A factory has a batch of hand sanitizer with 50% alcohol content. How much pure (100% concentration) alcohol should they include in a 600mL bottle to make a 68% mixture? mL

Question

A factory has a batch of hand sanitizer with  50% alcohol content. How much pure  (100% concentration) alcohol should they include in a  600mL bottle to make a  68% mixture?  mL

Bytelearn · Accepted Answer

Denote amount of alcohol: Let's denote the amount of pure alcohol to be added as x mL. The initial volume of the hand sanitizer is 600 mL with 50% alcohol content. We want to find the value of x that will result in a final mixture with 68% alcohol content.Calculate initial alcohol amount: The initial amount of alcohol in the 600 mL hand sanitizer is 50% of 600 mL. We calculate this as follows: (50/100) * 600 = 0.50 * 600 = 300 mL of alcohol.Calculate total alcohol amount: After adding x mL of pure alcohol, the total volume of the mixture will be 600 mL + x mL. The total amount of alcohol in the new mixture will be the initial 300 mL plus the x mL of pure alcohol added.Set up equation: We want this new mixture to have a 68% alcohol concentration. Therefore, the total amount of alcohol (300 mL + x mL) should be 68% of the total volume (600 mL + x mL). We can set up the following equation to represent this relationship: (300 + x) = 0.68 * (600 + x).Distribute and simplify: Now we solve for x. First, distribute the 0.68 on the right side of the equation: 300 + x = 0.68 * 600 + 0.68 * x.Isolate x: Simplify the equation by performing the multiplication: 300 + x = 408 + 0.68x.Subtract to isolate x: To isolate x, we subtract 0.68x from both sides of the equation: 300 + x - 0.68x = 408 + 0.68x - 0.68x.Simplify equation: This simplifies to: 300 + 0.32x = 408.Subtract to isolate x: Subtract 300 from both sides to isolate the term with x: 0.32x = 408 - 300.Calculate difference: Calculate the difference: 0.32x = 108.Divide to solve for x: Divide both sides by 0.32 to solve for x: x = 108 / 0.32.Perform division: Perform the division to find the value of x: x = 337.5 mL.

A factory has a batch of hand sanitizer with $50\%$ alcohol content. How much pure $(100\%$ concentration) alcohol should they include in a $600\text{mL}$ bottle to make a $68\%$ mixture? $\newline$ $\text{mL}$

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A factory has a batch of hand sanitizer with $50\%$ alcohol content. How much pure $(100\%$ concentration) alcohol should they include in a $600\text{mL}$ bottle to make a $68\%$ mixture? $\newline$ $\text{mL}$