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put this solution on YOUR website! A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution?
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Answer 11589 by venugopalramana(581) About Me on 2005-12-23 11:03:10 (Show Source):
A chemist needs 10 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How much of each of the two solutions should she mix to obtain her desired solution?
LET X LITRES OF 20% SOLUTION BE REQUIRED...SINCE TOTAL IS 10 LITRES ,WE HAVE TO ADD 10-X LITRES OF 70% SOLUTION..SO TAKING A SALT BALANCE
SALT IN X LITRES OF 20 % SOLUTION......=X*20/100=0.2X
SALT IN 10-X LITRES OF 10 % SOLUTION.. =(10-X)*70/100=7-0.7X
SALT IN 10 LITRES OF FINAL 50 % SOLUTION =10*50/100=5
HENCE
0.2X+7-0.7X=5
0.7X-0.2X=7-5=2
0.5X=2
X=2/0.5=20/5=4 LITRES OF 20 % SWOLUTION AND 10-4=6 LITRES OF 70% ARE NEEDED TO BE MIXED TO GET 10 LITRES OF 50% SOLUTION
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SEE THIS EXAMPLE WHICH IS SIMILAR TO YOUR PROBLEM AND TRY.IF YOU STILL HAVE DIFFICULTY PLEASE COME BACK
In a chemistry class, 7 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of a 10% solution are needed?
A. 4.5L LET X LITRES OF 10% SOLUTION BE NEEDED.IT HAS X*10/100 LITRES OF SILVER IODIDE.
B. 2.5L 7 LITRES OF 4% SILVER IODIDE SOLUTION HAS 7*4/100=28/100 LITRES OF SILVER IODIDE
C. 7.0L TOTAL SOLUTION =X+7 LITRES…….IT HAS TOTAL 10X/100+28/100=(10X+28)/100 LITRES OF SILVER IODIDE
D. 3.5L FINAL CONCENTRATION =6 %…SO 6/100=(10X+28)/(X+7)*100..OR…6=(10X+28)/(X+7)...CROSS MULTIPLYING..
6(X+7)=(10X+28)…OR…6X+42=10X+28..OR….10X-6X=42-28=14…OR….4X=28..OR..X=7 LITRES.C IS THE ANSWER
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