Test Index

Solutions Part 1

Section: Chemistry

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Question : 83 of 100

Marks: +1, -0

At

80^{\circ}\mathrm{C}

, the vapour pressure of pure liquid '

A

520\ \mathrm{mm\ Hg}

and that of pure liquid '

B

1000\ \mathrm{mm\ Hg}

. If a mixture solution of '

A

B

80^{\circ}\mathrm{C}

1\ \mathrm{atm}

pressure, the amount of '

A

' in the mixture is (

1\ \mathrm{atm}=760\ \mathrm{mm\ Hg}

52 mol percent
34 mol percent
48 mol percent
50 mol percent

Solution:

At 1 atmospheric pressure the boiling point of mixture is

80^{\circ}\mathrm{C}

.

At boiling point the vapour pressure of mixture,

P_T=1

atmosphere

=760\ \mathrm{mm\ Hg}

.

Using the relation,

P_T=P_A^{\circ} X_A+P_B^{\circ} X_B

, we get

P_T=520 X_A+1000 (1-X_A)

\{P_A^{\circ}=520\ \mathrm{mm\ Hg},.

P_B^{\circ}=1000\ \mathrm{mm}\ H\ g, X_A+X_B=1\}

or

\;\; 760=520 X_A+1000-1000 X_A

or

480 X_A=240

or

\;\; X_A=\;\frac{240}{480} = \;\frac{1}{2}

or

\;\; 50\ \mathrm{mol}

percent

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