Chapter 5: Problem 163
At \(80^{\circ} \mathrm{C}\), the vapour pressure of pure liquid 'A' is 520 \(\mathrm{mm} \mathrm{Hg}\) and that of pure liquid 'B' is \(1000 \mathrm{~mm} \mathrm{Hg}\). If a mixture solution of 'A' and 'B' boils at \(80^{\circ} \mathrm{C}\) and \(\mathrm{I}\) atm pressure, the amount of 'A' in the mixture is ( \(1 \mathrm{~atm}=\) \(760 \mathrm{~mm} \mathrm{Hg}\) ). (a) \(52 \mathrm{~mol}\) per cent (b) 34 mol per cent (c) 48 mol per cent (d) \(50 \mathrm{~mol}\) per cent
Short Answer
Step by step solution
Key Concepts
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