A mixture of 0.119 g of boron trifluoride (BF3) gas and 0.007 g of hydrogen sulfide (H2S) gas is held in a 1.74 L vessel at 19.24 ∘C. What are the mole fractions of the individual gases?

molar mass BF3 = 67.805 g/mole

molar mass CO2 = 44.009 g/mole

moles BF3 present = 13.5 g x 1 mole/67.805 g = 0.1991 moles

moles CO2 present = 12.8 g x 1 mole/44.009 g = 0.2908 moles

TOTAL moles present = 0.1991 + 0.2908 = 0.4899 moles

Mole fraction BF3 = 0.1991/0.4899 = 0.406 (to 3 sig. figs.)

Mole fraction CO2 = 0.2908/0.4899 = 0.594 (to 3 sig. figs.)

144

Explanation:

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