SOLUTION: A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid. How many cubic centimeters of a 15% solution of HCl acid should be mixed with the 60 cm^3 of 40% ac
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid. How many cubic centimeters of a 15% solution of HCl acid should be mixed with the 60 cm^3 of 40% ac
Log On
Question 1207794: A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid. 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? How much of the 25% solution is there?
You can put this solution on YOUR website! .
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?
~~~~~~~~~~~~~~~~~~~~~~~~
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.
------------------
It is a standard and typical mixture word problem for two ingredients.
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.