Question

Derive the formula to calculate the normality of an acid of density d gm/ml containing x % by weight. The eq. wt of the acid is E.

Answer 1

1. Here acid is present x % by weight that means 'x' gm acid is present in 100 gm solution.
2. Using the given density (d gm/ml) of solution and amount of acid (x% by weight) we can easily find out the volume of solution.
3. Normality is a unit of concentration which means that in a solution what gm equivalent amount of solute is present in one-liter volume of that solution.
4. To convert the unit of volume from ‘ml’ to ‘liter’ we use the data, 1 litre = 1000 ml.

Complete step by step answer:
We have to know that the unit of percentage strength in terms of $\dfrac{{Weight}}{{Weight}}$, which expressed as $\dfrac{{Weight\,of\,solute(g)}}{{Weight\,of\,solution(g)}} \times 100$
So x % by weight of acid means x gm acid (solute) is present in 100 gm of solution.
Therefore the solution is taken 100 gm.
We know that density is the mass of unit volume of a substance
$Density = \dfrac{{Mass}}{{Volume}}$
From the given data of density (d gm/ml) of solution, we can get the volume of solution which is $= \dfrac{{Mass}}{{Density}}$
$= \dfrac{{100gm}}{{d\,gm/ml}}$
$= \dfrac{{100}}{{d\,}}ml$
So the volume of solution is $\dfrac{{100}}{{d\,}}ml$ the equivalent weight of acid is E which is given we know that normality, denoted by N of any solution is $\dfrac{{Number\,of\,gm\,equivalent\,of\,solute}}{{Volume\,of\,solution\,(in\,litre)}}$
$= \dfrac{{\dfrac{{Given\,mass\,of\,acid}}{{Equivalent\,weight\,of\,acid}}}}{{Volume\,of\,solution\,(in\,litre)}}$
Putting the values,
$= \dfrac{{\dfrac{x}{E}}}{{\dfrac{{100}}{d} \times 1000}}$
$= \dfrac{{x \times d \times 1000}}{{E \times 100}}$
$= \dfrac{{10xd}}{E}$

Therefore the normality of the solution expressed by $E$ equivalent of acid, $x$ weight of volume d density of solution is $\dfrac{{10xd}}{E}$.

Note: Students should be careful about that if we take weight of solute in milligram units then we take volume of solution in milliliter units.
There have another formula to express percentage strength in terms of $\dfrac{{Weight}}{{Weight}}$ which is $\dfrac{{Weight\,of\,solute(g)}}{{Volume\,of\,solution\,(ml)}} \times 100$.