Question

40 gm of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ (mol. Wt. = 375 gm/mol) sample containing some inert impurities in acidic medium is completely reacted with 125 ml of 33.6 V of ${{H}_{2}}{{O}_{2}}$. What is the percentage purity of the sample?
(A) 28.12%
(B) 70.31%
(C) 85%
(D) None of the above

Answer 1

Hint: To solve the given problem, we just need to have a thorough idea of normality. A new concept of volume strength is introduced here, so focus on it as it will help you solve the illustration. Also, for a while ignore option (D) and focus on other options as only one answer would be correct.

Complete step by step solution:
Let us solve the given illustration directly, we just need to know the basics of normality as explained below:
Normality-Normality is the number of gram equivalents or mole equivalents of solute present in one litre of a solution.
Mole equivalent is the number of moles who are actually the reactive units in the reaction. Similar goes with the gram equivalent.
It is expressed as,
\[N=M\times n\]
where, n is the ratio of molar mass to the equivalent mass of the component.
Normality can simply be expressed as,
\[N=\dfrac{{{n}_{g}}}{V}\]
where,
V = volume of solution in litres
${{n}_{g}}$ = number of gram equivalents i.e. it is the ratio of given mass to the equivalent mass of a compound.
The given problem requires volume strength concept, so let us make a look towards it;
Volume Strength-It is the term referring to the volume of oxygen gas liberated from the total volume of ${{H}_{2}}{{O}_{2}}$ solution.
In terms of normality, volume strength for ${{H}_{2}}{{O}_{2}}$ is generally expressed as,
\[Vol.strength=Normality\times 5.6\]
Now, solving following illustration we will get the required answer as;
Given that,
Given mass of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ = 40 gm
Molar mass of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ = 375 gm/mol
Volume strength of ${{H}_{2}}{{O}_{2}}$ = 33.6 V
Volume of ${{H}_{2}}{{O}_{2}}$ solution = 125 ml
So, we can say that,
Normality of ${{H}_{2}}{{O}_{2}}$ = $\dfrac{33.6}{5.6}=6N$
Thus,
Equivalent moles of ${{H}_{2}}{{O}_{2}}$ is given as,
\[Eq.moles=6N\times \dfrac{125ml}{1000ml}=0.75\]
 Now, for $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$,
\[\begin{align}
& Ba{{\left( Mn{{O}_{4}} \right)}_{2}}\to B{{a}^{2+}}+2Mn{{O}_{4}}^{-} \\
& \\
\end{align}\]
Ba is an electropositive element hence, does not get reduced. Only $Mn{{O}_{4}}^{-}$ (Mn having oxidation state as +7) gets reduced in the acidic medium. Thus,
\[2Mn{{O}_{4}}^{-}+16{{H}^{+}}+10{{e}^{-}}\to 2M{{n}^{2+}}+8{{H}_{2}}O\]
Thus, 10 electrons are required for the reduction.
Equivalent mass of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ = $\dfrac{375gm/mol}{10}=37.5gmmo{{l}^{-1}}e{{q}^{-1}}$
Equivalent moles of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ = $\dfrac{40gm}{37.5gmmo{{l}^{-1}}e{{q}^{-1}}}\times x$
where, x is the percent purity of the sample having $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ and some inert impurities in it.
As, $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ is completely reacted with ${{H}_{2}}{{O}_{2}}$ in acidic medium thus;
Equivalent moles of $Ba{{\left( Mn{{O}_{4}} \right)}_{2}}$ = Equivalent moles of ${{H}_{2}}{{O}_{2}}$
$\dfrac{40gm\times x}{37.5gmmo{{l}^{-1}}e{{q}^{-1}}}=0.75$
Thus,
$x=\dfrac{0.75\times 37.5}{40}=0.703$
We can say that,
Purity of sample = $x\times 100=0.703\times 100=70.3%$

Therefore, option (B) is correct.

Note: Do note to use the units properly and do not get confused in concepts of molarity and normality. Though they are interrelated, their multiplying factors differ. Here, an outdated concept is introduced known as volume strength. Do note to have a proper knowledge of it before we use it.