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The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
It is given that:
_18-11-08)_Utpal_12_Chemistry_2_12_html_322b825.gif){kind=small}
= 450 mm of Hg
_18-11-08)_Utpal_12_Chemistry_2_12_html_36d5f3ea.gif){kind=small}
= 700 mm of Hg
ptotal = 600 mm of Hg
From Raoult’s law, we have:
_18-11-08)_Utpal_12_Chemistry_2_12_html_1936cfc1.gif)
Therefore, total pressure, _18-11-08)_Utpal_12_Chemistry_2_12_html_612b76d6.gif){kind=small}
_18-11-08)_Utpal_12_Chemistry_2_12_html_7d57b158.gif)
Therefore, _18-11-08)_Utpal_12_Chemistry_2_12_html_7eb2c6c5.gif){kind=small}
= 1 − 0.4
= 0.6
Now, _18-11-08)_Utpal_12_Chemistry_2_12_html_fb4393e.gif){kind=small}
= 450 × 0.4
= 180 mm of Hg
_18-11-08)_Utpal_12_Chemistry_2_12_html_m5097ea18.gif){kind=small}
= 700 × 0.6
= 420 mm of Hg
Now, in the vapour phase:
Mole fraction of liquid A_18-11-08)_Utpal_12_Chemistry_2_12_html_m1928f4d.gif){kind=small}
_18-11-08)_Utpal_12_Chemistry_2_12_html_cbee892.gif)
= 0.30
And, mole fraction of liquid B = 1 − 0.30
= 0.70
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