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Ipatiy [6.2K]
3 years ago
12

150 lb of NaCl are initially dissolved in 500 gal of a NaCl solution in a container. Water is continuously added to the containe

r at a rate of 5 gal/min and the container content, which is kept uniform by mixing, is also withdrawn out of the container at 5 gal/min. Determine how much NaCl will be in the tank at the end of 1.5 hours.
Chemistry
1 answer:
puteri [66]3 years ago
5 0

Answer:

After 90 minutes there will be 60.5 lb NaCl in the tank

Explanation:

Initially, you have 150 lb/500 gal = 0.300 lb/gal NaCl

After 1 minute, the volume will be the same but you will have less NaCl in the solution.

After the first minute, the mass of NaCl you will lose will be:

0.3 lb/gal · 5 gal = 1.5 lb

Then, the concentration of the solution will be:

(150 lb - 1.5 lb)/ 500 gal = 0.297 lb/gal

After the second minute:

mass of NaCl lost = 0.297 lb/gal · 5 gal = 1.485 lb

concentration of the solution = (148.5 lb - 1.485 lb) / 500 gal = 0.29403 lb/gal

After the third minute:

mass of NaCl lost = 0.294 lb/gal · 5 gal = 1.47015 lb

concentration of the solution = 147.015 lb - 1.47015 lb / 500 gal = 0.2910897 lb/gal

Notice that the solution is being diluted by 1% every minute. Then, the concentration of the solution after every minute will be 99% of the previous solution. For example, after the first minute, the concentration of the solution  will be 99% of the original solution (99% of 0.3 lb/gal = 0.297 lb/gal). Mathematically, we can express it as follows:

0.3  lb/gal · 0.99 = 0.297 lb/gal

After the second minute:

0.297 lb/gal · 0.99 = 0.29403 lb/gal

Since 0.297 lb/gal = 0.3  lb/gal · 0.99

we can write for the second minute:

0.3  lb/gal · 0.99 · 0.99 = 0.29403 lb/gal

And for the third minute:

0.3  lb/gal · 0.99 · 0.99 · 0.99 = 0.2910897 lb/gal

And for the nth minute:

0.3 lb/gal · 0.99ⁿ = concentration of the solution after n minutes

Now, we can find the concentration of the solution after 90 min:

0.300 lb/gal · 0.99^90 = 0.121 lb/gal

The mass of NaCl after 90 minutes will be:

0.121 lb/gal · 500 gal = 60.5 lb

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<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

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3.301 mass % of potassium hydroxide solution.

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Mass of solvent = (100 - 3.301) g = 96.699 g

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W_{solvent} = Mass of solvent = 96.699 g

Putting values in above equation, we get:

\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m

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