Q. How much of 80% alcohol should b mixed with 30ml of 95% alcohol in order to prepare 90% alcohol?
To determine the amount of 80% alcohol needed to mix with 30ml of 95% alcohol in order to prepare a solution of 90% alcohol, we can use a mixture equation.
Let's assume "x" represents the amount of 80% alcohol needed to be mixed.
Now, let's calculate the quantity of alcohol in each solution.
For the 30ml of 95% alcohol, the amount of alcohol in it is:
Amount of alcohol = 30ml × (95/100) = 2850/100 = 28.5ml
For the "x" ml of 80% alcohol, the amount of alcohol in it is:
Amount of alcohol = x × (80/100) = 0.8x
Now, let's build the equation based on the total amount of alcohol in the final 90% solution:
Total amount of alcohol = amount of alcohol from 95% alcohol + amount of alcohol from 80% alcohol
30ml + 0.8x = 0.9(x + 30)
Simplify the equation:
30 + 0.8x = 0.9x + 27
Rearrange the equation:
0.8x - 0.9x = 27 - 30
-0.1x = -3
Now, solve for "x":
x = (-3) / (-0.1) = 30
So, you would need to mix 30ml of 80% alcohol with 30ml of 95% alcohol to prepare a 90% alcohol solution.
.90(V+30ml)=30ml*.95+.80*V
solve for Volume V