SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 90l of 56% solution. What amounts should be used?

Question 729919: Pure acid is to be added to a 10% acid solution to obtain 90l of 56% solution. What amounts should be used?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Pure acid is to be added to a 10% acid solution to obtain 90l of 56% solution. What amounts should be used?
-------
Equation:
acid + acid = acid
x + 0.10(90L-x) = 0.56*90
------
100x + 10*90L - 10x = 56*90
----
90x = 46*90
x = 46 L (amt. of pure acid to add)
==================
Cheers,
Stan H.
===================
RELATED QUESTIONS
pure acid is to be added to a 10%acid solution to obtain 90l of 89% solution . what... (answered by vleith)

pure acid is to be added to a 10% acid solution to obtain 90L of 44% solution what... (answered by josgarithmetic)

pure acid is to be added to a 10%acid solution to obtain 90l of 22% solution . what... (answered by josmiceli)

pure acid is to be added to a 10% acid solution to obtain 90L of 42% solution what... (answered by addingup)

pure acid is to be added to a 10% acid solution to obtain 90l of 75% solution? what... (answered by Fombitz,ewatrrr,josmiceli)

pure acid is to be added to a 10% acid solution to obtain 90L of 32% soulution. what... (answered by ikleyn,greenestamps)

pure acid is to be added to a 10% acid solution to obtain 90L of 19% solution. what... (answered by jorel1380)

Pure acid is to be added to a 10% solution to obtqin 90L of 14% solution. What amounts of (answered by josgarithmetic)

pure acid is to be added to 10% acid solution to obtain 90L of 53% solution. what amount... (answered by stanbon)