Question

A gaseous product of a reaction is collected at 280k and 0.95 atm. Given R=0.0821L.atm, what is the molar mass of gas, in grams per mole, if 5.49 g of gas occupies 4.92 L? 133 g/mol 27g/mol 0.04g/mol 24g/mol

Answer 1

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

\[
PV = nRT
\]

Where:
- \( P \) is the pressure in atm
- \( V \) is the volume in liters
- \( n \) is the number of moles of gas
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin

Since we want to find the molar mass, we can express the number of moles \( n \) in terms of the mass of the gas and its molar mass \( M \):

\[
n = \frac{m}{M}
\]

Substituting this into the ideal gas equation:

\[
PV = \frac{m}{M}RT
\]

Rearranging for \( M \):

\[
M = \frac{mRT}{PV}
\]

Now, plug in the given values:

- \( m = 5.49 \, \text{g} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 280 \, \text{K} \)
- \( P = 0.95 \, \text{atm} \)
- \( V = 4.92 \, \text{L} \)

Now we can calculate \( M \):

\[
M = \frac{(5.49 \, \text{g}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (280 \, \text{K})}{(0.95 \, \text{atm}) \times (4.92 \, \text{L})}
\]

Calculating the numerator:

\[
5.49 \times 0.0821 \times 280 = 128.99908 \, \text{g·L·atm/(K·mol)}
\]

Calculating the denominator:

\[
0.95 \times 4.92 = 4.674 \, \text{atm·L}
\]

Now divide the numerator by the denominator:

\[
M = \frac{128.99908}{4.674} \approx 27.56 \, \text{g/mol}
\]

Thus, rounding to two decimal places, the molar mass of the gas is approximately:

\[
M \approx 27 \, \text{g/mol}
\]

According to the options provided, the closest answer is:

**27 g/mol**.