Question

What is the $\dfrac{N}{{10}}$ $HCl$ solution?

Answer 1

Normality is used to express the concentration of any solution represented as $\left( {\rm N} \right)$ . Normality is defined as a ratio of gram equivalent of initial solute taken ${N_{\left( {eq} \right)}}$ to volume of solution considered in litres ${V_{\left( l \right)}}$ .
$\left( {\rm N} \right)$ $= \dfrac{{{N_{\left( {eq} \right)}}}}{{{V_{\left( l \right)}}}}$ .

Complete Step By Step Answer:
Equivalent weight of any acid is calculated by taking the ratio of molecular weight of acid to acidity of the acid.
$Eq = \dfrac{M}{n}$
Where $Eq$ ,is equivalent weight of the compound
$M$ , is molecular weight of compound
$n$ , is basicity of acid
As we know, the atomic mass of hydrogen is $\left( 1 \right)$ and atomic mass of chlorine is $\left( {35.5} \right)$ . Molecular mass of hydrogen chloride will become
$HCl$ $= 1 + 35.5$
$HCl$ $= 36.5$
From the structure it is clear that acid has only one hydrogen atom to release therefore, it is monoacidic in nature with a value of $n$ is $\left( 1 \right)$ .
Now put the values in the above formula to calculate equivalent weight of $HCl$
$Eq = \dfrac{{36.5}}{1}$
From calculating the above equation, the equivalent weight of $HCl$ is $36.5$ .
Gram equivalent mass of compound is calculated by dividing mass of solute by equivalent mass of solute.
${N_{\left( {eq} \right)}} = \dfrac{M}{{Eq}}$
For one mole of $HCl$ value of gram equivalent mass will be:
${N_{\left( {eq} \right)}} = \dfrac{{36.5}}{{36.5}}$
After calculating the above equation, we finally get the gram equivalent mass of one mole or $36.5$ grams of $HCl$ is $\left[ 1 \right]$ .
When we dissolve $\left( {35.5} \right)$ grams of $HCl$ in one litre of solution, normality of solution will become-
$\left( {\rm N} \right)$ $= \dfrac{1}{1}$
$\left( {\rm N} \right)$ $= 1$
When one mole of hydrogen chloride is dissolved in one litre of solution the normality of solution becomes $1{\rm N}$ .
But when we take $3.65g$ of $HCl$ , value of gram equivalent mass will be:
${N_{\left( {eq} \right)}} = \dfrac{{3.65}}{{36.5}}$
After solving we get –
${N_{\left( {eq} \right)}} = 0.1$
So, when we dissolve $3.65g$ of $HCl$ in one litres of solution the normality of solution will become-
$\left( {\rm N} \right)$ $= \dfrac{{0.1}}{1}$
After calculating the above equation, we get
$\left( {\rm N} \right)$ $= 0.1{\rm N}$
$\Rightarrow$ $\dfrac{N}{{10}}$ $HCl$ solution is prepared by adding $3.65g of$ HCl $ in one litres of solution.

Note:
Normality of solution depends upon the temperature of operating condition. Normality of solution is also expressed with the help of molarity of compound with the use of acidity and basicity of compound.