Scuba divers when go in deep sea face a very painful situation known as "bends", the formation of ${\mathrm{N}}_2$ bubbles around the joints due to decrease in solubility of ${\mathrm{N}}_2$ as a result of reduced pressure, when the diver rises up towards sea surface. This situation is avoided by-

The oxygen dissolved in water exerts a partial pressure of $20 \mathrm{kPa}$ in the vapour above water. The molar solubility of oxygen in water is ______ $\times {10}^{-5} \mathrm{mol} {\mathrm{dm}}^{-3}.$

(Round off to the Nearest Integer).

[Given : Henry's law constant $={\mathrm{K}}_{\mathrm{H}}=8.0\times {10}^4 \mathrm{kPa}$ for ${\mathrm{O}}_2.$ Density of water with dissolved oxygen$=1.0 \mathrm{kg} {\mathrm{dm}}^{-3}$]

${\mathrm{CO}}_2$ gas is bubbled through water during a soft drink manufacturing process at $298 \mathrm{K}$. If ${\mathrm{CO}}_2$ exerts a partial pressure of $0.835 \mathrm{bar}$ then $\mathrm{x} \mathrm{m} \mathrm{mol}$ of ${\mathrm{CO}}_2$ would dissolve in $0.9 \mathrm{L}$ of water. The value of $\mathrm{x}$ is _______. (Nearest integer)

(Henry's law constant for ${\mathrm{CO}}_2$ at $298 \mathrm{K}$ is $1.67\times {10}^3 \mathrm{bar}$)

Henry's constant (in kbar) for four gases $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$  and $\delta$ in water at $298 \mathrm{K}$ is given below :

$\begin{array}{ccccc}& \mathrm{\alpha }& \mathrm{\beta }& \mathrm{\gamma }& \mathrm{\delta }\\ {\mathrm{K}}_{\mathrm{H}}& 50& 2& 2\times {10}^{-5}& 0.5\end{array}$

(density of water $={10}^3 \mathrm{k} {\mathrm{gm}}^{-3}$ at $298 \mathrm{K}$ ) This table implies that :

From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is