# What is the molar mass of 31.3 g of gas exerting a pressure of 5.71 atm on the walls of a 3.04 L container at 297 K?

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Meave60 Share
Mar 9, 2018

The molar mass of the gas is $\text{43.9 g/mol}$.

#### Explanation:

Use the ideal gas law equation to determine moles. Then divide the given mass by the moles to get molar mass.

$P V = n R T ,$

where:

$P$ is pressure, $V$ is volume, $n$ is moles, $R$ is the gas constant, and $T$ is the temperature in Kelvins.

Organize the data:

Known

$P = \text{5.71 atm}$

$V = \text{3.04 L}$

$R = \text{0.0820575 L atm K"^(-1) "mol"^(-1)}$

$T = \text{297 K}$

Unknown

n"

Calculate moles

Rearrange the ideal gas law equation to isolate $n$. Plug in the known values and solve.

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

n=((5.71color(red)cancel(color(black)("atm")))xx(3.04color(red)cancel(color(black)("L"))))/((0.0820575color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1))xx(297color(red)cancel(color(black)("K"))))="0.712254 mol" (I am keeping some guard digits to reduce rounding errors.)

Determine molar mass

Divide the given mass by the calculated moles.

M=("31.3 g")/("0.712254 mol")="43.9 g/mol" (rounded to three significant figures)

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Mar 9, 2018

The molar mass of the gas is $43.946$ grams per mole or $43.9$ grams per mole using significant figures.

#### Explanation:

Use the ideal gas law $P V = n R T$.

$P$ is the pressure in atmospheres (atm)

$V$ is the volume in liters (L)

$n$ is the number of moles

$R$ is the universal gas constant ($.08206$ $L \cdot a t m \cdot {K}^{- 1} \cdot m o {l}^{- 1}$)

$T$ is the temperature in kelvin

Notice that number of moles is $\frac{g r a m s}{m o l a r m a s s}$ or $\frac{g}{x}$ where x is the molar mass of the gas.

Plug it into the ideal gas law for $n$.

$P V = \frac{g}{x} R T$

Rearrange and isolate x.

$x = \frac{g R T}{P V}$

Plug everything in (excluding units)

$x = \frac{31.3 \cdot .08206 \cdot 297}{5.71 \cdot 3.04}$

$x = 43.946 \frac{g}{m o l}$

(If you cancel out all the units, you will find that you will get $\frac{g}{m o l}$ which is the correct unit for molar mass.)

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