Question

A solution of 8% Boric acid is to be diluted by adding a 2% Boric acid solution to it. The resulting solution is to be more than 4% but less than6% boric acid. If there are $\text{640 litres}$ of the 8% solution, how many litres of the 2% solution needs to be added.

Answer 1

This problem can be solved from the knowledge of percentage composition of the solution. The percentage composition of the solution can be expressed as the weight of the solute in grams present in $\text{100 mL}$ of the solution.

Complete step by step answer:
As per the given question, $\text{640 litres}$ of the 8% Boric acid solution is present. Let the volume of the 2% solution of boric acid be equal to $\text{x Litres}$.
Therefore the total volume of the solution = $\left( \text{640 + x} \right)\text{ litres}$
Boric acid in the 8 % Boric acid solution = $640\times \dfrac{8}{100}=51.2$litres
Boric acid in the 2 % Boric acid solution = $\text{x}\times \dfrac{2}{100}=0.02\text{x}$litres
Net volume of the acid = $51.2+0.02\text{x}$ litres
The resulting mixture has more than 4% but less than6% boric acid therefore,
$\dfrac{4}{100}\left( 640+\text{x} \right)$<$51.2+0.02\text{x}$ < $\dfrac{6}{100}\left( 640+\text{x} \right)$
Or, $\left( \text{0}\text{.04 }\times\text{ 640} \right)\text{+ 0}\text{.04x}$ < $51.2+0.02\text{x}$ < $\left( \text{0}\text{.06 }\times\text{ 640} \right)\text{+ 0}\text{.06x}$
Or, $\left( 25.6-51.2 \right)\text{+ 0}\text{.04x-0}\text{.02x}$ < 0 < $\left( 38.4-51.2 \right)\text{+ 0}\text{.06x-0}\text{.02x}$
Or, $0.02\text{x}-25.6$ < 0 < $0.04\text{x}-12.8$
Or, $0.02\text{x}-25.6$ < 0, considering the first case.
X < 1280.
And, 0 <$0.04\text{x}-12.8$, considering the second condition,
320 < x
Therefore, the amount of the 2% Boric acid solution that needs to be added must be greater than 320 litres but lesser than 1280 litres.

Note:
There are different other units in which the concentration of the solution can be expressed. The mole fraction expresses the moles of the solute as the fraction of the number of the moles of the solute to the sum of the moles of the solute and the moles of the solvent.