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20 May, 10:24

in a mixture of helium and chlorine, occupying a volume of 12.8 l at 605.6 mmhg and 21.6 oc, it is found that the partial pressure of chlorine is 143 mmhg. what is the total mass of the sample?

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  1. 20 May, 11:49
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    Mass of sample = 8.483 g

    Explanation:

    Given data;

    Volume of mixture = 12.8 L

    Pressure = 605.6 mmHg (605.6 / 760 = 0.797 atm)

    Temperature = 21.6 °C (21.6 + 271.15 = 294.8 K)

    Partial pressure of chlorine = 143 mmHg (143/760 = 0.19 atm)

    Solution:

    First of all we will determine the number of moles of mixture.

    PV = nRT

    n = PV/RT

    n = 0.797atm * 12.8L / 0.0821 atm. dm³ mol⁻¹ K⁻¹ * 294.8 K

    n = 10.202 / 24.2031

    n = 0.422 mol

    partial pressure of chlorine is 0.19 atm so mole fraction is,

    mole fraction = 0.19/0.797

    mole fraction = 0.24

    moles of chlorine = 0.24 * 0.422 = 0.1013 mol

    moles of helium = moles of mixture - moles of chlorine

    moles of helium = 0.422 - 0.1013

    moles of helium = 0.3207 mol

    Mass of chlorine = moles * molar mass

    Mass of chlorine = 0.1013 mol * 71 g/mol

    Mass of chlorine = 7.2 g

    Mass of helium = moles * molar mass

    Mass of helium = 0.3207 mol * 4 g/mol

    Mass of helium = 1.283 g

    Mass of sample = mass of chlorine + mass of helium

    Mass of sample = 7.2 g + 1.283 g

    Mass of sample = 8.483 g
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