Question

A solution of $8\% $ boric acid is to be diluted by adding $2\% $ boric acid solution to it. The resulting mixture is to be more than $4\% $ but less than $6\% $ boric acid. If there are $640$ liters of the $8\% $ solution, how many liters of $2\% $ solution will have to be added?

Answer 1

Hint: Percent solution is the solution expressed in unit $\% $. It may be percentage by weight, percentage by volume, molar concentration. In this question nothing is given so we will take percentage by volume.
Formula used: $\dfrac{{x \times volume}}{{100}}$ where volume is total volume of solution and $x$ is percentage given.

Complete step by step answer:
Percentage by volume of a solution is defined as volume of a solute dissolved per $100$ units of solution. $8\% $ Solution of boric acid is the solution which is given to us initially. By $8\% $ solution we mean $\dfrac{{8 \times volume}}{{100}}$ , Volume of $8\% $ solution is $640$ liter.
$2\% $ Solution of boric acid is added to $8\% $ solution of boric acid. By $2\% $ solution we mean $\dfrac{{2 \times volume}}{{100}}$ we have to find the volume of this $2\% $ solution.
Concentration of resulting solution must be less than $6\% $ (by $6\% $ solution we mean $\dfrac{{6 \times volume}}{{100}}$) and greater than $4\% $ ( by $4\% $ solution we mean $\dfrac{{4 \times volume}}{{100}}$).
Concentration of the final solution lies in the range $4\% - 6\% $. The final solution is the sum of the initial volume and volume of $2\% $ solution used.
Let volume of $2\% $ solution is x (we have to find this)
In $8\% $ solution volume is $640$ liter (given)
In $4\% $ solution volume is x+$640$ liter (original solution was $640$ liter and x liter solution is added so total volume is x+$640$ liter)
In $6\% $ solution volume is x+$640$ liter
Net solution should be less than $6\% $ so equation becomes
$\dfrac{2}{{100}}x + \dfrac{{8 \times 640}}{{100}} < \dfrac{{6 \times (x + 640)}}{{100}}$
This equation can be written as,
$2x + 5120 < 6x + 3840$
Solving this we get,
$320 < x$
This is equation $1$
Also solution should be more than $4\% $ so equation becomes
$\dfrac{2}{{100}}x + \dfrac{{8 \times 640}}{{100}} > \dfrac{{4 \times (x + 640)}}{{100}}$
This equation can be written as,
$2x + 5120 > 4x + 2560$
Solving this we get,
$1280 > x$
This is equation $2$ , combining both equations $1\& 2$
$320 < x < 1280$
Therefore x must lie in this range.
Additional information: Other related terms are molarity, normality, mass percent etc. Molarity is the measure of number of moles present per liter of solution. It can be calculated as:
Molarity $ = \dfrac{{m \times 1000}}{{M \times V}}$where $m$ is given mass of substance, $M$ is molar mass of substance and $V$ is volume of solution. Units of molarity are $mol{L^{ - 1}}$ .
Normality $\left( N \right)$ is also known as the equivalent concentration of solution. Normality is a measure of gram equivalent weight per liter of solution. Equivalent weight of an element is its gram atomic weight divided by its valence. Gram atomic weight is also known as atomic mass. Normality of a solution can be calculated as:
$Normality = \dfrac{m}{{E \times V}}$ Where$m$ weight of solute in grams, $E$ is equivalent weight of substance and $V$ is volume of solution in liter.
Mass percent is among one of the ways of representing concentration of an element in a compound. Mass percent can be calculated as:
 Mass percent $ = \dfrac{{g \times 100}}{G}$ where $g = $ gram of solute and $G = $ gram of solution.

Note:
Mass percent is mass of solute dissolved per $100$ ml of solution.
Percentage by volume means volume of solute dissolved per $100$ units of solution.