$K_H$ is Henry's constant and has the unit

The $K_{H}$ values for some gases at the same temperature '$T$' are given below:

Gas	$K_{H} / \text{kbar}$
$Ar$	$40.3$
$CO_{2}$	$1.67$
$HCHO$	$1.83 \times 10^{-5}$
$CH_{4}$	$0.413$

where $K_{H}$ is Henry's Law constant in water. The order of their solubility in water is:

Calculate the solubility of a gas having partial pressure $0.15 \ bar$ at $25^{\circ} C$.
$[K_{H} = 0.16 \ mol \ dm^{-3} \ bar^{-1}]$

Air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportions of $20\%$ and $79\%$ by volume at $298\ K$. The water is in equilibrium with air at a pressure of $10\ atm$. At $298\ K$,if the Henry's law constants for oxygen and nitrogen are $3.30 \times 10^7\ mm$ and $6.51 \times 10^7\ mm$,respectively,the mole ratio of gases $(O_2 : N_2)$ dissolved in water is about

What is the solubility of a gas in water at $25^{\circ} C$ if the partial pressure is $0.18 \ atm$? $(K_{H} = 0.16 \ mol \ dm^{-3} \ atm^{-1})$

The Henry's law constant for the solubility of $N_2$ gas in water at $298 \ K$ is $1.0 \times 10^5 \ atm$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10 \ moles$ of water at $298 \ K$ and $5 \ atm$ pressure is