Chapter 12 - Sections 12.1-12.8 - Exercises - Cumulative Problems - Page 592: 104

$0.602L$

We know that $n_{gas}=\frac{P_{gas}V_{gas}}{RT}$ $\implies n_{gas}=\frac{0.95395atm\space mol.K(1.65L)}{(0.08206L.atm)(298.15K)}=0.064334mol$ Now $S_{gas}=k_HP_{gas}$ $S_{gas}=(0.112M/atm)(0.95395atm)=0.10684M$ We can find the required volume as $V_{soln}=\frac{n_{gas}}{S_{gas}}$ $V_{soln}=\frac{0.064334mol}{0.10684mol/L}=0.602L$