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What is the mass of oxalic acid, ${H_2}{C_2}{O_4}$ , which can be oxidized by $C{O_2}$ by $100mL$ of $MnO_4^ - $ solution, $10mL$ of which is capable of oxidizing $50mL$ of $1.00N$ ${I^ - }$ to ${I_2}$ ?
A. $2.25g$
B. $52.2g$
C. $25.2g$
D. $22.5g$

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Last updated date: 12th Apr 2024
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MVSAT 2024
Answer
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Hint: We need to realize that oxalic corrosive is having the equation ${H_2}{C_2}{O_4}$ . It is likewise called crab corrosive. It is really a white glasslike strong which is by and large found to get broken up in arrangement and colourless. The equation used to ascertain the load for oxalic corrosive must be given underneath,
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $

Complete answer:
We need to know, the arrangement that is the $100mL$ of $MnO_4^ - $ , at that point it is inquired as to whether we add $10mL$ in $50mL$ of $10N$ ${I^ - }$ , at that point it will be completely changed over into , so we can compose its condition,
$M{n^{ + 7}} + 5{e^ - } \to M{n^{ + 2}}$
Then,
${I^ - } + {e^ - } \to {I_2}$
We can write the expression as,
${N_1}{V_1} = {N_2}{V_2}$\[\]
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $
${N_2}$ = Normality of ${I^ - }$
${V_2}$ = Volume of ${I^ - }$
To calculate the normality ${N_1}$ ,
Applying given values in the above expression,
${N_1} \times 10 = 50 \times 10$
Then,
${N_1} = 5$
Now, we have to calculate mass of the ${H_2}{C_2}{O_4}$ by the use of the value of molar mass of ${H_2}{C_2}{O_4}$ .The molar mass of ${H_2}{C_2}{O_4}$ = $90g/mol$ .
Where, we can see that carbon in oxalic corrosive in $ + 3$ state, and it is evolving to, so the valency of this is $2$ , and reciprocals are something similar.
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Applying given values in the above equation,
$5 \times 10 = \dfrac{{w \times 2}}{{90}} \times 100$
Then, to find out the weight,
$w = \dfrac{{5 \times 10 \times 90}}{{2 \times 100}} = \dfrac{{4500}}{{200}} = 22.5$
Therefore,
The weight is $22.5g$

Hence, the option (D) is correct.

Note:
We need to establish ordinariness too. In this way, we ought not to get confused regarding ordinariness and molarity. At the point when ordinariness is the quantity of the same solute disintegrated per liter of arrangement. $N$ is represented as normality, and $M$ is represented as molarity.
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