A mixture of ${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ weighing $0.24 \mathrm{g}$ on being treated with $\mathrm{KI}$ in acid solution liberates just sufficient ${\mathrm{I}}_2$ to react with $60 \mathrm{mL}$ of $0.1 \mathrm{N} {\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ solution. Calculate the percentage of $\mathrm{Cr}$ and $\mathrm{Mn}$ in the mixture.

Federal regulations set an upper limit of $50$parts per million $\left(\mathrm{ppm}\right)$ of ${\mathrm{NH}}_3$ in the air in a work environment (that is, $50 \mathrm{mL} {\mathrm{NH}}_3$ per ${10}^6 \mathrm{mL}$ of air). The density of ${\mathrm{NH}}_3\left(\mathrm{g}\right)$ at room temperature is $0.771 \mathrm{g} {\mathrm{L}}^{-1}$. Air from a manufacturing operation was drawn through a solution containing $100 \mathrm{mL}$ of $0.0105 \mathrm{M}$ $\mathrm{HCl}$. The ${\mathrm{NH}}_3$ reacts with $\mathrm{HCl}$ as follows:

${\mathrm{NH}}_3\left(\mathrm{aqueous}\right)+\mathrm{HCl}\left(\mathrm{aqueous}\right)\to {\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{aqueous}\right)$

After drawing air through the acid solution for $10$ minutes at a rate of $10 \mathrm{litres} {\min }^{-1}$, the acid was titrated. The remaining acid required $13.1 \mathrm{mL}$ of $0.0588 \mathrm{M} \mathrm{NaOH}$ to reach the equivalence point.

$\left(\mathrm{a}\right)$ How many grams of ${\mathrm{NH}}_3$ were drawn into the acid solution?

$\left(\mathrm{b}\right)$ How many $\mathrm{ppm}$ of ${\mathrm{NH}}_3$ were in the air?

$\left(\mathrm{c}\right)$ Is this manufacturer in compliance with regulations?

In alkaline condition, ${\mathrm{KMnO}}_4$ reacts as follows:

$2{\mathrm{KMnO}}_4+2\mathrm{KOH}\to 2{\mathrm{K}}_2{\mathrm{MnO}}_4+{\mathrm{H}}_2\mathrm{O}+\mathrm{O}$,

The equivalent weight for ${\mathrm{KMnO}}_4$ is