#### Answer

$P_{CO}$ = 6.33 atm

#### Work Step by Step

1. Find k:
$$molarity = \frac{0.0354 \times 10^{-3} L \space CO\times \frac{1 \space mol}{22.4 \space L}}{10^{-3} \space L \space water} = 1.58 \times 10^{-3} M \space CO$$
$$ k = \frac{C}{P_{gas}} = \frac{1.58 \times 10^{-3} \space M \space CO}{1 \space atm}$$
2. Find the necessary pressure:
$$P_{gas} = \frac{C}{k} = \frac{0.0100 \space M \space CO}{1.58 \times 10^{-3} \space M \space CO/atm} = 6.33 \space atm$$