At 20^ C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20^ above an equimolar mixture of benzene and methyl benzene is _____________ 10 -2 . (Nearest int

At 20 C the vapour pressure of benzene is 70 torr and that o: null (Chemistry)

Vapour pressure of pure benzene,

p_A^o

= 70 Torr

Vapour pressure of pure methyl benzene,

p_B^o

= 20 Torr

This mixture is equimolar, so number of moles of benzene,

n_A = number of moles of methyl benzene, n_B

Mole fraction of benzene in vapour phase, y_A =

{{{p_A}} \over {{p_T}}}

.... (i)

where p_A is pressure of benzene in mixture.

{p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}

p_T is total pressure,

{p_T} = p_A^o{\chi _A} + p_B^o{\chi _B} = {p_A} + {p_B}

(pressure of methyl benzene)

{p_T} = {{p_A^o} \over 2} + {{p_B^o} \over 2} = {{p_A^o + p_B^o} \over 2}

Putting in above equation (i),

{y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}

y_A = 0.78 = 78

\times

10^{$-$
2}

Explanation

The electronic configuration of Pt atomic number 78 is

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