A gaseous product of a reaction is collected at 280K and 0.95 atm. Given R=0.0821L⋅atmmol⋅K

, what is the molar mass of the gas, in grams per mole, if 5.49 g of gas occupies 4.92 L?(1 point)
Responses

0.04 g/mol
0.04 g/mol

133 g/mol
133 g/mol

24 g/mol
24 g/mol

27 g/mol

1 answer

To find the molar mass of the gas, we can use the ideal gas law equation:

\[ PV = nRT \]

where:

  • \( P \) is the pressure (in atm)
  • \( V \) is the volume (in liters)
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant (\(0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\))
  • \( T \) is the temperature (in Kelvin)

First, let's calculate the number of moles (\( n \)) of the gas using the given conditions:

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

We rearrange the ideal gas law to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting in the values we have:

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

Calculating the right-hand side:

  1. Calculate the numerator: \[ 0.95 \times 4.92 = 4.674 \]

  2. Calculate the denominator: \[ 0.0821 \times 280 = 22.988 \]

Now, substituting these values into the equation for \( n \):

\[ n = \frac{4.674}{22.988} \approx 0.203 , \text{mol} \]

Now we can find the molar mass (\( M \)) of the gas using the formula:

\[ M = \frac{\text{mass}}{n} \]

We know that the mass of the gas is 5.49 g, so we can calculate the molar mass:

\[ M = \frac{5.49 , \text{g}}{0.203 , \text{mol}} \approx 27.0 , \text{g/mol} \]

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

The correct answer is 27 g/mol.