Question

A $5.0mL$ solution of ${{H}_{2}}{{O}_{2}}$ liberates $0.508g$ of Iodine from an acidified $KI$ solution. The strength of ${{H}_{2}}{{O}_{2}}$ solution, in terms of volume strength at STP will be
A) 2.24 V
B) 4.48V
C) 8.96 V
D) None of these

Answer 1

Hint: We should have the basic idea of molarity, normality and volume strength
Volume strength is the measure of expressing concentration of ${{H}_{2}}{{O}_{2}}$in terms of volume of oxygen formed during the decomposition reaction of${{H}_{2}}{{O}_{2}}$.

Complete step by step solution:
-Molarity is defined as the number of moles of solute present in 1 litre of the solution.
-Normality is defined as the gram equivalent of the solute particles in one litre of solution.
The reaction happening here is,
\[{{H}_{2}}{{O}_{2}}+2{{I}^{-}}+2{{H}^{+}}\to 2{{H}_{2}}O+{{I}_{2}}\]
Molecular mass of iodine molecule =254g/mol
And let’s calculate the moles of iodine.
\[Moles\text{ }of\text{ }iodine\text{ }=~\dfrac{Given\,weight}{Molecular\,mass}\]
\[Moles\text{ }of\text{ }iodine\text{ }=~\dfrac{0.508}{254}moles\]
And we know the equation, ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
In this case, ${{M}_{1}}$ = Molarity of ${{H}_{2}}{{O}_{2}}$
${{V}_{1}}$= Volume of ${{H}_{2}}{{O}_{2}}$
${{M}_{2}}$=Molarity of iodine
${{V}_{2}}$= volume of iodine
Here, ${{V}_{1}}$=5 ml and ${{V}_{2}}$=1000 ml
And now let’s calculate the molarity of${{H}_{2}}{{O}_{2}}$,${{M}_{1}}$
Molarity $\times $ volume = number of moles $\times $1000 (As volume is in ml)
Substitute the values in the equation,
${{M}_{1}}\times 5=\dfrac{0.508}{254}\times 1000$
${{M}_{1}}=0.4M$
Thus we got the molarity of${{H}_{2}}{{O}_{2}}$, which is 0.4M
Now let’s find the normality of the solution.
The equation of normality we use here is,
$Normality=n-factor\times Molarity$
Here $n-factor =2$
n-factor can be explained for acids it is the number of ${{H}^{+}}$ ions replaced by the acids in a reaction and in for base it is number of $O{{H}^{-}}$ ions replaced in a reaction .
Hence the normality, which is represented as N
$N=2\times 0.4$
$N=0.8$
Now volume strength of ${{H}_{2}}{{O}_{2}}$ at STP is equal to normality multiplied by a value 5.6.ie
Volume strength of ${{H}_{2}}{{O}_{2}}$=$N\times 5.6$=$0.8\times 5.6$
Volume strength of ${{H}_{2}}{{O}_{2}}$= 4.48 Volume

So the correct option is option (B).

Note: The volume strength of $1N$ solution of ${{H}_{2}}{{O}_{2}}$ at STP is $5.6L$
There is a chance of getting confused between molality and molarity.
Molality is the number of moles of solute per kilogram of the solvent and molarity is the number of moles of solute present in 1 litre of the solution.