Question

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 $25^{o}C$. The pressure of water vapour at $25^{o}C$ is

• A. 18 mm Hg
• B. 20 mm Hg
• C. 22 mm Hg
• D. 24 mm Hg

Answer 1

Answer: (D)

24 mm Hg

Answer 2

Volume of 0.0168 mol of $O_{2}$ at STP

$= 0.0168 \times 22400cc = 376.3cc$

$V_{1} = 376.3cc$, $P_{1} = 760mm$, $T_{1} = 273K$

$V_{2} = 428cc$, $P_{2} = ?$, $T_{2} = 298K$

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$ gives $P_{2} = 730mm$ (approx.)

∴ Pressure of water vapour$= 754 - 730 = 24$mm Hg