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^(@)C`. The pressure of water vapour at `25^(@)C` isA. 18 mm HgB. 20 mm HgC. 22 mm HgD. 245 mm Hg.

The volume of 0.0168 mol of `O_(2)` obtained by decomposition of `KClO

1.	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^(@)C`. The pressure of water vapour at `25^(@)C` isA. 18 mm HgB. 20 mm HgC. 22 mm HgD. 245 mm Hg.
Answer» Correct Answer - D Pressure of dry gas, P=nRT/V `=0.0168xx(0.0821xx1000xx760" ml "mm)xx(298)/(428)` 730 mm Pressure of moist gas=754 mm. Hence, pressure of water vapour =(754-730)mm=24 mm. Alternatively, volume of `0.0168" mol"` of `O_(2)` at STP `=0.0168xx22400=376.3" ml"` Thus, `V_(1)=376.3" ml", P_(1)=760" mm", T_(1)=273" K"` `V_(2)=428" ml",P_(2)=?,T_(2)=298" K"`. Calculate `P_(2)`.

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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^(@)C`. The pressure of water vapour at `25^(@)C` isA. 18 mm HgB. 20 mm HgC. 22 mm HgD. 245 mm Hg.

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