A mixture of pure ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and pure ${\mathrm{KMnO}}_4$ weighing $0.561 \mathrm{g}$ was treated with excess of $\mathrm{KI}$ in acid medium. Iodine liberated required $100 \mathrm{mL}$ of $0.15 \mathrm{N}$ hypo solution for exact oxidation. What is the percentage of each in the mixture? The reactions involved are:

$2{\mathrm{KMnO}}_4+8{\mathrm{H}}_2{\mathrm{SO}}_4+10\mathrm{KI}\to 6{\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{I}}_2$

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+6\mathrm{KI}+7{\mathrm{H}}_2{\mathrm{SO}}_4\to 4{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+7{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2$

${\mathrm{I}}_2+2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\to 2\mathrm{NaI}+{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$

Under what pressure must an equimolar mixture of ${\mathrm{Cl}}_2$ and ${\mathrm{PCl}}_3$ be placed at $250°\mathrm{C}$ in order to obtain $80\%$ conversion of ${\mathrm{PCl}}_3$ into ${\mathrm{PCl}}_5\text{?} {\mathrm{K}}_{\mathrm{p}}$ for ${\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_5\left(\mathrm{g}\right)$ is $1.78$.

Round off your answer up to two places of decimal.

$0.45 \mathrm{g}$ of an organic compound $\left(\mathrm{A}\right)$, on ignition, gives $0.905 \mathrm{g} {\mathrm{CO}}_2$ and $0.185 \mathrm{g}$ ${\mathrm{H}}_2\mathrm{O}. 0.35 \mathrm{g} \left(\mathrm{A}\right)$ on boiling with ${\mathrm{HNO}}_3$ and adding ${\mathrm{AgNO}}_3$ solution gives $0.574 \mathrm{g}$ of $\mathrm{AgCl}$. The vapour density of $\left(\mathrm{A}\right)$ is $87.5 .$ $\left(\mathrm{A}\right)$ on hydrolysis with $\mathrm{Ca} (\mathrm{OH}{)}_2$ yields $\left(\mathrm{B}\right)$, which on mild reduction gives an optically active compound $\left(\mathrm{C}\right)$. On heating $\left(\mathrm{C}\right)$ with ${\mathrm{I}}_2$ and $\mathrm{NaOH}$, iodoform is produced along with $\left(\mathrm{D}\right)$. With $\mathrm{HCl}$, $\left(\mathrm{D}\right)$ gives a solid, which is markedly more soluble in hot water than in cold. Identify $\left(\mathrm{A}\right), \left(\mathrm{B}\right), \left(\mathrm{C}\right)$ and $\left(\mathrm{D}\right)$.

Two organic compounds containing $\mathrm{C}=62.1\%, \mathrm{H}=10.34\%, \mathrm{O}=27.6\%$, with $\mathrm{KCN}$ and ${\mathrm{H}}_2{\mathrm{SO}}_4$, gave compounds, which on hydrolysis gave two isomeric monobasic acids with molecular mass $104.$ Name the compounds and also the third isomer.

The compound $\left(\mathrm{X}\right)$, with molecular formula ${\mathrm{C}}_5{\mathrm{H}}_8\mathrm{O}$, does not react appreciably with Lucas reagent at room temperature, but gives a precipitate with ammonium silver nitrate. With an excess of $\mathrm{MeMg} \mathrm{Br}, 0.42 \mathrm{g}$ of $\left(\mathrm{X}\right)$ gives $224 \mathrm{mL}$ of ${\mathrm{CH}}_4$ at $\mathrm{STP}$. The treatment of $\left(\mathrm{X}\right)$ with ${\mathrm{H}}_2$ in the presence of $\mathrm{Pt}$ catalyst, followed by boiling with an excess $\mathrm{HI}$, gives $\mathrm{n}-$pentane. Suggest the structure for $\left(\mathrm{X}\right)$.