$5 \mathrm{g}$ of bleaching powder was suspended in water and volume was made up to half a litre. $20 \mathrm{mL}$ of this suspension when acidified with acetic acid and treated with excess of $\mathrm{KI}$ solution, liberated ${\mathrm{I}}_2$, which required $20 \mathrm{mL}$ of a $\mathrm{N}/10$ hypo solution for titration. Calculate the percentage of available chlorine in bleaching powder.

A $10 \mathrm{g}$ mixture of ${\mathrm{Cu}}_2\mathrm{S}$ and $\mathrm{CuS}$ was treated with $200 \mathrm{mL}$ of $0.75 \mathrm{M} {\mathrm{MnO}}_4^{-}$ in acid solution producing ${\mathrm{SO}}_2, {\mathrm{Cu}}^{2+}$ and ${\mathrm{Mn}}^{2+}$. The ${\mathrm{SO}}_2$ was boiled off and the excess ${\mathrm{MnO}}_4^{-}$ was titrated with $175 \mathrm{mL}$ of $1 \mathrm{M} {\mathrm{Fe}}^{2+}$ solution. Calculate the percentage of $\mathrm{CuS}$ in the original mixture.

A $0.518 \mathrm{g}$ sample of limestone is dissolved and then, $\mathrm{Ca}$ is precipitated as ${\mathrm{CaC}}_2{\mathrm{O}}_4$. After filtering and washing the precipitate, it requires $40 \mathrm{mL}$ of $0.25 \mathrm{N}$ ${\mathrm{KMnO}}_4$ solution acidified with ${\mathrm{H}}_2{\mathrm{SO}}_4$ to titrate it. What is the percentage of $\mathrm{CaO}$ in limestone?

${\mathrm{MnO}}_4^{-}+{\mathrm{CaC}}_2{\mathrm{O}}_4+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CaSO}}_4+{\mathrm{Mn}}^{2+}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

$20 \mathrm{mL}$ of $\frac{\mathrm{M}}{60}$ solution of ${\mathrm{KBrO}}_3$ was added to a certain volume of ${\mathrm{SeO}}_3^{2-}$ solution. ${\mathrm{Br}}_2$ evolved was removed by boiling and the excess of ${\mathrm{KBrO}}_3$ was back titrated with $5.1 \mathrm{mL}$ of $\frac{\mathrm{M}}{25}$ solution of ${\mathrm{NaAsO}}_2$. Calculate the weight of ${\mathrm{SeO}}_3^{2-}$ in the solution.

give the answer up to three places of decimal.

Concentrated $\mathrm{HCl}$ solution is $37.0\% \mathrm{HCl}$ and has a density of $1.19 \mathrm{g} {\mathrm{mL}}^{-1}$. A dilute solution of $\mathrm{HCl}$ is prepared by diluting $4.50 \mathrm{mL}$ of this concentrated $\mathrm{HCl}$ solution to $100 \mathrm{mL}$ with water. Then, $10 \mathrm{mL}$ of this dilute $\mathrm{HCl}$ solution reacts with an ${\mathrm{AgNO}}_3$ solution. Calculate the volume of $0.108 \mathrm{M} {\mathrm{AgNO}}_3$ solution required to precipitate all the chloride as $\mathrm{AgCl}\left(\mathrm{s}\right)$.

Give the answer up to two decimal places.

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?