A chemist has 10 liters of a solution that is 10 percent nitric acid b

Question

A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

JeffTargetTestPrep · Accepted Answer

We see that 0.1 x 10 = 1 liter is nitric acid. We can let w = amount of water that must be added to dilute the solution to 4%:

4/100 = 1/(10 + w)

4(w + 10) = 100

w + 10 = 25

w = 15

Alternate Solution:

We have 10 liters of 10% nitric acid. To it we add x liters of water (which has 0% nitric acid), and the result is (10 + x) liters of 4% nitric acid. We can express this in an equation as:

10(0.10) + x(0) = (10 + x)(0.04)

1 + 0 = 0.4 + 0.04x

0.6 = 0.04x

60 = 4x

15 = x

Answer: A

broall · Answer

In 10 liters of that solution, there is 1 liter nitric acid and 9 liters water.

In the new solution that contains 4% acid or 1 liter acid. This means the total volume of the new solution is 1 / 4% = 25 liter.
The volume of water need to be added is 25 - 10 = 15 liters.

The answer is A.

gracie · Answer

A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

let w=liters of water to be added
.10*10=.04(10+w)
w=15 liters
A

Benacher07 · Answer

Here, 10% of 10 liters is 1 liter, which will become 4% :
If 1L ==>4%
Then 24 ==> 96%

And since we already have 9 liters of water we still need to add 15 liter to reach 24 : 24-9 =15
Hence answer : A

abar7 · Answer

We can solve this by ratios as well.

\frac{Original content of Acid}{Original Content of water+Water to be added}=\frac{Acid Content Required}{Water Content Required}

\frac{.1*10}{.9*10+x}= \frac{4%}{96%}

\frac{1}{9+x}=\frac{1}{24}

24=9+x
Hence, x=15.
Therefore, 15 liters of water must be added.

AkshdeepS · Answer

LeoN88 · Answer

Total= 10 lt
Acid= 1
Water= 9
Water to add= x

1/(10+x) = 4/100

so x is 15.

coylahood · Answer

Using the weighted average method we can get the weights of the different solutions as follows:

Weight 1/Weight 2= (C2-Average)/(Average - C1)
Where C1 and C2 represent the concentrations of the first and second solutions respectively.
C1 = 10%;
C2 = 0% (Note that concentration of nitric acid in water is 0%);
Average = 4%
therefore W1/W2=(0-4)/(4-10) = -4/-6 = 2/3.
Therefore for every 2 liters of the 10% nitric acid solution, 3 liters of water should be added.
Since there are 10 liters of the 10% nitric acid solution, 15 liters of water must be added to reduce the concentration of nitric acid in the solution to 4%.

mm229966 · Answer

In 10 liters - 1 liter acid and 9 liters water.
We're now adding water, meaning we're not changing that 1 liter acid. SO in the new mix 4% would be 1 liter and the whole new mix 25 liter; 25-10 = 15 litres - A.

AnushkaGovil123 · Answer

can someone explain this by mixture diagram?

TIA

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A chemist has 10 liters of a solution that is 10 percent nitric acid b

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A chemist has 10 liters of a solution that is 10 percent nitric acid b

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