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28 June, 01:29

A 0.334 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen?

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  1. 28 June, 03:33
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    We can use the ideal gas law equation for the above reaction to find the number of moles present

    PV = nRT

    P - pressure - 1.41 atm x 101325 Pa/atm = 142 868 Pa

    V - 109 x 10⁻⁶ m³

    R - 8.314 Jmol⁻¹K⁻¹

    T - 398 K

    substituting the values in the equation

    142 868 Pa x 109 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 398 K

    n = 4.70 x 10⁻³ mol

    number of moles = mass present / molar mass

    molar mass = mass / number of moles

    = 0.334 g / 4.70 x 10⁻³ mol = 71.06 g/mol

    halogens exist as diatomic molecules

    Therefore atomic mass - 71.06 / 2 = 35.5

    halogen with 35.5 g/mol is Cl

    unknown halogen is Cl
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