Question

What is the volume of 0.140 g of `C_2H_2F_4` vapor at 1.06 atm and 14.2°C?

Answer 1

Approx. `31*mL`....

Explanation:

We use Ideal Gas equation....

`V=(nRT)/P-=((0.140*g)/(102.04*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx287.35*K)/(1.06*atm)=??*L`