If a sample of 12.0 grams of fluorine gas at 45.00C has a pressure of 0.544 atm, what is the volume of the container? R = 0.082

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If a sample of 12.0 grams of fluorine gas at 45.00C has a pressure of 0.544 atm, what is the volume of the container?

R = 0.0821 L atm/mol K

15.1 L

4.29 L

30.3 L

2.14 L

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  1. Answer: Volume of the container is 30.3 L.

    Explanation:

    Given: Mass = 12 g

    Temperature = [tex]45^{o}C[/tex] = (45 + 273) K = 318 K

    Pressure = 0.544 atm

    R = gas constant = 0.0821 L atm/mol K

    Moles is the mass of substance divided by its molar mass.

    Hence, moles of fluorine (molar mass = 18.99 g/mol) are as follows.

    [tex]Moles = \frac{mass}{molar mass}\\= \frac{12.0 g}{18.99 g/mol}\\= 0.631 mol[/tex]

    Formula used to calculate the volume of container is as follows.

    PV = nRT

    where,

    P = pressure

    V = volume

    n = no. of moles

    R = gas constant

    T = temperature

    Substitute the value into above formula as follows.

    [tex]PV = nRT\\0.544 atm \times V = 0.631 mol \times 0.0821 L atm/mol K \times 318 K\\V = 30.3 L[/tex]

    Thus, we can conclude that volume of the container is 30.3 L.

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