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A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid.
(a) How many cubic centimeters of a 15% solution of HCl acid should be mixed with
the 60 cm^3 of 40% acid to obtain a solution of 25% HCl?
(b) How much of the 25% solution is there?
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Your equation is incorrect.
Let x be the volume of the 15% solution to add, in cubic centimeters.
The volume of the pure HCl in 60 cm^3 of the 40% HCl acid is 0.4*60 = 24 cm^3.
The volume of the final solution is (60+x) cm^3,
and the volume of the pure HCl acid in (60+x) cm^3 of the final 25% HCl acid is 0.25*(60+x) cm^3.
We make the setup equation by equating the volumes of pure HCl acid in
ingredients and in final mixture
0.4*60 + 0.15*x = 0.25*(60+x).
It is a proper setup equation.
You can take it and continue your solution from this point to find x.
At this point, I completed my explanation on how to make a proper setup equation for this problem.
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It is a standard and typical mixture word problem for two ingredients.
There is entire bunch of introductory lessons covering various types of mixture problems for two ingredients
- Mixture problems
- More Mixture problems
- Solving typical word problems on mixtures for solutions
- Word problems on mixtures for antifreeze solutions
- Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts
- Word problems on mixtures for dry substances like candies, dried fruits
- Word problems on mixtures for dry substances like soil and sand
- Word problems on mixtures for alloys
- Typical word problems on mixtures from the archive
in this site.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.