Asked by rimple
how can i solve this?prepare 150ml of a 7.5%dextrose sol.you have D5%W and D10%W. how much of each sol. is needed?
I assume the 7.5 dextrose is going to be on a w/w basis also.
This is really simple, no algebra needed. Since 7.5 is halfway between 5 and 10, use equal volumes of the two stock solutions.
Now lets make it harder: Prepare a 9 percent solution.
Let F be the volume of five percent solution
T be the volume of ten percent.
F+T=150
.05F + .10T=.9(150)
Then solve this system of two equations for F, and T.
I assume the 7.5 dextrose is going to be on a w/w basis also.
This is really simple, no algebra needed. Since 7.5 is halfway between 5 and 10, use equal volumes of the two stock solutions.
Now lets make it harder: Prepare a 9 percent solution.
Let F be the volume of five percent solution
T be the volume of ten percent.
F+T=150
.05F + .10T=.9(150)
Then solve this system of two equations for F, and T.
Answers
Answered by
mirela
alligation formula.
higher conc 10%
desire conc 7.5%
lower conc 5%
10-7.5 equal 2.5
7.5- 5 equal 2.5
total (2.5+2.5 )=5
2.5:5*150ml=75 ml of D10%W
2.5:5*150ml= 75ml of D5%W
higher conc 10%
desire conc 7.5%
lower conc 5%
10-7.5 equal 2.5
7.5- 5 equal 2.5
total (2.5+2.5 )=5
2.5:5*150ml=75 ml of D10%W
2.5:5*150ml= 75ml of D5%W
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