Question

50 mL of a chlorofluorocarbon gas effused through a porous barrier in 157 seconds. The same volume of argon effused in 76 seconds under the same conditions. What is the formula of the chlorofluorocarbon?

Answer 1

Answer:

$CCl_{2}FCF_{3}$

Explanation:

To solve this equation we can use Graham's Law of effusion to calculate the molar mass of the chloroflourocarbon

$\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2} } \\\\\sqrt{\frac{M_{1} }{39.948 } } = \frac{157 }{76}\\\\\\M_{1} = 170.477$

Which chloroflorocarbon has this molar mass?

C2Cl2F4