At $80\,^{\circ}C$,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 $80\,^{\circ}C$ and $1\, atm$ pressure,then the amount of '$A$' in the mixture is ......... $mol\, \%$. $(1\, atm = 760\, mm\, Hg)$
- A$52$
- B$34$
- C$48$
- D$50$
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