Answer:
[tex]V=3.79L[/tex]
Explanation:
Hello,
In this case, we are talking about an ideal gas problem in which 12.0 g of chlorine gas is at standard both pressure (1atm) and temperature (0°C), therefore, to compute the corresponding volume, one applies the ideal gas law, then converts from grams to moles considering the diatomic chlorine and subsequently solves for volume as shown below:
[tex]PV=nRT\\V=\frac{nRT}{P}=\frac{12.0gCl_2*\frac{1molCl_2}{70.9gCl_2} *0.082 \frac{atm*L}{mol*K} * 273.15K}{1atm} \\V=3.79L[/tex]
Best regards.