# Question #2cbc3

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to notice here is that the gas was collected **over water**, which implies that it is mixed with *water vapor*.

The given volume actually contains a *mixture* of the gas and of water vapor. As you know, the pressure of a gas mixture can be expressed in terms of the **partial pressures** of its constituent gaseous components - this is known as **Dalton's Law of Partial Pressures**.

Simply put, the **total pressure** of a gaseous mixture will be equal to the sum of the partial pressures of the gases that make up said mixture.

#color(blue)(|bar(ul(color(white)(a/a)P_"total" = sum_i P_icolor(white)(a/a)|)))#

Here

In your case, you have

#P_"total" = P_"water" + P_"gas"#

So, you know that water has a vapor pressure of

#P_"gas" = P_"total" - P_"water"#

#P_"gas" = "648 torr" - "27 torr" = "621 torr"#

Now, *one way* to go about solving for the volume of the gas at **STP** is to use the **ideal gas law** equation to solve for the **number of moles** of *dry gas* collected.

Once you know how many *moles* of gas you have, you an use the **molar volume of gas at STP** to determine the new volume of the gas.

So, the ideal gas law equation looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Rearrange the equation to solve for

#PV = nRT implies n = (PV)/(RT)#

Make sure that you convert the pressure from *mmHg* to *atm*, the volume of the gas from *milliliters* to *liters*, and the temperature from *degrees Celsius* to *Kelvin*. Plug in your values to get

#n = (621/760color(red)(cancel(color(black)("atm"))) * 89.0 * 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 + 27.2)color(red)(cancel(color(black)("K")))) = "0.002949 moles"#

You know that **STP** conditions are defined as a pressure of **one mole** of any ideal gas occupies

This means that

#0.002949color(red)(cancel(color(black)("moles"))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole")))))^(color(purple)("molar volume of a gas at STP")) = "0.06694 L"#

Expressed in *milliliters* and rounded to three **sig figs**, the answer will be

#V_(STP) = color(green)(|bar(ul(color(white)(a/a)"66.9 mL"color(white)(a/a)|)))#

**SIDE NOTE** *Many sources still list STP conditions as being a pressure of* *and a temperature of*

*Under these conditions for pressure and temperatue, one mole of any ideal gas occupies*

*If this is the value given to you for the molar volume of a gas at STP, simply redo the last calculation using* *instead of*