Answer:
[tex]m=8.28gCl_2[/tex]
Explanation:
Hello there!
In this case, according to the given information, it possible for us to realize that this problem is solved via the ideal gas equation:
[tex]PV=nRT[/tex]
By which we solve for the moles of chlorine gas as shown below:
[tex]n=\frac{PV}{RT}[/tex]
Next, we plug in the given pressure (in atm), the volume and standard pressure (273.15K) to obtain:
[tex]n=\frac{610/760atm*3.26L}{0.08206\frac{atm*L}{mol*K}*273.15K}\\\\n=0.117molCl_2[/tex]
Then, since the molar mass of chlorine gas (Cl2) is 70.9 g/mol, the mass of this gas turns out to be:
[tex]m=0.117molCl_2*\frac{70.9gCl_2}{1molCl_2} \\\\m=8.28gCl_2[/tex]
Regards!
