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This is a standard algebra word problem so I will try to give lots of details since it can be generalized to many situations. Let us call the NH3OH just N for short. You start with a given volume say V of that and it is 25 %, now you want to make it 10%. To make this simpler, I can going to assume a volume but we can go back and make that volume arbitrary after you see how to do the problem So we have 100 cc of 25% N. That mean to find out how much N we have ( how much of the actual chemical in the solution) we multiply .25x100=25 grams. Now we want 10 percent solution. Call the amount we need to add by the letter A So the final total amount of liquid is 100+A since we start with 100 cc and add A. Now A has NO N in it at all. So we have two equation 100+A=Total volume, call it TV and the amount of N in the total volume is 25 mg since it all come from the original No we want 10 percent of TV to equals 25 since that is all the available N we have. That is to way, we are adding water and a fixed amount of N and we need the TV x .1=25 So TV =250 ml. Since we started with 100, we must have added 150 cc of water. Let us check we have 250 ml of fluid with 25 mg of N and 25/250=.1 that is 10% . Now the only part that used the 100 cc was to figure out how many grams of N we had So we could call the initial volume IV and we have .25IV of N, the rest of the computation will now be the same. We could have TV = Initial volume plus what we added and if we are doing a dilution with water, the total amount of chemical in TV will always be the initial volume x the percent solution. We can further generalize to any percent solutions, do you see how?

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Q: How do you dilute 25 percent NH3OH TO 10 percent?
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