Titration of ${\mathrm{Na}}_3{\mathrm{PO}}_4$ with $\mathrm{HCl}$ to phenolphthalein end point produces ${\mathrm{Na}}_2{\mathrm{HPO}}_4$; subsequent titration to methyl orange end point produces ${\mathrm{NaH}}_2{\mathrm{PO}}_4$. Find the $\mathrm{mg}$ of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$ present in a sample which required $25\mathrm{ml},0.12 \mathrm{NHCl}$ solution to reach the phenolophthalein end point and an additional $35\mathrm{ml}$ of the same acid to reach the methyl orange end point.

Consider the reaction,

${\mathrm{aFeS}}_2+{\mathrm{bO}}_2\to {\mathrm{cFe}}_2{\mathrm{O}}_3+{\mathrm{dSO}}_2$

Which is correct for the above reaction?

What is the weight of sodium bromate and molarity of solution necessary to prepare $85.5 \mathrm{mL}$ of $0.672 \mathrm{N}$ solution, when the half-cell reaction is ${\mathrm{BrO}}_3^{-}+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to {\mathrm{Br}}^{-}+3{\mathrm{H}}_2\mathrm{O}$?

Give answer after multiplying with 100 and rounding off to the nearest integer value.

The ${\mathrm{Mn}}_3{\mathrm{O}}_4$ formed on strong heating of a sample of ${\mathrm{MnSO}}_4· 4{\mathrm{H}}_2\mathrm{O}$ was dissolved in $100 {\mathrm{cm}}^3$ of $0.1 \mathrm{N} {\mathrm{FeSO}}_4$ containing dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$. The resulting solution reacted completely with $50 {\mathrm{cm}}^3$ of ${\mathrm{KMnO}}_4$ solution. $25 {\mathrm{cm}}^3$ of this ${\mathrm{KMnO}}_4$ solution required $30 {\mathrm{cm}}^3$ of $0.1 \mathrm{N} {\mathrm{FeSO}}_4$ solution for complete reaction. Calculate the amount of ${\mathrm{MnSO}}_4·4{\mathrm{H}}_2\mathrm{O}$ in the sample.

Give answer after multiplying with 100 and rounding off to the nearest integer value.