Question

How many ml of both 75% and 5% stock must be combined to prepare 16 ounces of a 25% product?

Answer: [a] ml of 75% stock (#ml) Round to the nearest whole number

Answer: [b] ml of 5% stock (#ml)  Round to the nearest whole number

Answer 1

A x% solution means if we take total volume of 100 of mL of that solution, then the volume of solute present should be x mL

Now given one solution has concentration = 75%

And another solution has concentration = 5%

Required final concentration (M2) = 25%

And final volume (V2) = 16 ounces

= 16 * 29.5 mL

= 472‬ mL

That means total volume of solute in this solution = 472‬ mL * 25%

= 118 mL

Lets say we have to take x mL of 75% solution.

So volume of solute in 75% solution = 75% * x

= 0.75x mL

And

Lets say we have to take y mL of 5% solution.

So volume of solute in 25% solution = 5% * y

= 0.05y mL

So we can say-

0.75x + 0.05y = 118 mL

Now for the 75% solution, lets take

initial concentration of solution (M1 =) = 75%

Final concentration of solution (M2 =) = 25%

Final volume of solution (V2) = 118 mL

So initial volume required (V1) = M2V2 / M1

= 25% * 118 mL / 75%

= 39.33 mL

So putting this value of x in the equation-

0.75x + 0.05y = 118 mL

0.75 * 39.33 mL + 0.05y = 118 mL

29.5 mL + 0.05y = 118 mL

0.05y = 118 mL - 29.5 mL

= 88.5 mL

y = 88.5 mL / 0.05

= 1,770‬ mL

So-

Answer: [39.33 = 40] ml of 75% stock (#ml) Round to the nearest whole number

Answer: [1,770] ml of 5% stock (#ml) Round to the nearest whole number