SOLUTION: How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid
Algebra.Com
Question 489271: How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How much pure acid must be added to 60 ml of a 35% solution to produce a mixture that is 80% acid?
ml of pure acid/ml of mixture = concentration of acid
---------------------------
Equation:
acid + acid = acid
0.35*60 + x = 0.80(60+x)
---
Multiply thru by 100 to get:
35*60 + 100x = 80*60 + 80x
-----------------
20x = 45*60
x = 15 ml (amt. of pure acid needed)
============================
Cheers,
Stan H.
RELATED QUESTIONS
What quantity of pure acid must be added to 300 mL of a 40% acid solution to produce a... (answered by ewatrrr)
What quantity of pure acid must be added to 300 mL of a 50% acid solution to produce a... (answered by Fombitz)
What quantity of pure acid must be added to 600 mL of a 30% acid solution to produce a... (answered by josgarithmetic)
What quantity of pure acid must be added to 400 mL of a 40% acid solution to produce a... (answered by richwmiller)
What quantity of pure acid must be added to 400 mL of a 50% acid solution to produce a... (answered by josgarithmetic)
How much pure acid must be added to 50mL of 25% acid solution to produce a mixture that... (answered by mananth)
How much pure acid must be added to 50ml of 25% acid solution to produce a mixture that... (answered by mananth)
how much pure acid must be added to 500 ml of an 8% acid solution to make a 20% acid... (answered by jorel1380)
What quantity of pure acid must be added to 400 mL of a 25% acid solution to produce a... (answered by nerdybill)