Question

the volume of 0.0168 mole of O2 obtained by decomposition of Kclo3 and collected by displacement of water is 428ml at a pressure of 754 mm Hg at 25 degree Celsius . The pressure of water vapour at 25 degree Celsius is​

Answer 1

Answer:

24mmHg make me brainlyest

Answer 2

Given:

The volume of 0.0168 mole of O2 obtained by decomposition of Kclo3 and collected by displacement of water is 428ml at a pressure of 754 mm Hg at 25 degree Celsius .

To find:

The pressure of water vapour at 25 degree Celsius is​

Solution:

From given, we have,

the volume of 0.0168 moles of O2 obtained by decomposition of Kclo3 and collected by displacement of water is 428ml at a pressure of 754 mm Hg at 25 degree Celsius.

Using the Gas law equation, we have,

n = PV/RT

n = [(753/760) × 0.428] / [0.0821 × 298] = 0.01736 mol

Now   consider,

n = nw  + nO2

nw = 0.01736 - 0.0168 = 0.00056 mol

The pressure of the vapour = χ{water vapour} × P

= nw/n × P

= 0.00056/0.01736 × 754

= 24.32 mm Hg

∴ The pressure of water vapour at 25 degree Celsius is​ 24 mm Hg