Question

Question #e6a5f

Answer 1

Here's what I got.

Explanation:

A wet gas is simply a term used to denote gases that are collected over water and, as a result, are mixed with water vapor.

In this regard, wet oxygen is a mixture of dry oxygen and water vapor.

So, you need to collect a volume of wet oxygen gas that would be equivalent to a `"337 mL"` of dry oxygen gas at `"1.00 atm"`.

The trick here is to realize that the total pressure of the oxygen gas - water vapor mixture will be equal to the atmospheric pressure measured in the lab.

`P_"lab" = overbrace(P_"dry oxygen" + P_"water vapor")^(color(purple)("the pressure of the wet oxygen gas"))`

In order to be able to find the volume of dry oxygen gas, you need to know the vapor pressure oif water at `15^@"C"`. The listed value is equal to

`P_"water" = "12.73 torr"`

http://www.endmemo.com/chem/vaporpressurewater.php

This means that at `"744 torr"`, the pressure of the dry oxygen gas would be

`P_"dry oxygen" = "744 torr" - "12.73 torr" = "731.27 torr"`

Assuming that the temperature remains constant, you can use Boyles' Law to find the volume of wet oxygen gas that must be collected at `"744 torr"` in order to produce that volume of dry oxygen

`color(blue)(P_1V_1 = P_2V_2)" "`, where

`P_1` - the pressure of the dry oxygen that's part of the oxygen - water vapor mixture
`V_1` - the volume of the wet oxygen gas
`P_2`, `V_2` - the pressure and volume of the dry oxygen gas

Rearrange to get

`V_1 = P_2/P_1 * V_2`

Plug in your values to get - remember that you have

`"1 atm " = " 760 torr"`

`V_1 = (760color(red)(cancel(color(black)("torr"))))/(731.27color(red)(cancel(color(black)("torr")))) * "337 mL" = color(green)("350. mL")`

Alternatively, you can actually use the ideal gas law equation to get the same result.

`color(blue)(PV = nRT)`

The trick this time is to realize that you must have equal numbers of moles of oxygen gas in both cases. So, you need

`n_"dry oxygen" = (P_2 * V_2)/(RT)`

`n_"dry oxygen" = (1 color(red)(cancel(color(black)("atm"))) * 337 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 15.0)color(red)(cancel(color(black)("K"))))`

`n_"dry oxygen" = "0.014245 moles"`

When collected over water, the pressure of the gas will be equal to `"731.27 torr"`. This means that the same amount of gas would be present in a volume of

`V_1 = (n_"dry oxygen" * RT)/P_1`

`V_1 = (0.014245 color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 15.0)color(red)(cancel(color(black)("K"))))/(731.27/760color(red)(cancel(color(black)("atm"))))`

`V_1 = color(green)("350. mL")`

The answer is rounded to three sig figs.