Question

Can anyone solve this? From a 95% ethanol solution, prepare a 125 ml of a 75% ethanol solution.

Answer 1

We'll call the amount of 95% ethanol `x`

Explanation:

What we want:
`125ml,75%harr125*0.75ml` of pure ethanol

What we use:
`x ml,95%harrx*0.95`ml of pure ethanol

`0.95*x=125*0.75->x=(125*.75)/0.95->`

`x~~98.7ml` of the 95% stuff, and add water to 125ml