At $80\,^oC,$ the vapour pressure of pure liquid $'A'$ is $520\, mm\,Hg$ and that of pure liquid $'B'$ is $1000\, mm\,Hg.$ If a mixture solution of $'A'$ and $'B'$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $'A'$ in the mixture is ........... $mol$ percent $(1\, atm = 760\, mm\,Hg).$
- A$52$
- B$34$
- C$48$
- D$50$
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