# Question 2cbc3

Mar 23, 2016

$\text{66.9 mL}$

#### 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.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{\text{total}} = {\sum}_{i} {P}_{i} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here ${P}_{i}$ represents the partial pressure of gas $i$.

${P}_{\text{total" = P_"water" + P_"gas}}$

So, you know that water has a vapor pressure of $\text{27 torr}$ at ${27.2}^{\circ} \text{C}$. This means that the partial pressure of the gas will be

${P}_{\text{gas" = P_"total" - P_"water}}$

${P}_{\text{gas" = "648 torr" - "27 torr" = "621 torr}}$

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Rearrange the equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

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 $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L}$.

This means that $0.002949$ moles will occupy a volume of

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

${V}_{S T P} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{66.9 mL} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE Many sources still list STP conditions as being a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions for pressure and temperatue, one mole of any ideal gas occupies $\text{22.4 L}$.

If this is the value given to you for the molar volume of a gas at STP, simply redo the last calculation using $\text{22.4 L}$ instead of $\text{22.7 L}$.