State Henry’s law and explain why are the tanks used by scuba divers filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen)?

Henry's law constant for oxygen and nitrogen dissolved in water at 298 K are 2.0 xx 10^(9) Pa and 5.0 xx 10^(9) Pa , respectively . A sample of water at a temperature just above 273 K was equilibrated with air (20% oxygen and 80% nitrogen ) at 1 atm. The dissolved gas was separated from a sample of this water and the dried. Determine the composition of this gas.

According to the Avogadro's law , equal number of moles of gases occupy the same volume at identical conditions of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro's law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of nitrogen (N_(2)) and 20% by volume of oxygen (O_(2)) . If air is taken at STP , then its 1 mol would occupy 22.4 L . 1 mol of air would contain 0.8 mol of N_(2) and 0.2 mol of O_(2) hence the mole fraction of N_(2) and O_(2) are given by X_(n_(2))=0.8, X_(o_(2))=0.2 If air is treated as a solution of O_(2) and N_(2) , then % w/w of oxygen is:

According to the Avogadro's law , equal number of moles of gases occupy the same volume at identical conditions of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro's law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of nitrogen (N_(2)) and 20% by volume of oxygen (O_(2)) . If air is taken at STP , then its 1 mol would occupy 22.4 L . 1 mol of air would contain 0.8 mol of N_(2) and 0.2 mol of O_(2) hence the mole fraction of N_(2) and O_(2) are given by X_(n_(2))=0.8, X_(o_(2))=0.2 Volume occupied by air at STP containing exactly 11.2 gm f nitrogen:

According to the Avogadro's law , equal number of moles of gases occupy the same volume at identical conditions of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro's law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of nitrogen (N_(2)) and 20% by volume of oxygen (O_(2)) . If air is taken at STP , then its 1 mol would occupy 22.4 L . 1 mol of air would contain 0.8 mol of N_(2) and 0.2 mol of O_(2) hence the mole fraction of N_(2) and O_(2) are given by X_(n_(2))=0.8, X_(o_(2))=0.2 Density of air at STP is: