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

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

Answer 1

Answer: (D)

Applying, $PV = nRT$ for dry gas
$P = ?, V = 428 \,mL = 0.428\, L, n = 0.0168\, mol$
$R = 0.0821 \,L$ atm $K^{-1}$ mol$^{-1}, T= 25^{\circ}C = (273 + 25) K = 298 \,K$
$P = \frac{0.0168 \times 0.0821 \times 298}{0.428} = 0.96\, atm$
$= (0.96 \times 760)\, mm \,Hg = 730\, mm\, Hg$
$P_{\text{moist gas}} = p_{\text{dry gas}} + P_{\text{water vapour}}$
$P_{\text{water vapour}} = 754 - 730 = 24\, mm\, Hg$