JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
Consider a binary solution of two volatile liquid components 1 and \(2 . x_1\) and \(y_1\) are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of \(\frac{1}{x_1}\) vs \(\frac{1}{y_1}\) are given respectively as what?
- A \(\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}\)
- B \(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}\)
- C \(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}\)
- D \(\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}\)
Step-by-step Solution
Detailed explanation
Given mole fraction of liquid 1 in liquid and vapour phase is \(x_1\) and \(y_1\) respectively.…
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