contestada

A chemist needs 10 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20% and 50%. How many liters of each solution will satisfy each condition? a) Use 2 liters of the 50% solution. b) Use as little as possible of the 50% solution. c) Use as much as possible of the 50% solution.

Respuesta :

nandhini123

Answer:

a) 1 litre of  10% solution and 7 litre of 20% solution

b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) 3.75 litres of 50% solution and 6.25 litres of 20% solution

Explanation:

Given:

chemist needs = 10 liters of a 25% acid solution

Concentration of three solutions that are to be mixed = 10%, 20% and 50%.

Solution:

A) Use 2 liters of the 50% solution

Let us mix this with 10% and 20% solution

They will have to equal 8 litres

Let x=20% solution

Then (8-x) =10%

So the equation becomes,

10%(8-x)+ 20%x+50%(2)=25(10)

(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)

0.8-0.1x+0.2x+1.0=2.5

0.2x-0.1x=2.5-0.8-1.0

0.1x=0.7

[tex]x=\frac{0.7}{0.1}[/tex]

x= 7

so, 8-x = 8 -7= 1 litre of  10% solution and 7 litre of 20% solution

B)Use as little as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+20%(10-x)=25%(10)

0.50x+0.2(10-x)=0.25(10)

0.5x+2.0-0.2x=2.5

0.3x=2.5-2.0

0.3x=0.5

[tex]x=\frac{0.5}{0.3}[/tex]

x=1.67  

now (10-x)=(10-1.67)=8.33

so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) ) Use as much as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+10%(10-x)=25%(10)

0.50x+0.1(10-x)=0.25(10)

0.5x+1.0-0.1x=2.5

0.4x=2.5-1.0

0.4x=1.5

[tex]x=\frac{1.5}{0.3}[/tex]

x=3.75

Now, (10-x)=(10- 3.75)=6.26

So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution

The amount of liters of each solution to satisfy each given condition are;

A) 8.33 liters of 20% solution

B) 6.25 liters of 10% solution

C) 1 liter of the 10% solution

What is the required volume of solution?

A) Use as little as possible the 50% solution.

Mix it with 20% solution only

Let x be the amount of 50% solution

Thus;

(10 - x) = 20% solution

equation:

0.50x + 0.20(10 - x) = 0.25(10)

0.5x + 2 - 0.2x = 2.5

0.3x = 2.5 - 2

0.3x  = 0.5

x = 0.5/0.3

x = 1.67 liters of 50% solution required

Thus; 10 - 1.67 = 8.33 liters of 20% solution

B) Use as little as possible of the 50% solution;

Mix it with the 10% solution only

Let x be amount of 50% solution

Thus;

(10 - x) = 10% solution

equation:

0.50x + 0.10(10 - x) = 0.25(10)

0.5x + 1 - 0.1x = 2.5

0.4x = 2.5 - 1

x = 1.5/0.4

x = 3.75 liters of 50% solution required. Thus;

10 - 3.75 = 6.25 liters of 10% solution

C) Use 2 liters of the 50% solution

Mix it with 10% and the 20% and they will have to equal 8 liters.

Let x be the amount of 20% solution

Thus;

8 - x = 10% solution

Equation:

0.20x + 0.10(8 - x) + 0.50(2) = 0.25(10)

0.20x + 0.8 - 0.10x + 1 = 2.5

0.2x - 0.1x + 1.8 = 2.5

0.1x = 2.5 - 1.8

0.1x = 0.7

x = 0.7/0..1

x = 7 liters of 20% solution

Thus; 8 - 7 = 1 liter of the 10% solution

Read more about volume of solution at; https://brainly.com/question/25736513

Otras preguntas