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Q.

The volume of 0.0168 mole of 02 obtained by thermal decomposition of KClO3 and collected by displacement of water is 428 mL at a pressure of 754 mmHg at 25°C. Thepressure of water Vapour at 25°C is

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a

18 mm Hg

b

20 mm Hg

c

24 mm Hg

d

30 mm Hg

answer is C.

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Detailed Solution

pV = nRTpatm=nVRT=0.0168×0.0821×2980.428=0.9603atmpatm=730mmHg ∴ptotal =patm +pvapour  754=730+pvapour  Vvapour =24mmHg
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