Answer:
0,727 mL
Explanation:
We add x mL of the 95% ethanol solution to 1 mL: C1=0,95 and V1=x mL
As we already had 1 ml of solution, now we'll have (1 + x) mL of the 40% ethanol solution: C2=0,4 and V2=1+X mL
Now, using the C1V1=C2V2 equation:
(0,95)x=0,4(x+1)
0,95x=0,4x + 0,4
0,95x-0,4x=0,4
0,55x=0,4
x=0,4/0,55=0,727 mL
Therefore, we need to add 0,727 mL of a 95% ethanol solution to the enzyme solution to obtain a final concentration of 40% ethanol.