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A container holds 10 liters of a solution which is 20% acid.
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Updated on: 05 Dec 2018, 19:56
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A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?
Re: A container holds 10 liters of a solution which is 20% acid.
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05 Dec 2018, 13:25
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nadeem790 wrote:
A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?
a) 5 b) 10 c) 20 d) \(33 \frac{1}{3}\) e) 50
Let's keep track of the acid
A container holds 10 liters of a solution which is 20% acid. 20% of 10 liters = 2 liters So, there are 2 liters of ACID to start
6 liters of pure acid are added to the container 2 + 6 = 8 So, there are now 8 liters of acid in the RESULTING solution.
What percent of the resulting mixture is acid? We added 6 liters to 10 liters, to get a RESULTING solution with volume = 16 liters 8/16 = 1/2 = 50%
Re: A container holds 10 liters of a solution which is 20% acid.
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05 Dec 2018, 19:59
Expert Reply
nadeem790 wrote:
A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?
a) 5
b) 10
c) 20
d) \(33 \frac{1}{3}\)
e) 50
20% acid in 10 liter of solution means \(10*\frac{20}{100}=2\)..
When we add 6 liters of acid.. Acid becomes 2+6=8 Total solution becomes : 10+6=16
so acid as % of resulting mixture = \(100*\frac{8}{16}=50\)%