Question

The Normality and volume strength of a solution made by mixing 1L each of 5.6 vol and 11.2 vol ${{H}_{2}}{{O}_{2}}$ solution is:
a.) 1N,5.6 vol
b.) 1.5N,5.6vol
c.) 1.5N, 8.4 vol
d.) 1N, 8.4 vol

Answer 1

Volume strength is 5.6 times of Normality. When solutions of two different normalities are mixed, the resultant normality is $=\dfrac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}$.For hydrogen peroxide, 1N is equal to 5.6 vol, so 5.6 vol is equal to 1N and 11.2 vol is equal to 2N. Volume strength is used to express the concentration of hydrogen peroxide.

Complete step by step answer:
-Volume strength refers to the number of volumes of oxygen evolved from 1 volume of hydrogen peroxide.
As normality is no. of equivalent weights present in 1 liter of solution so in 1N solution of hydrogen peroxide, 1 equivalent weight of hydrogen peroxide is present in 1 liter of solution.
-1 equivalent weight of hydrogen peroxide is 0.5 moles of hydrogen peroxide.
-0.5 moles of hydrogen peroxide forms 0.25 moles of oxygen.
-If reaction takes place at STP, 1 mole of oxygen will occupy volume of 22.4L so 0.25 moles of oxygen will occupy volume of 5.6L.
So, Volume strength= $5.6\times$Normality
5.6 Vol=1N and 11.2 vol will be 2N.
When solution is made by mixing 1L each of 5.6vol and 11.2 vol ${{H}_{2}}{{O}_{2}}$ solution, then resultant Normality of solution will be

& =\dfrac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}} \\\ & =\dfrac{1\times 1+2\times 1}{1+1} \\\ & =\dfrac{3}{2} \\\ & =1.5N \\\ \end{aligned}$$ Volume strength of resulting solution will be $1.5\times 5.6=8.4$Vol The Normality and volume strength of a solution made by mixing 1L each of 5.6 vol and 11.2 vol ${{H}_{2}}{{O}_{2}}$ solution is 1.5N, 8.4 vol **So, the correct answer is “Option C”.** **Note:** Volume strength refers to the number of volumes of oxygen evolved from 1 volume of hydrogen peroxide. If reaction takes place at STP, 1 mole of oxygen will occupy volume of 22.4L so 0.25 moles of oxygen will occupy volume of 5.6L. So, Volume strength= $5.6\times $Normality. When solutions of two different normalities are mixed, the resultant normality is given by formula $=\dfrac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}$.