The volume of oxygen collected by the decomposition of potassium chlorate at `24^(@)C` and atmospheric pressure of 760 mm Hg is 128 mL. Calculate the mass of oxygen gas obtained. The pressure of the water vapour at `24^(@)C` is `22.4 mm Hg`.

Options

A. `0.123 g`
B. `0.163 g`
C. `0.352 g`
D. `1.526 g`

Answer

B

Explanation

From the total pressure and the vapour pressure of water, we can calculate the partial pressure of `O_2`.
`p_(O_2)= p_(r )-P_(H_2 O)= 760 -22.4= 737.6 mm Hg`
From the ideal gas equation we write `w= (pVM)/(RT)= (0.9705 xx 0.128 xx 32.0)/(0.0821 xx 2997)= 0.163g`.