General chem question!!?

The radius of a xenon atom is 1.3×10−8cm. A 100-mL flask is filled with Xe at a pressure of 1.5 atm and a temperature of 293 K. Calculate the fraction of the volume that is occupied by Xe atoms.

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  • Dr W
    Lv 7
    3 weeks ago
    Favorite Answer

    a couple of things you need to know

    .. (1) volume fraction = volume atoms / volume flask

    .. (2) V sphere = 4/3*π *r³ 

    .. (3) PV = nRT ---> n = PV/(RT)

    so that volume fraction is

    ..4/3 *π *(1.3e-8cm)³.. .. 6.022e23 atoms.. ... ... ... ..1.5atm*0.100L

     ------- ----- ------ ----- x ------ ------ ------ -- x ----- ----- ----- ----- ----- ----- ----- ----- = 

    .. .. ... ...atom.. ... ... ... ... .. 1 mol.. ... .. .. .0.08206Latm/molK * 293K * 100cm³

    .. = 3.46e-4 cm³ atom / cm³ flask

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  • Anonymous
    3 weeks ago

    Calculate the moles of Xenon in the flask using the ideal gas equation.  Atoms = moles * Avogadro's number.

    Calculate the volume of 1 atom, assume the atom is a sphere.  Volume/atom * # of atoms = volume occupied by the atoms.   Simple division after that.

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  • Fern
    Lv 7
    3 weeks ago

    volume of one Xe atom:

    4/3(pi)r^3 = 4/3(3.14)(1.3 x 10^-8cm = 9.2 x 10^-24 cm3

    moles of Xe present:

    PV = nRT

    n= 1.5 atm(0.100 Liter)/0.0820(293K)

    n = 6.24 x 10^-3 moles

    number of Xe atoms: 6.24 x 10^-3 moles Xe x 6.02 x 10^23 atoms/mole Xe = 3.75 x 10^21 atoms

    Volume of Xe atoms:

    3.75 x 10^21 atoms x 9.2 x 10^-24 cm3/atom = 0.0345 cm3

    0.0345 cm3 x 1 mL/1cm3 = 0.0345 ml

    Fraction occupied by Xe atoms:

    0.0345 mL/100 mL = 0.000345

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  • 3 weeks ago

    1.3×10−8cm = 13–8 = 5 cm, which is a silly value for an atom.

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