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9 December, 03:15

In a mixture of helium and chlorine, occupying a volume of 14.6 l at 871.7 mmhg and 28.6oc, it is found that the partial pressure of chlorine is 355 mmhg. what is the total mass of the sample?

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  1. 9 December, 03:56
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    First, let's find the total moles.

    PV=nRT

    (871.7 mmHg) (1 atm/760 mmHg) (14.6 L) = n (0.0821 L-atm/mol-K) (28.6 + 273)

    Solving for n,

    n = 0.676 moles

    Assuming ideal gas behavior, the pressure fraction is also equal to mole fraction.

    Mole fraction of Chlorine = 355/871.7 = 0.407

    Mole fraction of Helium = 1 - 0.407 = 0.593

    Knowing Chlorine to be 35.45g/mol and Helium to be 4 g/mol.

    Mass of Chlorine = (0.407) (0.676) (35.45 g/mol) = 9.75 g

    Mass of Helium = (0.593) (0.676) (4 g/mol) = 1.6 g

    Total Mass = 9.75+1.6 = 11.35 g
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