Question

10 mL of ${{H}_{2}}{{O}_{2}}$solution (volume strength = x) required 10mL of $\dfrac{N}{0.56}\,MnO_{4}^{-}$ solution in acidic medium. Hence, x is:
(A) 0.56
(B) 5.6
(C) 0.1
(D) 10

Answer 1

Hint: To answer this question we should know that in acidic medium 1 molar equivalent of ${{H}_{2}}{{O}_{2}}$ will combine with 1 molar equivalent of $MnO_{4}^{-}$. Using this relation we will calculate the normality of given ${{H}_{2}}{{O}_{2}}$.
\[{{N}_{{{H}_{2}}{{O}_{2}}}}\,{{V}_{{{H}_{2}}{{O}_{2}}}}\,=\,{{N}_{MnO_{4}^{-}}}\,{{V}_{MnO_{4}^{-}}}\]

Complete step by step answer:
Let’s look at the solution of the given question:
It is given in the question that
Volume strength of ${{H}_{2}}{{O}_{2}}$= x
Volume of ${{H}_{2}}{{O}_{2}}$= 10mL
Normality of $MnO_{4}^{-}$= $\dfrac{N}{0.56}\,MnO_{4}^{-}$
Volume of $MnO_{4}^{-}$= 10mL
Now, we will use the normality equation to calculate the normality of ${{H}_{2}}{{O}_{2}}$
\[{{N}_{{{H}_{2}}{{O}_{2}}}}\,{{V}_{{{H}_{2}}{{O}_{2}}}}\,=\,{{N}_{MnO_{4}^{-}}}\,{{V}_{MnO_{4}^{-}}}\]
On putting the values given in the question
\[{{N}_{{{H}_{2}}{{O}_{2}}}}\,=\,\dfrac{10\times 1}{0.56\times 10}\]
\[{{N}_{{{H}_{2}}{{O}_{2}}}}\,=\,\dfrac{1}{0.56}\]
Now, we know that,
5.6 volume strength of ${{H}_{2}}{{O}_{2}}$= 1N of ${{H}_{2}}{{O}_{2}}$
Therefore, 1 volume strength of ${{H}_{2}}{{O}_{2}}$= $\dfrac{1}{5.6}$ N ${{H}_{2}}{{O}_{2}}$
So, x volume strength of ${{H}_{2}}{{O}_{2}}$= $\dfrac{x}{5.6}$N ${{H}_{2}}{{O}_{2}}$
Now, we will compare the two normalities of the given ${{H}_{2}}{{O}_{2}}$
\[\dfrac{x}{5.6}=\dfrac{1}{0.56}\]
\[x=\dfrac{5.6}{0.56}=\,10\]
Therefore, x = 10
So, the volume strength of given ${{H}_{2}}{{O}_{2}}$ is 10
Hence, the answer of the given question is option (D)

Additional Information:
Volume strength of ${{H}_{2}}{{O}_{2}}$ is defined as the volume of oxygen gas released at NTP on the decomposition of ${{H}_{2}}{{O}_{2}}$ into oxygen gas and water.
Volume strength = $normality\,\times \,5.6$

Note: Hydrogen peroxide is used as a mild antiseptic to prevent infection. It contains peroxide linkage. In peroxide linkage two oxygen bonds are linked together via a single bond.