The volume of0.0168 mol of `O_(2)` obtained by decomposition of `KClO_(3)` and collected by dispfacement of water is 428 mL at a pressure of754 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. 24 mm Hg

Answer

D

Explanation

Volume of 0.0168 moles at STP
`=0.0168xx22400=376.3` mL.
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` or `(760xx376.3)/273=(P_(2)xx428)/298` or `P_(2)=730mm`
Pressure of water= 754 - 730=24 mm