If O _ 2 gas is bubbled through water at 303K , the number of...

$p = {K}_{H} \times x$

$0 . 920 = 46 . 82 \times {10}^{3}$ bar $\times \dfrac{mol of {O}_{2}}{mol of {H}_{2}O}$

$0 . 920 = 46 . 82 \times {10}^{3} \times \dfrac{molof{O}_{2}}{\dfrac{1000}{18}}$

$0 . 920 = 46 . 82 \times 18 \times {n}_{{O}_{2}}$

$\dfrac{0 . 920}{46 . 82 \times 18} = {n}_{{O}_{2}}$

$⇒ 1 . 09 \times {10}^{- 3} = {n}_{{0}_{2}}$

$⇒ mmolof{O}_{2} = 1millimole$