Question

$\text{NaClO}$solution reacts with ${{H}_{2}}S{{O}_{3}}$ as,
$NaClO+{{H}_{2}}S{{O}_{3}}\to NaCl+{{H}_{2}}S{{O}_{4}}$
A solution of$NaClO$used in the above reaction contained $15g$of $NaClO$per liter. The normally of solution would be [AMU$1999$]
A.$0.8$
B.$0.6$
C.$0.2$
D.$0.33$

Answer 1

Hint: Normality of any solution is defined by the number of grams of equivalents present per liter of solution. In this problem we just have the mass of $NaClO$per liter, hence first we have to calculate the grams equivalent of $NaClO$and then put the value in the respective formula of normality that is provided below.

Formula Used:Equivalent weight$=\dfrac{M}{Basicity}$
Where $M$denotes the molar mass of the solute and the basicity of acid means the total number of replaceable ${{H}^{+}}$ions present in an acidic solution.
Normality,$N=$$\dfrac{w\times 1000}{eq.\,weight\,\times V\,(ml)}$
Here,$w=$ weight of the component.
$V=$Volume of solution in $ml$

Complete answer:In this problem, the solute is $NaClO$ dissolved in the solvent ${{H}_{2}}S{{O}_{3}}$and forms sodium chloride ($NaCl$) and sulphuric acid (${{H}_{2}}S{{O}_{4}}$). The overall reaction can be written as:
$NaClO+{{H}_{2}}S{{O}_{3}}\to NaCl+{{H}_{2}}S{{O}_{4}}$
The molar mass of $NaClO$,${{M}_{NaClO}}=$(Atomic weight of $Na$)$+$(Atomic weight of $Cl$)$+$(Atomic weight of oxygen)
$\therefore {{M}_{NaClO}}=23+35.5+16=74.5\,gm/mol$
In this solution, Sulphuric acid${{H}_{2}}S{{O}_{4}}$has two replaceable ${{H}^{+}}$ions per mole, so the basicity of acid$=2$
${{H}_{2}}S{{O}_{4}}\rightleftharpoons 2{{H}^{+}}+SO_{4}^{2-}$
Now we can calculate the equivalent weight of $NaCl{{O}_{4}}$by using the values of molar mass and basicity of acid as the formula of equivalent weight$=\dfrac{{{M}_{NaCl{{O}_{4}}}}}{Basicity}=\dfrac{74.5}{2}=37.25\,gm$.
The normality of the solution,$N=\dfrac{{{W}_{NaCl{{O}_{4}}}}\times 1000}{eq.\,weight\,\times V\,(ml)}$ ………..(i)
Given the weight of $NaCl{{O}_{4}},{{W}_{NaCl{{O}_{4}}}}=15\,gm$
And Volume of the solution $=1\,Litre=1000\,ml$
Putting these values in equation (i),
$N=\dfrac{15\,\times 1000}{37.25\,\,\times 1000}=0.403$
So, from the option, the calculated value is very nearly equal to $0.33$, so it would be the correct option.

Thus, option (D) is correct.

Note: In chemistry, especially in physical and analytical chemistry, the mole concept is a fundamental concept. It has huge applications such as molarity, normality, and mole fraction, etc are used in redox reactions, acid-base chemical reactions, precipitation reactions, etc