Explanation:
a) Pressure of the gas a STP = P = 1 atm
Temperature f the gas at STP = T = 273.15 K
Moles of given gas i.e. [tex]CCl_2F_2[/tex] at STP = n
[tex]n=\frac{14.6 g}{121 g/mol}=0.1207 mol[/tex]
Volume occupied by the given gas = V
[tex]PV=nRT[/tex] (ideal gas equation)
[tex]V=\frac{nRT}{P}=\frac{0.1207 mol\times 0.0821 atm l/mol K\times 273.15 K}{1 atm}=2.71 L[/tex]
The volume occupied by 14.6 grams of [tex]CF_2Cl_2[/tex] is 2.71 liters.
b) Pressure of the gas a STP = P = 1 atm
Temperature f the gas at STP = T = 273.15 K
Moles of given gas i.e. [tex]CH_3CH_2F[/tex] at STP = n
[tex]n=\frac{14.6 g}{48g/mol}=0.3042 mol[/tex]
Volume occupied by the given gas = V
[tex]PV=nRT[/tex]
[tex]V=\frac{nRT}{P}=\frac{0.3042mol\times 0.0821 atm l/mol K\times 273.15 K}{1 atm}=6.82 L[/tex]
The volume occupied by 14.6 grams of [tex]CH_3CH_2F[/tex] is 6.82 liters.
