Question

A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68%

Answer 1

Let the first solution contains 50% acid Let second solution contains 80% acid Let the first solution be x ml and second solution y ml added x + y = 100 ..... (1)

The acid content in the mixture of y 50% of x + 80% of y

$\frac{50x}{100}+\frac{80y}{100}=68\Rightarrow \frac{x}{2}+\frac{4y}{5}=68$ $\Rightarrow \frac{5x+8y}{10}=68\Rightarrow 5x+8y=\mathrm{680...}\left(2\right)$ $\left(1\right)\times 5\Rightarrow 5x+5y=\mathrm{500....}\left(3\right)$ $\left(2\right)\Rightarrow 5x+8y=\mathrm{680....}\left(4\right)$ $\left(3\right)-\left(4\right)gives$ $-3y=180\Rightarrow y=\frac{-180}{-3}=60$ $Substituteyvaluein\left(1\right)$ $\therefore$ The quantity of first solution = 40 ml $\therefore$ The quantity of second solution = 60 ml