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

27 g/mol
27 g/mol

133 g/mol
133 g/mol

24 g/mol

3 answers

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

\[ PV = nRT \]

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 (\( 0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \))
  • \( T \) is the temperature in Kelvin

First, we need to find the number of moles \( n \) using the values given.

Given:

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

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

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

Substituting 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 this:

  1. Calculate \( PV \): \[ PV = 0.95 \times 4.92 = 4.674 , \text{L} \cdot \text{atm} \]

  2. Calculate \( RT \): \[ RT = 0.0821 \times 280 = 22.988 , \text{L} \cdot \text{atm} / \text{mol} \]

  3. Calculate \( n \): \[ n = \frac{4.674}{22.988} \approx 0.203 , \text{mol} \]

Now that we have the number of moles, we can calculate the molar mass \( M \):

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

Given the mass of the gas is \( 5.49 , \text{g} \):

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

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

Final Answer: 27 g/mol

are you sure

Let's double-check the calculations carefully.

We want to find the molar mass \( M \) of the gas using the Ideal Gas Law \( PV = nRT \).

Given Values:

  • Pressure \( P = 0.95 , \text{atm} \)
  • Volume \( V = 4.92 , \text{L} \)
  • Temperature \( T = 280 , \text{K} \)
  • Ideal Gas Constant \( R = 0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)
  • Mass of gas \( m = 5.49 , \text{g} \)

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

Using the formula: \[ n = \frac{PV}{RT} \]

Substituting the values:

  1. Calculate \( PV \): \[ PV = 0.95 , \text{atm} \times 4.92 , \text{L} = 4.674 , \text{L} \cdot \text{atm} \]

  2. Calculate \( RT \): \[ RT = 0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \times 280 , \text{K} = 22.988 , \text{L} \cdot \text{atm} / \text{mol} \]

  3. Now calculate \( n \): \[ n = \frac{4.674 , \text{L} \cdot \text{atm}}{22.988 , \text{L} \cdot \text{atm} / \text{mol}} = 0.203 , \text{mol} \text{ (approximately)} \]

Step 2: Calculate the molar mass \( M \)

Now using the molar mass formula: \[ M = \frac{\text{mass}}{n} \]

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

Conclusion

After verifying the calculation, the molar mass of the gas is approximately 27 g/mol.

So yes, the answer is indeed 27 g/mol.