contestada

How many mL of 75% alcohol should be mixed with 10% of 1000 cc alcohol to prepare 30% of 500 mL alcohol solution? a. 346.16 mL b. 234.43 mL c. 153.84 mL d. 121.12 mL e.

Respuesta :

MathPhys

Answer:

C. 153.84 mL

Step-by-step explanation:

Let's say x is the volume of 75% solution and y is the volume of 10% solution.

Sum of the volumes:

x + y = 500

Sum of the alcohol amounts:

0.75x + 0.10y = 0.30(500)

0.75x + 0.10y = 150

Solve the system of equations using either substitution or elimination.  I'll use substitution.

y = 500 - x

0.75x + 0.10 (500 - x) = 150

0.75x + 50 - 0.10x = 150

0.65x = 100

x = 153.84

You need 153.84 mL of 75% solution.

Cricetus

"153.84 mL" of 75% alcohol should be added. A further explanation is provided below.

Let,

  • 75% alcohol used be "x".
  • 10% alcohol used be "y".

then,

→ [tex]x+y = 500[/tex]

         [tex]y = (500-x)[/tex]

now,

→ [tex]75(x)+10(500-x) = 500\times 30[/tex]

                [tex]65x+5000=15000[/tex]

                           [tex]65x=15000-5000[/tex]

                           [tex]65x=10000[/tex]

                               [tex]x = \frac{10000}{65}[/tex]

                                  [tex]= \frac{2000}{13}[/tex]

                                  [tex]= 153.84 \ mL[/tex]

Thus the above response i.e., "option c" is correct.

Learn more:

https://brainly.com/question/12925084

Otras preguntas