Question

For $x$ gram of an acid furnishes 0.5 mole ${H_3}{O^ + }$ in aqueous solution. What will be the normality of acid if $y$ gram dissolves in a 500 ml solution?
$(1)\dfrac{{y \times 1000}}{{x \times 500}}$
$(2)\dfrac{y}{x}$
$(3)\dfrac{y}{{2x \times 500}}$
$(4)\dfrac{{y \times x}}{{5000}}$

Answer 1

Normality is an important concentration unit in chemistry. Normality is defined as the number of gram equivalents of solute present per litre of solution. It is denoted by the letter ‘N’. The unit of normality is eq/L or meq/L or N. The volume of solution must be in litres in the formula. For example the Normality of Hydrochloric acid solution is expressed as 0.1 N.

Complete answer:
Normality=Number of gram equivalents of solute/Volume of solution in litres
Where number of gram equivalents=Weight of solute/Equivalent mass of solute
The formula for Equivalent mass =Molecular mass/n
Where n is the valency factor which is equal to the amount of ${H^ + }$ and $O{H^ - }$ions produced by the acids and bases respectively. In case of salt it is the total positive or negative charge.
Number of moles is given by the formula given mass/Molecular mass
Given-
In the question we have been given mass of an acid $= x$ $gram$; Volume of solution $= 500ml$
Number of moles of ${H_3}{O^ + }$ $= 0.5mole$; Amount of solute dissolved $= y$ $gram$
Since ${H_3}{O^ + }$ is Monovalent therefore Molecular mass=Equivalent mass
$0.5$ Mole ${H_3}{O^ + }$ $= x$ $grams$
So $0.5mole = \dfrac{x}{{Eq.Mass}}$ since here molecular and equivalent mass is same)
So now $Eq.Mass = \dfrac{x}{{0.5}}$
$Normality = \dfrac{{y \times 1000}}{{\dfrac{x}{{0.5}} \times 500}} = \dfrac{y}{x}$ (We have actually divided 500 ml by 1000 to convert it into litres so it will come in the numerator)
Thus option (2) is correct.

Note:
Normality is an important concept to study Redox reactions, Acid-Base reactions, titrations and precipitation reactions. It has huge applications in the medical field and chemistry laboratory. However the value of Normality changes with temperature since in its formula we have the concentration terms. And also the calculations require a proper well defined value of the Equivalence factor.