Question

You need 440 mL of a 30% alcohol solution. On hand, you have a 40% alcohol mixture. How much of the 40% alcohol mixture and pure water will you need to obtain the desired solution

Answer 1

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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