Question

Calculate the mass (m) of the oxalic acid which can be oxidized to $C{O_2}$ by $100ml$ of an $Mn{O_4}^ - $solution,$10ml$of which is capable to oxidizing $50ml$of $1N$${I^ - }\,to\,{I_2}$.
$Mn{O_4}\, + \,8{H^ + }\, + \,5{e^ - }\, \to \,M{n^{2 + }}\, + \,4{H_2}O$
${H_2}{C_2}{O_{4\,}}\, \to \,2C{O_2}\, + \,2{H^ + }\, + \,2{e^ - }$
$2{I^ - }\, \to \,{I_2}\,\, + \,2{e^ - }$
What is the value of $10m$?

Answer 1

Hint: In this question, $100ml$ of $Mn{O_4}^ - $ solution is given in which $10ml$oxidizes the $50ml$ of $1N$${I^ - }\,to\,{I_2}$.So, here we are calculating the value of $10m$.

Complete step by step answer:
Normality is described as the number of gram or mole equivalents of solute present in one litre of a solution.
Normality is mainly used in three common situations:
-To determine the concentrations in acid-base chemistry.
-Used in precipitation reactions to measure the number of ions which are likely to precipitate in a given reaction.
-It is used in redox reactions to find the number of electrons which a reducing or an oxidizing agent can donate or accept.
How to Calculate the Normality?
-The first tip that we can follow is to gather information about the equivalent weight of the reacting substance or the solute.
-The second step involves calculating the no. of gram equivalent of solute.
-We can remember that the volume is to be calculated in litres.
-Finally, normality is calculated using the formula and replacing the values.
According to the Question:
Firstly, we can write the balanced chemical equation
$2KMn{O_{4\,}}\,\, + \,\,5{H_2}{C_2}{O_4}\,\, + \,\,3{H_2}S{O_4}\,\, \to \,\,2MnS{O_4}\,\, + \,\,10C{O_2}\,\, + \,\,{K_2}S{O_4}\,\, + \,\,8{H_2}O$
By applying the Formula
$(KMn{O_4})\,\,\,{N_1}\,{V_1}\,\,\,\, = \,\,\,{N_2}\,{V_2}\,\,\,\left( {{I_2}} \right)$
Put the value of ${V_1}\, = \,10$ and ${V_2}\, = \,50$where $N$is the Normality
So, ${N_1}\, \times \,10\, = \,1\, \times \,50$
So , ${N_1}\, = \,5N$
Then , apply the n- factor for $KMn{O_4}$, $7 - 2\, = \,5$
Calculating the number of moles of $KMn{O_4}$
Moles of $KMn{O_4}$$ = \,5 \div \,5\, = 1$
$2\,mol\,KMn{O_4}\, = \,5\,mol\,{H_2}{C_2}{O_4}$
$1\,mol\,KMn{O_4}\, = \,\,2.5mol\,{H_2}{C_2}{O_4}$
$100\,ml\,or\,0.1l\, = \,0.1\, \times \,2.5\, = \,0.25\,moles$
So, finally we get the number of moles $ = \,0.25\,moles$
So , Mass of Oxalic Acid (${H_2}{C_2}{O_4}$) is $0.25\, \times \,90\, = 22.5\,gm$

Note: While calculating the normality, the substitution of the value must be accurate .In this question, main point is that we can always remember that when we have two volumes given in the question then apply the formula ${N_1}\,{V_{1\,}}\, = \,{N_2}\,{V_2}$ and while calculating the number of moles always use unitary method.