# Question #8b6ce

Nov 9, 2015

By using the Ideal Gas Law equation ($P V = n R T$)

#### Explanation:

Look at the given, you were given the Pressure (P), Temperature (T), Volume (V) and mass (m) of an unknown gas.

P = 79.97 kPa
V = 4.167 L
T = ${30}^{\circ} C$
m= 20.83 g

Based on the given, you can definitely use the Ideal Gas Law equation to find the molar mass of the gas, where R is a constant equal to $0.0821 \frac{L \cdot a t m}{m o l \cdot K}$.

But first, we need to convert units:
P = $79.97 \cancel{\text{kPa}}$ x $\frac{0.00986923 a t m}{1 \cancel{\text{kPa}}}$ = $0.789 a t m$
V = $4.167 L$
T = ${30}^{\circ} C$ + $273.15 K$ = $303.15 K$
m= $\text{20.83 g}$

Now we are ready to compute.

$P V = n R T$

$\left(0.789 a t m\right) \left(4.167 L\right)$ = $n$ ($0.0821 \frac{L \cdot a t m}{m o l \cdot \cancel{K}}$) ($303.15 \cancel{K}$)

$\text{3.288 atm} \cdot L$ = $n$ ($24.8583 \frac{L \cdot a t m}{m o l}$)

n = $\left(3.288 \cancel{\text{atm"* cancel L)/(24.8583 cancel L* cancel (atm)/"mol}}\right)$ = 0.1322 mol

Since molar mass is defined as weight of substance over the number of moles substance or

MM = $\left(\text{g substance")/("mol substance}\right)$

then

MM = $\left(\text{20.83 g")/("0.1322 mol}\right)$

MM = $\textcolor{red}{157.56 \frac{g}{\text{mol}}}$

Therefore, the molar mass of the unknown gas is 157.56 g/mol.