Correct option is (A)

The reaction given is a disproportionation reaction where iodine (${\mathrm{I}}_2$) undergoes both oxidation and reduction. Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

In the process, iodine gets oxidized from 0 oxidation state (in ${\mathrm{I}}_2$) to +5 oxidation state (in ${\mathrm{IO}}_3^{-}$), and reduced from 0 oxidation state (in ${\mathrm{I}}_2$) to -1 oxidation state (in ${\mathrm{I}}^{-}$).

The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction. In this case, the n-factor for ${\mathrm{IO}}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for ${\mathrm{I}}^{-}$, it's 1 (as iodine goes from 0 to -1).

Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction). Hence, the molar ratio of ${\mathrm{IO}}_3^{-}$ to ${\mathrm{I}}^{-}$ must be 1:5.

So, the balanced reaction would be:

${\mathrm{IO}}_3^{-}+6{\mathrm{H}}^{+}+5{\mathrm{I}}^{-}\to 3{\mathrm{I}}_2+3{\mathrm{H}}_2\mathrm{O}$

This equation yields 3 moles of ${\mathrm{I}}_2$, but the original equation needs to produce 6 moles of ${\mathrm{I}}_2$, so the entire equation is multiplied by 2:

$2{\mathrm{IO}}_3^{-}+12{\mathrm{H}}^{+}+10{\mathrm{I}}^{-}\to 6{\mathrm{I}}_2+6{\mathrm{H}}_2\mathrm{O}$

This tells us that to get 6 moles of ${\mathrm{I}}_2$, you need 10 moles of ${\mathrm{I}}^{-}$. So, $x=10$

Correct option is 3

Milieq of acid $A=$ Milieq of base $M(OH{)}_2$

$\begin{array}{rl}& (\mathrm{M}\times \mathrm{V}\times \mathrm{n}-{\text{ Factor }}_A=(\mathrm{M}\times \mathrm{V}\times \mathrm{n}-\text{ Factor }{)}_{\mathrm{M}(\mathrm{OH}{)}_2}\\ \\ & [\mathrm{n}-\text{ Factor of }\mathrm{M}(\mathrm{OH}{)}_2=2]\\ \\ & 0.1\times 10\times \mathrm{n}\text{-Factor }=0.05\times 30\times 2\\ \\ & (\mathrm{n}-\text{ Factor }{)}_{\mathrm{A}}=3\end{array}$

Hence basicity of acid $A$ is 3.

Correct option is 10

Molar mass of Na₂CO₃.xH₂O

= 23 × 2 + 12 + 48 + 18x

= 46 + 12 + 48 + 18x

= (106 + 18x)

As n_factor in dissolution will be determined from net cationic or anionic charge; which is 2

Eq wt = $\frac{M}{2}$ = (53 + 9x)

volume = 100 ml = 0.1 Litre

Normality =

$\Rightarrow$ 0.1= $\frac{\frac{1.43}{53+9x}}{0.1}$

$\Rightarrow$ 53 + 9x = 143

$\Rightarrow$ 9x = 90

$\Rightarrow$ x = 10

Correct option is (C)

$\therefore$ Equivalent of KMnO₄ = Eq of FeC₂O₄ + Eq of Fe₂(C₂O₄)₃ + FeSO₄

$\Rightarrow$ moles of KMnO₄ $\times$ 5 = 3 $\times$ 1 + 6 $\times$ 1 + 1 $\times$ 1

$\Rightarrow$ moles of KMnO₄ = $\frac{10}{5}$ = 2

Correct option is (D)

The reducing agent loses electron during redox reaction i.e. oxidises itself.

$\left(1\right)H_2\stackrel{-1}{O_2}+2H^{+}+2e^{-}\stackrel{}{⟶}2H_2\stackrel{-2}{O}(\mathrm{Re}d.)$

$\left(2\right)H_2\stackrel{-1}{O_2}\stackrel{}{⟶}\stackrel{0}{O_2}+2H^{+}+2e^{-}(Ox.)$

$\left(3\right)H_2\stackrel{-1}{O_2}+2e^{-}\stackrel{}{⟶}2\stackrel{-2}{O}{H}^{-}(Red.)$

$\left(4\right)H_2{O}_2^{-1}+2O{H}^{-}\stackrel{}{⟶}\stackrel{0}{O_2}+H_{2}O+2e^{-}\left(Ox.\right)$

Correct option is (C)

After balancing the reaction we get

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}$ + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + $8H_{2}O$

$\therefore$ $x$ = 2, $y$ = 5 and z = 16