Question

you need a 35% alcohol Solution on hand you have a 270 mL of a 30% alcohol mixture you also have 80% alcohol mixture how much of the 80% mixture will you need to add to obtain the desired solution

Answer 1

Explanation:

Let 'x' be the amount of 80% alcohol solution needed.

The amount of alcohol in the 30% alcohol solution is:

$270\cdot\frac{30}{100}=270\cdot0.3=81ml$

The amount of alcohol in the 80% alcohol solution is:

$x\cdot\frac{80}{100}=x\cdot0.8$

The sum of these amounts is the amount of alcohol resulting of the mixture. We need the 35% of the final mixture be alcohol. This is:

$(270+x)\frac{35}{100}=(270+x)0.35$

So we have to solve the following equation for x:

$\begin{gathered} 270\cdot\frac{30}{100}+x\cdot\frac{80}{100}=(270+x)\frac{35}{100} \\ 81+0.8x=(270+x)\cdot0.35 \end{gathered}$

Solving we have:

$\begin{gathered} 81+0.8x=270\cdot0.35+0.35x \\ 81+0.8x-0.35x=94.5 \\ 0.45x=94.5-81 \\ x=\frac{94.5-81}{0.45} \\ x=30 \end{gathered}$

The total amount of the 35% alcohol solution obtained is:

$x+270=30+270=300mL$

Answer:

You will need 30 mL of the 80% alcohol solution.

The total amount of the 35% solution you'll obtain is 300mL