The volume of 0.0168 mol of `O_(2)` obtained by decomposition of `KClO_(3)` and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at `25^(@)C`. The pressure of water vapour at `25^(@)C` is

Options

A. 18 mm Hg
B. 20 mm Hg
C. 22 mm Hg
D. 245 mm Hg.

Answer

D

Explanation

Pressure of dry gas, P=nRT/V
`=0.0168xx(0.0821xx1000xx760" ml "mm)xx(298)/(428)`
730 mm
Pressure of moist gas=754 mm. Hence, pressure of water vapour =(754-730)mm=24 mm.
Alternatively, volume of `0.0168" mol"` of `O_(2)` at STP
`=0.0168xx22400=376.3" ml"`
Thus, `V_(1)=376.3" ml", P_(1)=760" mm", T_(1)=273" K"`
`V_(2)=428" ml",P_(2)=?,T_(2)=298" K"`.
Calculate `P_(2)`.