 tons of water.
This also means that you have 900,000 kg of ethylene glycol and 1,600,000 kg of water.

Water has a density of 1 kg /L, from which you can calculate the volume with the equation
1kg/L = 1600000 / v; giving us a volume of 1,600,000 L.

As for ethylene glycol, the density is 1.11kg /L.
1.11kg/L = (900000 + 555555.55 kg) / v; giving us a volume of: 1,311,306.31 L.

And a total volume of: 1,600,000L + 1,311,306.31L = 2,911,306.31 L
Just convert from liters to gallons as:

Then it would be able to fill an olympic pool