Answer to Question #274234 in General Chemistry for carl

Question

If the oxygen gas sample in #1 passed through a drier that decreased the pressure of the mixture to 750 Torr and the pressure exerted by the water vapour is only 80% of the saturated vapour pressure at the given temperature, what would be the pressure of the dry oxygen?

Answer 1

passed through a drier that decreased the pressure of the mixture to 750 torr and the pressure exerted by the water vapour is only 80 percent of the saturated vapour pressure at the given temperature, what would be the pressure of the dry oxygen? p(total)=p1+p2. p2=p(total)-p1 = 758-31.8 = 726,2 torr.