At 80°C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B'…
Chemistry · NEET · NTA Exams — Solutions
At 80°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°C and 1 atm pressure the amount of 'A' in the mixture is:
(1 atm = 760 mm Hg)
- 52 mole per cent
- 34 mole per cent
- 48 mole per cent
- 50 mole per cent
Answer
(B) 34 mole per cent
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