Chemistry 9th Edition

$1.14\times 10^{-3}\frac{mol}{L}$
We know that $C=kP$......eq(1) This can be rearranged as: $k=\frac{C}{P}$ We plug in the known values to obtain: $k=\frac{8.21\times 10^{-4}}{0.790}=1.04\times 10^{-3}\frac{mol}{atm.L}$ We plug in the known values in equation (1) to obtain: $C=\frac{1.04\times 10^{-3}mol}{L.atm}\times 1.10atm=1.14\times 10^{-3}\frac{mol}{L}$