Since my old brain doesn't math like it did even a few years back, can someone tell me what the equivalent dilution is for 70% iso if 99% iso is 10:1? Final concentration of that would be 9.1% (
thanks Google!). And 70% is already diluted part of the way there.
No one? Okay, I found this:
How do I Make 70 Percent Isopropyl Alcohol?
which says that to find dilution rates we need this formula:
C1_V1=C2_V2, where C1 and V1 is the starting concentration and volume of the solution and C2 and V2 is the final concentration and volume of the dilution.
I don't know what the underscores mean in this equation but I think it means multiplication from the example they give:
Solve the equation C1_V1=C2_V2 for the unknown variable of V2. Substitute known variables: 100_V1=70_500, V1=35000/100, V1= 350.
So if we want a 9.1% concentration of 946 ml (32 ounces or a standard spray bottle), we know the following:
C1 = 70
V1 = ?
C2 = 9.1
V2 = 946
Which makes our equation:
70(V1) = (9.1)(946)
70(V1) = 8609
V1 = 8609/70
V1 = 122.98 ml
That would mean we need 123 ml of 70% iso mixed with 823 ml of water. Right?
I also want to check with
@newty...did you do 1 part iso and 10 parts water or 1 part iso in
9 parts water? That would change the calculations above.
I love when you all indulge me this detailed shit!