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?(2 points) Responses 24 g/mol 24 g/mol 0.04 g/mol 0.04 g/mol 27 g/mol 27 g/mol 133 g/mol

1 answer

To find the molar mass of the gas, we can first use the Ideal Gas Law, which is represented by the formula:

\[ PV = nRT \]

Where:

  • \( P \) = pressure (in atm)
  • \( V \) = volume (in liters)
  • \( n \) = number of moles of gas
  • \( R \) = ideal gas constant
  • \( T \) = temperature (in Kelvin)

We can rearrange this to find \( n \):

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

Given:

  • \( P = 0.95 \) atm
  • \( V = 4.92 \) L
  • \( R = 0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)
  • \( T = 280 , \text{K} \)

Plugging in the values:

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

Calculating the denominator:

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

Now plug in the values:

\[ n = \frac{(0.95)(4.92)}{22.996} \] \[ n = \frac{4.674}{22.996} \approx 0.203 , \text{mol} \]

Next, we need to find the molar mass (\( M \)) of the gas using the formula:

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

Given the mass of the gas is 5.49 g, we calculate:

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

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