110ml of the 40% alcohol mixture and pure water will need to obtain the desired solution
Given that 0.30 * 440 = 132 mL of pure alcohol needed
132 mL = 0.40 * x mL of 0.4
x = 330 mL of 70%
440 - 330 = 110 mL of H2O
Water content differs
If you have 440 mL of a 30% alcohol mixture, then (440mL)*(0.30) = 132mL of that solution is alcohol.
Similarly, if you have x mL of a 40% alcohol mixture, then (x mL)(0.40) = 0.40x mL of that solution is alcohol.
When you combine these two liquids, you will have (440 + x) mL of liquid and you want it to be 40% alcohol.
Hence 110ml of the 40% alcohol mixture and pure water will need to obtain the desired solution
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