Question

A gaseous product of a reaction is collected at 280K and 0.95 atm. Given
R
=
0.0821
L
⋅
a
t
m
m
o
l
⋅
K
, what is the molar mass of the gas, in grams per mole, if 5.49 g of gas occupies 4.92 L?(2 points)
Responses

27 g/mol
27 g/mol

0.04 g/mol
0.04 g/mol

24 g/mol
24 g/mol

133 g/mol
133 g/mol

Answer 1

To find the molar mass of the gas, we can use the Ideal Gas Law equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles of gas
- \( R \) = Ideal gas constant (\( 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \))
- \( T \) = Temperature (in Kelvin)

First, we need to calculate the number of moles \( n \) of the gas using the given pressure, volume, and temperature:

### Step 1: Calculate the number of moles (n)

Given:
- \( P = 0.95 \, \text{atm} \)
- \( V = 4.92 \, \text{L} \)
- \( T = 280 \, \text{K} \)

Substituting in the Ideal Gas Law:

\[
n = \frac{PV}{RT} = \frac{(0.95 \, \text{atm})(4.92 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(280 \, \text{K})}
\]

### Step 2: Calculate

First, calculate the denominator:

\[
0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \times 280 \, \text{K} = 22.996 \, \text{L} \cdot \text{atm} / \text{mol}
\]

Now, calculate \( n \):

\[
n = \frac{(0.95)(4.92)}{22.996} \approx \frac{4.684}{22.996} \approx 0.203 \, \text{mol}
\]

### Step 3: Calculate Molar Mass

Using the mass of the gas (5.49 g) and the number of moles \( n \):

\[
\text{Molar Mass} = \frac{\text{mass}}{n} = \frac{5.49 \, \text{g}}{0.203 \, \text{mol}} \approx 27.0 \, \text{g/mol}
\]

### Conclusion

Therefore, the molar mass of the gas is approximately **27 g/mol**.