$\mathrm{x}$ millimoles of ${\mathrm{KMnO}}_4$ reacts completely with $\mathrm{y}$ millimoles of ${\mathrm{MnSO}}_4$ in the presence of fluoride ions, to give ${\mathrm{MnF}}_4$ quantitatively. Then

The amount of wet $\mathrm{NaOH}$ containing $15\%$ water required to prepare $70$ litres of $0.5 \mathrm{N}$ solution is

$28{{\mathrm{NO}}_3}^{–}+3{\mathrm{As}}_2{\mathrm{S}}_3+4{\mathrm{H}}_2\mathrm{O}\to 6{{\mathrm{AsO}}_4}^{3–}+28\mathrm{NO}+9{{\mathrm{SO}}_4}^{2–}+8{\mathrm{H}}^{+}.$

What is the equivalent mass of ${\mathrm{As}}_2{\mathrm{S}}_3$ in the above reaction (molecular mass of ${\mathrm{As}}_2{\mathrm{S}}_3=\mathrm{M}$)?

$125 \mathrm{mL}$ of $63\% (\mathrm{w}/\mathrm{v})$ ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4.2{\mathrm{H}}_2\mathrm{O}$ solution is made to react with $25 \mathrm{mL}$ of $40\% (\mathrm{w}/\mathrm{v})$ $\mathrm{NaOH}$ solution.

The resulting solution is (ignoring hydrolysis of ions)

How many litres of ${\mathrm{Cl}}_2$ at $\mathrm{STP}$ will be liberated by the oxidation of $\mathrm{NaCl}$ using $10 \mathrm{g} {\mathrm{KMnO}}_4$ in an acidic medium? (Atomic weight of $\mathrm{Mn}=55$, $\mathrm{K}=39$)

One gram of ${\mathrm{Na}}_3{\mathrm{AsO}}_4$ is boiled with excess of solid $\mathrm{Kl}$ in the presence of strong $\mathrm{HCl}$. The iodine evolved is absorbed in $\mathrm{KI}$ solution and titrated against $0.2 \mathrm{N}$ hypo solution. Assuming the reaction to be ${{\mathrm{AsO}}_4}^{3–}+2{\mathrm{H}}^{+}+2{\mathrm{I}}^{–}\to {{\mathrm{AsO}}_3}^{3–}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2$, calculate the volume of hypo consumed [atomic weight of $\mathrm{As}=75$].

If $10 \mathrm{g}$ of ${\mathrm{V}}_2{\mathrm{O}}_5$ is dissolved in an acid and is reduced to ${\mathrm{V}}^{2+}$ by zinc metal, how many moles of ${\mathrm{I}}_2$ are reduced by the resulting solution, if it is further oxidised to ${\mathrm{VO}}^{2+}$ ions?
[Assume no change in the ${\mathrm{Zn}}^{2+}$ ions] (Atomic masses are $\mathrm{V}=51, \mathrm{O}=16, \mathrm{I}=127$)