Given below are two statements:
Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$).
Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below:
Statement I:
In the oxalic acid vs $ KMnO_4 $ titration (in the presence of dilute $ H_2SO_4 $), the solution needs to be heated initially to $ 60^\circ C $. However, no heating is required in the Ferrous Ammonium Sulphate (FAS) vs $ KMnO_4 $ titration under similar conditions.
This statement is true.
In the titration of oxalic acid with $ KMnO_4 $, heating to around $ 60^\circ C $ is required to initiate the reaction because it is relatively slow at room temperature.
In contrast, the reaction between $ KMnO_4 $ and FAS is fast even at room temperature, so no initial heating is necessary.
Statement II:
In oxalic acid vs $ KMnO_4 $ titration, the initial formation of $ Mn^{2+} $ takes place at high temperature, which then acts as a catalyst (autocatalyst) for the further reaction. In the case of FAS vs $ KMnO_4 $, heating may cause oxidation of $ Fe^{2+} $ to $ Fe^{3+} $ by atmospheric oxygen, introducing errors in the experiment.
This statement is also true.
In the oxalic acid titration, the $ Mn^{2+} $ ions formed during the reaction act as an autocatalyst, increasing the reaction rate after some time. Since the initial formation of $ Mn^{2+} $ is slow, heating is required to speed up the process.
In the FAS titration, heating may lead to aerial oxidation of $ Fe^{2+} $ to $ Fe^{3+} $, which can interfere with accurate titration results.
Conclusion:
Both Statement I and Statement II are true.
Final Answer:
The final answer is $ \ \text{Both Statement I and Statement II are true} $.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Given below are two statements:
Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$).
Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)