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2.36 of liquid A was dissolved in of liquid B . The vapour pressure of pure liquid B was found to be $500 \; Torr.$ Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure the solution is $475 \; Torr.$

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For calculating partial vapour pressure we need to calculate mole fractions of components.

So number of moles of liquid A :

$= \frac{100}{140} = 0.714$

and moles of liquid B :

$= \frac{1000}{180} = 5.556$

Mole fraction of A (x_A) :

$= \frac{0.714}{0.714+5.556} = 0.114$

and mole fraction of B (x_B) :

$= \frac{5.556}{0.714+5.556} = 0.866$

Now, P_total = P_A + P_B

or $P_{total} = P_A^{\circ}x_A\ + P_B^{\circ}x_B$

or $475 = P_A^{\circ}\times0.114\ + 500\times0.886$

or $P_A^{\circ} = 280.7\ torr$

Thus vapour pressure in solution due to A =

$= 280.7 \times0.114 = 32\ torr$

Posted by

Devendra Khairwa

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