# Question 4e4a9

Jul 31, 2017

$106.67$ ml of water to be added.

#### Explanation:

Let $x$ ml of water is added to dilute the acd solution.

Acid content in 35% acid solution is $80 \cdot 0.35 = 28$ ml

After adding , total quantity of 15% acid solution is $80 + x$ ml

Acid content in 15%# acid solution is $\left(80 + x\right) \cdot 0.15$ ml

Quantity of acid remains same before and after dilution.

$\therefore \left(80 + x\right) \cdot 0.15 = 28 \mathmr{and} 80 + x = \frac{28}{0.15} = 186 \frac{2}{3}$

$x = 186 \frac{2}{3} - 80 = 106 \frac{2}{3} \approx 106.67$ ml

$106.67$ ml of water to be added [Ans]