The volume of 0.0168 mol of O2 obtained by decomposition of KClO3 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

Question

The volume of 0.0168 mol of O2 obtained by decomposition of KClO3 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 (A) 18 mm Hg (B) 20 mm Hg (C) 22 mm Hg (D) 24 mm Hg

Tardigrade Admin · Accepted Answer

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°C = (273 + 25) K = 298 K P = (0.0168 × 0.0821 × 298/0.428) = 0.96 atm = (0.96 × 760) mm Hg = 730 mm Hg P textmoist gas = p textdry gas + P textwater vapour P textwater vapour = 754 - 730 = 24 mm Hg

Q. The volume of $0.0168$ mol of $O_{2}$ obtained by decomposition of $K Cl O_{3}$ and collected by displacement of water is $428 m L$ at a pressure of $754$ mm $H g$ at $2 5^{\circ} C$ . The pressure of water vapour at $2 5^{\circ} C$ is

$18$ mm Hg

$20$ mm Hg

$22$ mm Hg

$24$ mm Hg

Q. The volume of 0.0168 mol of O2​ obtained by decomposition of KClO3​ and collected by displacement of water is 428mL at a pressure of 754 mm Hg at 25∘C. The pressure of water vapour at 25∘C is

18 mm Hg

20 mm Hg

22 mm Hg

24 mm Hg

Applying, PV=nRT for dry gas P=?,V=428mL=0.428L,n=0.0168mol R=0.0821L atm K−1 mol−1,T=25∘C=(273+25)K=298K P=0.4280.0168×0.0821×298​=0.96atm =(0.96×760)mmHg=730mmHg Pmoist gas​=pdry gas​+Pwater vapour​ Pwater vapour​=754−730=24mmHg

Q. The volume of $0.0168$ mol of $O_{2}$ obtained by decomposition of $K Cl O_{3}$ and collected by displacement of water is $428 m L$ at a pressure of $754$ mm $H g$ at $2 5^{\circ} C$ . The pressure of water vapour at $2 5^{\circ} C$ is

$18$ mm Hg

$20$ mm Hg

$22$ mm Hg

$24$ mm Hg