A mixture of oxygen and hydrogen is analysed by passing it over hot copper oxide through a drying tube. Hydrogen reduces $\mathrm{CuO}$ according to the equation, $\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$; the copper formed is then oxidised by oxygen as:

$\mathrm{Cu}+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{CuO}$
$100 {\mathrm{cm}}^3$ of the mixture measured at $25°\mathrm{C}$ and $750 \mathrm{mm}$ yields $84.5 {\mathrm{cm}}^3$ of dry oxygen measured at $25°\mathrm{C}$ and $750 \mathrm{mm}$, after passing over $\mathrm{CuO}$ and a drying agent. What is the mole per cent of ${\mathrm{H}}_2$ in the mixture?

When $100 \mathrm{mL}$ of an ${\mathrm{O}}_2-{\mathrm{O}}_3$ mixture was passed through turpentine, there was a reduction in volume by $20 \mathrm{mL}$. If $100 \mathrm{mL}$ of such a mixture is heated, what will be the increase in volume (in ml) ?

[Hint: ${\mathrm{O}}_3$ is absorbed by turpentine]

$5.22\times {10}^{-4} \mathrm{mol}$ of a gas containing ${\mathrm{H}}_2, {\mathrm{O}}_2$ and ${\mathrm{N}}_2$ exerted a pressure of $67.4 \mathrm{mm}$ in a certain standard volume. The gas was passed over a hot platinum filament which combined ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ into ${\mathrm{H}}_2\mathrm{O}$ which was frozen out. When the gas was returned to the same volume, the pressure was $14.3 \mathrm{mm}$. Extra oxygen was added to increase the pressure to $44.3 \mathrm{mm}$. The combustion was repeated, after which the pressure read $32.9 \mathrm{mm}$. What was the mole fraction of ${\mathrm{H}}_2, {\mathrm{O}}_2$ and ${\mathrm{N}}_2$ in the gas sample?

[Hint: $2$ volumes of ${\mathrm{H}}_2$ combine with $1$ volume of ${\mathrm{O}}_2$. In the first combustion, ${\mathrm{H}}_2$ is in excess, while in the second one, ${\mathrm{O}}_2$ is in excess]