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Equal moles of benzine and tolune are mixed the v.p of benzine and tolune in pure state are 700 and 600 mmHg respectively.the mole friction of benzine in vapour state is?

Equal moles of benzine and tolune are mixed the v.p of benzine and tolune in pure state are 700 and 600 mmHg respectively.the mole friction of benzine in vapour state is?

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
Suppose the moles of benzene and toluene are x.
Hence their mole fraction X=x/2x = ½
The vapour pressure of benzene in pure state P(a) =700 mm of Hg
The vapour pressure of toluene in pure state P(b) =600 mm of Hg
Now total pressure P (t) = P(a) X(a) + P(b) X(b)
                                            = 700 * ½  + 600 * ½
                                            =350 + 300
                                    P(t) = 650 mm of Hg
Now from Dalton’s law, P(a) X(a)=P(t) Y(a)         where Y(a)= mole fraction of benzene in vapour state
                                           700 * ½ = 650 * Y(a)
                                            Y(a) = 0.538 
ankit singh
askIITians Faculty 614 Points
3 years ago
Suppose the moles of Benzene & Toluene are x.
 
Hence, their mole fraction=2xx=21
 
Vapour pressure of Benzene in pure state P(a)=700mmofHg
 
and that of Toluene P(b)=600mmofHg
 
Now total pressureP(t)=P(a)×(a)×P(b)×(b)
 
=700×21+600×21=650mmofHg
 
Now, P(a)×(a)=P(t)Y(a) where Y(a)=molefractionofC6H6invapourstate
 
700(21)=650(Y(a))
 
Y(a)=650350=0.5380.54
 
Hence, the correct option is D

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