$500$ mL of $0.2$ $M$ $MnO_4^-$ solution in basic medium when mixed with $500$ mL of $1.5$ $M$ $KI$ solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard $x$ $M$ thiosulphate solution in presence of starch. If $300$ mL of thiosulphate was consumed, then the value of $x$ is . . . . . . .

Metal sulphate $(A) \xrightarrow{\text{Heat}}$ Oxide $(B) +$ Gas $(C) +$ Gas $(D) \xrightarrow{Cr_2O_7^{2-} / H^{+}}$ Green solution $\xrightarrow[\text{Excess}]{Na_2O_2}$ Yellow solution $(E)$. Compound $A, B, C, D$ and $E$ are respectively:

$100 \ mL$ of $0.1 \ M$ $Fe^{2+}$ solution was titrated with $\frac{1}{60} \ M$ $Cr_2O_7^{2-}$ solution in acid medium. What is the volume (in $L$) of $Cr_2O_7^{2-}$ solution consumed?

For the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$,which option represents twice the equivalent weight of the oxidizing agent?

$KMnO_4$ oxidises $C_2H_2O_4$ to form $CO_2$. In which of the following,the reaction will be faster?

Copper in an alloy is estimated by dissolving in conc. nitric acid. In this process,copper is converted to cupric nitrate with the evolution of nitric oxide $(NO)$. The mixture when treated with potassium iodide forms cupric iodide,which is unstable and decomposes to cuprous iodide and iodine. The amount of copper in the alloy is estimated by titrating the liberated iodine with sodium thiosulphate. The reactions are:
$a \,Cu + b \,HNO_3 \rightarrow c \,Cu(NO_3)_2 + d \,NO + e \,H_2O$
$f \,CuI_2 \rightarrow g \,Cu_2I_2 + h \,I_2$
$i \,Na_2S_2O_3 + j \,I_2 \rightarrow k \,Na_2S_4O_6 + l \,NaI$
(Fill in the blanks)
$(a)$ The coefficients are: $a=\ldots, b=\ldots, c=\ldots, d=\ldots$ and $e=\ldots$.
$(b)$ The coefficients are: $f=\ldots, g=\ldots$ and $h=\ldots$.
$(c)$ The coefficients are: $i=\ldots, j=\ldots, k=\ldots$ and $l=\ldots$.
$(d)$ If $2.54 \,g$ of $I_2$ is evolved from a $2.0 \,g$ sample of the alloy,what is the percentage of copper in the alloy? (Atomic weights of iodine and copper are $127$ and $63.5$,respectively).