Answer
$1.14\times 10^{-3}\frac{mol}{L}$
Work Step by Step
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:
$C=\frac{1.04\times 10^{-3}mol}{L.atm}\times 1.10atm=1.14\times 10^{-3}\frac{mol}{L}$
