Calculate the solubility of gas in water at $1.2 \text{ atm}$ and $25^{\circ} C$ if Henry's law constant is $0.45 \text{ mol dm}^{-3} \text{ atm}^{-1}$ at $25^{\circ} C$.

Assertion : When a salt such as $NaCl$ dissolves,the $Na^{+}$ and $Cl^{-}$ ions leaving the crystal lattice acquire far greater freedom.
Reason : In thermodynamic terms,the formation of solution occurs with a favourable change in free energy,i.e.,$\Delta H$ has a high positive value and $T\Delta S$ a low negative value.

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

Gas	$K_{H} \ (kbar)$
$\alpha$	$50$
$\beta$	$2$
$\gamma$	$2 \times 10^{-5}$
$\delta$	$0.5$

(Density of water $= 10^{3} \ kg \ m^{-3}$ at $298 \ K$). This table implies that:

If $O_2$ gas is bubbled through water at $303 \; K$,the number of millimoles of $O_2$ gas that dissolve in $1 \; L$ of water is (Nearest Integer). (Given: Henry's Law constant for $O_2$ at $303 \; K$ is $46.82 \; kbar$ and partial pressure of $O_2 = 0.920 \; bar$). (Assume solubility of $O_2$ in water is too small,nearly negligible).

How many grams of water are required to dissolve $4 \ g$ of sodium sulfate at $323 \ K$? The solubility of sodium sulfate at this temperature is $16\%$.

Observe the following data given in the table $(K_H = \text{Henry's law constant})$. The correct order of solubility of these gases is:

Gas	$K_H$ (kbar at $298 \ K$)
$CO_2$	$1.67$
$Ar$	$40.3$
$HCHO$	$1.83 \times 10^{-5}$
$CH_4$	$0.413$