Answer:
[tex]CCl_{2}FCF_{3}[/tex]
Explanation:
To solve this equation we can use Graham's Law of effusion to calculate the molar mass of the chloroflourocarbon
[tex]\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2} } \\\\\sqrt{\frac{M_{1} }{39.948 } } = \frac{157 }{76}\\\\\\M_{1} = 170.477\\[/tex]
Which chloroflorocarbon has this molar mass?
C2Cl2F4
