Question

Keiko needs 100mL of a 5% acid solution for a science experiment. She has available a 1% solution and a 6% solution. How many milliliters of the 1% solution and how many milliliters of the 6% solution should she mix to make the 5% solution?

Answer 1

Answer:

Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution

Explanation:

There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:

1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%

2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%

When clearing X and Y :

(X mL 6%    +   Y mL 1%      =    100 mL 5%) (-15,7)

X mL 94%  +   Y mL 99%   =   100 mL 95%

_______________________________

 -                      Y  0.83       =   16.5

                        Y                =   19.9 mL 1% solution

Replace Y in anyone equation and X = 80 mL 6% solution

I hope to see been helpful

Answer 2

20 ml of the 1% solution and 80 ml of the 6% solution should she mix to make the 100 ml of 5% solution

Further explanation

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. Concentration shows the amount of solute in a unit of the amount of solvent.

Molarity (M)

Molarity shows the number of moles of solute in every 1 liter of solution.

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = volume of the solution

For molarity a mixture can be formulated  :

$M_{mix}=\frac{total\:mole}{total\:volume}$

$M_{mix}=\frac{M1.V1+M2.V2}{V1+V2}$

Keiko needs 100 mL of a 5% acid solution

The available solutions are 1% and 6%

We can make a mathematical equation from the two statements above

• 1. V1 + V2 = 100

The total volume of the 1% and 6% solutions is 100 ml

• 2. The mixture concentration is 5%

5% = 1% .V1 + 6% .V2 /( V1 + V2 )

5% = 1% .V1 + 6% .V2 / 100

5 = 1% .V1 + 6% .V2 or

500 = V1 + 6V2

Equations 1 and 2 we do elimination and substitution

500 = V1 + 6V2

100  = V1 + V2

----------------------- -

400 =  5V2

V2 = 80 ml

V1 = 20 ml

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Keywords: reaction coefficients, potassium carbonate, electrolytes