Answer:
[tex]V=3.32 L[/tex]
Explanation:
Moles in the sample:
[tex]n=10 mL*\frac{0.708g}{mL}*\frac{1mol}{74g}[/tex]
[tex]n=0.096mol[/tex]
Max. volume to be in equilibrium:
[tex]V=\frac{n*R*T}{P}[/tex]
[tex]V=\frac{0.096mol*0.082 atm*L*mol^{-1}*K^{-1}*298K}{0.707atm}[/tex]
[tex]V=3.32 L[/tex]
If the volume is higher, the pressure will be less than the vapour pressure and the sample will be all gas and not L-V equilibrium.
