Asked by marisa
how do you solve a dilution problem, this is chemistry
Answers
Answered by
bobpursley
The standard way is to figure the moles you need in the new volume, subtract the moles you have (volume*Molarity), then add the grams of solute to the solution, add solvent to the specified volume, stir, label, and store appropriately.
Here is my favorite way.
Example1. You have 20 ml of 5MHCl, and want 500ml of .20M HCl.
ok, you are diluting it 5/.2or 25 times. you have to add one part original solution, and 24 parts water.
25 parts into 500ml is each part is 20ml. So add 20 ml of the original solution (1 Part), and 24 parts (480ml) water. STir, label, store appropriately.
Example2.
You have 4M CuSO4(aq). YOu want 315ml of .56M solution.
Ok, you are diluting it 4/.56=7.143 times. You must add on part original, and 6.143 water.
315ml into 7.143 parts is 44.1ml per part. So add 44.1ml original solution, add six 6.143 parts water (6.143*44.1=270.9 ml)
check: 270.9+44.1=315ml
Practice makes perfect.
Here is my favorite way.
Example1. You have 20 ml of 5MHCl, and want 500ml of .20M HCl.
ok, you are diluting it 5/.2or 25 times. you have to add one part original solution, and 24 parts water.
25 parts into 500ml is each part is 20ml. So add 20 ml of the original solution (1 Part), and 24 parts (480ml) water. STir, label, store appropriately.
Example2.
You have 4M CuSO4(aq). YOu want 315ml of .56M solution.
Ok, you are diluting it 4/.56=7.143 times. You must add on part original, and 6.143 water.
315ml into 7.143 parts is 44.1ml per part. So add 44.1ml original solution, add six 6.143 parts water (6.143*44.1=270.9 ml)
check: 270.9+44.1=315ml
Practice makes perfect.
Answered by
chiquita
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