When 0.0168 , mol of O_2 obtained from the de | vedclass

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(D) Using the ideal gas law $PV = nRT$ for dry $O_2$ gas:$n = 0.0168 \, mol$,$V = 428 \, mL = 0.428 \, L$,$T = 298 \, K$,$R = 0.0821 \, L \cdot atm \c

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$24$

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(D) Using the ideal gas law $PV = nRT$ for dry $O_2$ gas:$n = 0.0168 \, mol$,$V = 428 \, mL = 0.428 \, L$,$T = 298 \, K$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.$P_{dry} = \frac{nRT}{V} = \frac{0.0168 	imes 0.0821 	imes 298}{0.428} \approx 0.9607 \, atm$.Convert $P_{dry}$ to $mm \, Hg$: $0.9607 \, atm 	imes 760 \, mm \, atm^{-1} \approx 730.1 \, mm \, Hg$.Total pressure $P_{total} = P_{dry} + P_{water \, vapor}$.$P_{water \, vapor} = P_{total} - P_{dry} = 754 \, mm - 730.1 \, mm \approx 23.9 \, mm \approx 24 \, mm$.

List-$I$	List-$II$
$(A)$ Viscosity	$(I)$ Critical temperature
$(B)$ Ideal gas behaviour	$(II)$ Isobars
$(C)$ Liquefaction of gases	$(III)$ Compressibility factor
$(D)$ Charles' law	$(IV)$ $kg \ s^{-2}$
	$(V)$ $kg \ m^{-1} \ s^{-1}$

When $0.0168 \, mol$ of $O_2$ obtained from the decomposition of $KClO_3$ is collected over water at $25 \, ^\circ C$,the volume is $428 \, mL$ and the pressure is $754 \, mm$. What is the water vapor pressure at $25 \, ^\circ C$ in $mm$?

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